4278 lines
120 KiB
Markdown
4278 lines
120 KiB
Markdown
Page 516
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**Exercise Set 8.1**
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1. As in Example 8.1.2, the **congruence modulo $2$** relation $E$ is defined
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from $\mathbb{Z}$ to $\mathbb{Z}$ as follows: For every ordered pair
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$(m, n) \in \mathbb{Z} \times \mathbb{Z}$,
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$$ m E n \Leftrightarrow m - n \text{ is even} $$
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a. Is $0 E 0$? Is $5 E 2$? Is $(6, 6) \in E$? Is $(-1, 7) \in E$?
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_$0 E 0$:_
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Yes, $0 - 0 = 0$, and $0$ is even.
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_$5 E 2$:_
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No, $5 - 2 = 3$, and $3$ is not even.
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_$(6, 6) \in E$:_
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Yes, $6 - 6 = 0$, and $0$ is even.
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_$(-1, 7) \in E$:_
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Yes, $-1 - 7 = -8$, and $-8$ is even.
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b. Prove that for any even integer $n$, $n E 0$.
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**Proof:**
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Suppose $n \in 2\mathbb{Z}$, where $2\mathbb{Z}$ is the set of all even
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integers.
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By the definition for even, this means that $n = 2k$ for some integer $k$.
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By the definition for $E$, $n E 0$ if, and only if $n - 0$ is even.
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By substitution for $E$:
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$$ n - 0 = 2k - 0 $$
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$$ = 2k $$
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By the definition for even, this means that $n - 0$ is even, and therefore
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$n E 0$ is true.
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Q.E.D.
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2. Prove that for all integers $m$ and $n$, $m - n$ is even if, and only if,
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both $m$ and $n$ are even or both $m$ and $n$ are odd.
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_Hint:_ To prove a statement of the form $p \Leftrightarrow (q \vee r)$, you
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need to prove both (1)$p \to (q \vee r)$ and (2) $(q \vee r) \to p$. The easiest
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way to prove $p \to (q \vee r)$ is to prove the logically equivalent statement
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form $(p \wedge \neg q) \to r$. And the easiest way to prove $(q \vee r) \to p$
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is to prove the logically equivalent statement form
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$(q \to p) \wedge (r \to p)$. In this case, suppose $m$ and $n$ are any
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integers, and let $p$ be "$m - n$ is even," let $q$ be "both $m$ and $n$ are
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even," and let $r$ be "both $m$ and $n$ are odd."
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**Proof:**
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Suppose $m$ and $n$ are any integers.
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To prove that for all integers $m$ and $n$, $m - n$ is even if, and only if,
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both $m$ and $n$ are even or both $m$ and $n$ are odd, it must be shown first
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that if $m - n$ is even, then both $m$ and $n$ are even or both $m$ and $n$ are
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odd, then it must be shown second that if both $m$ and $n$ are even or both $m$
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and $n$ are odd, then $m - n$ is even.
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_Proof (first):_
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Suppose $m - n$ is even. To prove that both $m$ and $n$ must be even or both $m$
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and $n$ must be odd, all cases for where $m$ is even or odd and where $n$ is
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even or odd must be considered.
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_Case (both $m$ and $n$ are even):_
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Since both $m$ and $n$ are even, this means that $m = 2k$ and $n = 2p$ for some
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integers $k$ and $p$. Then:
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$$ m - n = 2k - 2p $$
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$$ = 2(k - p) $$
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Now, $k - p$ is an integer by the subtraction of integers. Therefore, by the
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definition of even, $m - n$ is even.
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_Case (both $m$ and $n$ are odd):_
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Since both $m$ and $n$ are odd, this means that $m = 2k + 1$ and $n = 2p + 1$
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for some integers $k$ and $p$. Then:
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$$ m - n = (2k + 1) - (2p + 1) $$
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$$ = 2k + 1 - 2p - 1 $$
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$$ = 2k - 2p $$
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$$ = 2(k - p) $$
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Now, $k - p$ is an integer by the subtraction of integers. Therefore, by the
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definition of even, $m - n$ is even.
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_Case ($m$ is even and $n$ is odd):_
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Since $m$ is even and $n$ is odd, $m = 2k$ and $n = 2p + 1$ for some integers
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$k$ and $p$. Then:
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$$ m - n = 2k - (2p + 1) $$
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$$ = 2k - 2p - 1 $$
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$$ = 2(k - p) - 1 $$
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Now, $k - p$ is an integer by the subtraction of integers. Thus, by the
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definition of odd, $m - n$ is odd, but by the supposition, $m - n$ is even. This
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is a contradiction.
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_Case ($m$ is odd and $n$ is even):_
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Since $m$ is odd and $n$ is even, $m = 2k + 1$ and $n = 2p$ for some integers
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$k$ and $p$. Then:
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$$ m - n = (2k + 1) - 2p $$
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$$ = 2k - 2p + 1 $$
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$$ = 2(k - p) + 1 $$
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Now, $k - p$ is an integer by the subtraction of integers. Thus, by the
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definition of odd, $m - n$ is odd, but by the supposition, $m - n$ is even. This
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is a contradiction.
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_Conclusion:_
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It can be concluded based off of all cases that when both $m$ and $n$ are even
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or both $m$ and $n$ are odd, $m - n$ is even.
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_Proof (second):_
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Suppose both $m$ and $n$ are both even or are both odd.
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In order to prove $m - n$ is even, both cases must be considered.
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_Case (both $m$ and $n$ are even):_
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Since both $m$ and $n$ are even, $m = 2k$ and $n = 2p$ for some integers $k$ and
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$p$. Then:
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$$ m - n = 2k - 2p $$
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$$ = 2(k - p) $$
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Now, $k - p$ is an integer by the subtraction of integers. Therefore, by the
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definition of even, $m - n$ is even.
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_Case (both $m$ and $n$ are odd):_
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Since both $m$ and $n$ are odd, $m = 2k + 1$ and $n = 2p + 1$ for some integers
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$k$ and $p$. Then:
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$$ m - n = (2k + 1) - (2p + 1) $$
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$$ = 2k + 1 - 2p - 1 $$
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$$ = 2k - 2p $$
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$$ = 2(k - p) $$
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Now, $k - p$ is an integer by the subtraction of integers. Therefore, by the
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definition of even, $m - n$ is even.
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_Conclusion:_
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In both cases, $m - n$ is even. Therefore it can be concluded that if both $m$
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and $n$ are even or if both $m$ and $n$ are odd, then $m - n$ is even.
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3. The **congruence modulo $3$** relation, $T$, is defined from $\mathbb{Z}$ to
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$\mathbb{Z}$ as follows: For all integers $m$ and $n$,
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$$ m T n \Leftrightarrow 3 | (m - n) $$
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a. Is $10 T 1$? Is $1 T 10$? Is $(2, 2) \in T$? Is $(8, 1) \in T$?
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_$10 T 1$:_
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Yes, since $3 | (10 - 1) = 3 | 9 = 3$
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_$1 T 10$:_
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Yes, since $3 | (1 - 10) = 3 | -9 = -3$
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_$(2, 2) \in T$:_
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Yes, since $3 | (2 - 2) = 3 | 0 = 0$
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_$(8, 1) \in T$:_
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No, since $3 | (8 - 1) = 3 \cancel{|} 7$.
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b. List five integers $n$ such that $n T 0$.
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$3$; $6$, $9$, $12$, $15$
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c. List five integers $n$ such that $n T 1$.
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$4$; $7$, $10$, $13$, $16$
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d. List five integers $n$ such that $n T 2$.
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$$ 3 | (n - 2) $$
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$5$, $8$, $11$, $14$, $17$
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e. Make and prove a conjecture about which integers are related by $T$ to $0$,
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which integers are related to $T$ to $1$, and which integers are related to $T$
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to $2$.
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_Hint:_ All integers of the form $3k + 1$, for some integer $k$, are related by
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$T$ to $1$.
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**Conjecture:**
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All integers of the form $3k$, for some integer $k$, are related by $T$ to $0$.
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All integers of the form $3p + 1$, for some integer $p$, are related by $T$ to
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$1$.
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All integers of the form $3m + 2$, for some integer $m$, are related to $T$ by
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$2$.
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4. Define a relation $P$ on $\mathbb{Z}$ as follows: For every ordered pair
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$(m, n) \in \mathbb{Z} \times \mathbb{Z}$,
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$$ m P n \Leftrightarrow m \text{ and } n \text{ have a common prime factor} $$
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a. Is $15 P 25$?
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Yes, because both $15$ and $25$ are divisible by $5$, which is a prime factor.
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b. Is $22 P 27$?
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No, because $22$ and $27$ have no common divisors.
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c. Is $0 P 5$?
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Yes, because both $0$ and $5$ are divisible by $5$, which is a prime factor.
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d. Is $8 P 8$?
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Yes, because both $8$ and $8$ are divisible by $2$, which is a prime factor.
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5. Let $X = \{a, b, c\}$. Recall that $\mathscr{P}(X)$ is the power set of $X$.
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Define a relation $\mathbf{S}$ on $\mathscr{P}(X)$ as follows: For all sets
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$A$ and $B$ in $\mathscr{P}(X)$,
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$$ A \mathbf{S}B \Leftrightarrow A \text{ has the same number of elements as } B $$
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a. Is $\{a, b\} \mathbf{S} \{b, c\}$?
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Yes, since both $\{a, b\}$ and $\{b, c}$ have the same number of elements,
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namely $2$ elements.
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b. Is $\{a\} \mathbf{S} \{a, b\}$?
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No, since $\{a\}$ has $1$ element and $\{a, b\}$ has $2$ elements, and
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$1 \neq 2$.
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c. Is $\{c\} \mathbf{S} \{b\}$?
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Yes, since both $\{c\}$ and $\{b\}$ have the same number of elements, namely $1$
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element.
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6. Let $X = \{a, b, c\}$. Recall that $\mathscr{P}(X)$ as follows: For all sets
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$A$ and $B$ in $\mathscr{P}(X)$,
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$$ A \mathbf{J} B \Leftrightarrow A \cap B \neq \emptyset $$
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a. Is $\{a\} \mathbf{J} \{c\}$?
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No, since $\{a\} \cap \{\c} = \emptyset$.
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b. Is $\{a, b\} \mathbf{J} \{b, c\}$?
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Yes, since $\{a, b\} \cap \{b, c\} = \{b\} \neq \emptyset$.
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c. Is $\{a, b} \mathbf{J} \{a, b, c\}$?
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Yes, since $\{a, b\} \cap \{a, b, c\} = \{a, b\} \neq \emptyset$.
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7. Define a relation $R$ on $\mathbb{Z}$ as follows: For all integers $m$ and
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$n$,
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$$ m R n \Leftrightarrow 5 | (m^2 - n^2) $$
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a. Is $1 R (-9)$?
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$$ 5 | ((1)^2 - (-9)^2) $$
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$$ 5 | (1 - 81) $$
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$$ 5 | (-80) = -16 $$
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Yes.
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b. Is $2 R 13$?
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$$ 5 | ((2)^2 - (13)^2) $$
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$$ 5 | (4 - 169) $$
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$$ 5 | (-165) = -33 $$
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Yes.
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c. Is $2 R (-8)$?
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$$ 5 | ((2)^2 - (-8)^2) $$
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$$ 5 | (4 - (64)) $$
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$$ 5 | (-60) = -12 $$
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Yes.
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d. Is $(-8) R 2$?
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$$ 5 | (64 - 4) $$
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$$ 5 | 60 = 12 $$
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Yes.
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8. Let $A$ be the set of all strings of _a_'s and _b_'s of length $4$. Define a
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relation $R$ on $A$ as follows: For every $s, t \in A$,
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$$ s R t \Leftrightarrow s \text{ has the same first two characters as } t $$
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a. Is _abaa_ $R$ _abba_?
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Yes, since _ab_ is the same first two characters of both _abaa_ and _abba_.
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b. Is _aabb_ $R$ _bbaa_?
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No, since _aa_ is the first two characters of _aabb_ and _bb_ is the first same
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two characters as _bbaa_, it can be concluded that _aabb_ and _bbaa_ do not have
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the same first two characters.
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c. Is _aaaa_ $R$ _aaab_?
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Yes, since _aa_ is the same first two characters of both _aaaa_ and _aaab_.
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d. Is _baaa_ $R$ _abaa_?
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No, since _ba_ and _ab_ are the first two characters of _baaa_ and _abaa_
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respectively.
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9. Let $A$ be the set of all strings of 0's, 1's, and 2's of length $4$. Define
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a relation $R$ on $A$ as follows: For every $s, t \in A$,
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$$ s R t \Leftrightarrow \text{ the same of the characters in } s \text{ equals the sum of the characters in } t $$
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a. Is 0121 $R$ 2200?
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$$ 0 + 1 + 2 + 1 = 4 = 2 + 2 + 0 + 0 $$
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Yes.
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b. Is 1011 $R$ 2101?
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$$ 1 + 0 + 1 + 1 = 3 = \neq 4 = 2 + 1 + 0 + 1 $$
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No.
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c. Is 2212 $R$ 2121?
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$$ 2 + 2 + 1 + 2 = 7 \neq 6 = 2 + 1 + 2 + 1 $$
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No.
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d. Is 1220 $R$ 2111?
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$$ 1 + 2 + 2 + 0 = 5 = 2 + 1 + 1 + 1 $$
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Yes.
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10. Let $A = \{3, 4, 5\}$ and $B = \{4, 5, 6\}$ and let $R$ be the "less than"
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relation. That is, for every ordered pair $(x, y) \in A \times B$,
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$$ x R y \Leftrightarrow x < y $$
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State explicitly which ordered pairs are in $R$ and $R^{-1}$.
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$$ R = \{(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) \} $$
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$$ R^{-1} = \{(4, 3), (5, 3), (6, 3), (5, 4), (6, 4), (6, 5) \} $$
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11. Let $A = \{3, 4, 5\}$ and $B = \{4, 5, 6\}$ and let $S$ be the "divides"
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relation. That is, for every ordered pair $(x, y) \in A \times B$,
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$$ x S y \Leftrightarrow x | y $$
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State explicitly which ordered pairs are in $S$ and $S^{-1}$.
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$$ S = \{(3, 6), (4, 4), (5, 5)\} $$
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$$ S^{-1} = \{(6, 3), (4, 4), (5, 5)\} $$
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12.
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a. Suppose a function $F: X \to Y$ is one-to-one but not onto. Is $F^{-1}$ (the
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inverse relation for $F$) a function? Explain your answer.
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No, if $F: X \to Y$ is one-to-one, but not onto, then its inverse relation
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$F^{-1}: Y \to X$ will have some elements in its domain that have not elements
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in the co-domain. More formally:
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$$ \exists y \in Y | (y, x) \notin F^{-1} $$
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which means $F^{-1}$ does not satisfy property 1 for being a function.
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b. Suppose a function $F: X \to Y$ is onto but not one-to-one. Is $F^{-1}$ (the
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inverse relation for $F$) a function? Explain your answer.
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No, if $F: X \to Y$ is onto, but not one-to-one, it follows that its inverse
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relation $F^{-1}: Y \to X$ will have at least one
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$y \in Y | (y, x_1) \in F^{-1} \wedge (y, x_2) \in F^{-1}$.
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This violates property 2 of the definition of a function.
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Draw the directed graphs of the relations defined in 13-18.
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13. Define a relation $R$ on $A = \{0, 1, 2, 3\}$ by
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$R = \{(0, 0), (1, 2), (2, 2)\}$.
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(Done by hand.)
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14. Define a relation $S$ on $B = \{a, b, c, d\}$ by
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$S = \{(a, b), (a, c), (b, c), (d, d)\}$.
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(Done by hand.)
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15. Let $A = \{2, 3, 4, 5, 6, 7, 8\}$ and define a relation $R$ on $A$ as
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follows: For every $x, y \in A$,
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$$ x R y \Leftrightarrow x | y $$
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(Done by hand.)
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16. Let $A = \{5, 6, 7, 8, 9, 10\}$ and define a relation $S$ on $A$ as follows:
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For every $x, y \in A$,
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$$ x S y \Leftrightarrow 2 | (x - y) $$
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(Done by hand.)
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17. Let $A = \{2, 3, 4, 5, 6, 7, 8\}$ and define a relation $T$ on $A$ as
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follows: For every $x, y \in A$,
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$$ x T y \Leftrightarrow 3 | (x - y) $$
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(Done by hand.)
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18. Let $A = \{0, 1, 3, 4, 5, 6\}$ and define a relation $V$ on $A$ as follows:
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For every $x, y \in A$,
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$$ x V y \Leftrightarrow 5 | (x^2 - y^2) $$
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(Done by hand.)
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Exercises 19-20 refer to unions and intersections of relations. Since relations
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are subsets of Cartesian products, their unions and intersections can be
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calculated as for any subsets. Given two relations $R$ and $S$ from $A$ to $B$,
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$$ R \cup S = \{(x, y) \in A \times B | (x, y) \in R \text{ or } (x, y) \in S\} $$
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$$ R \cap S = \{(x, y) \in A \times B | (x, y) \in R \text{ and } (x, y) \in S\} $$
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19. Let $A = \{2, 4\}$ and $B = \{6, 8, 10\}$ and define relations $R$ and $S$
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from $A$ to $B$ as follows: For every $(x, y) \in A \times B$,
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$$ x R y \Leftrightarrow x | y \text{ and } x S y \Leftrightarrow y - 4 = x $$
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State explicitly which ordered pairs are in $A \times B$, $R$, $S$, $R \cup S$,
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and $R \cap S$.
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$$ A \times B = \{(2, 6), (2, 8), (2, 10), (4, 6), (4, 8), (4, 10)\} $$
|
|
|
|
$$ R = \{(2, 6), (2, 8), (2, 10), (4, 8)\} $$
|
|
|
|
$$ S = \{(2, 6), (4, 8)\} $$
|
|
|
|
$$ R \cup S = \{(2, 6), (2, 8), (2, 10), (4, 8)\} = R $$
|
|
|
|
$$ R \cap S = \{(2, 6), (4, 8)\} = S $$
|
|
|
|
20. Let $A = \{-1, 1, 2, 4\}$ and $B = \{1, 2\}$ and define relations $R$ and
|
|
$S$ from $A$ to $B$ as follows: For every $(x, y) \in A \times B$,
|
|
|
|
$$ x R y \Leftrightarrow |x| = |y| \text{ and } x S y \Leftrightarrow x - y \text{ is even} $$
|
|
|
|
State explicitly which ordered pairs are in $A \times B$, $R$, $S$, $R \cup S$,
|
|
and $R \cap S$.
|
|
|
|
$$ A \times B = \{(-1, 1), (-1, 2), (1, 1), (1, 2), (2, 1), (2, 2), (4, 1), (4, 2)\} $$
|
|
|
|
$$ R = \{(-1, 1), (1, 1), (2, 2)\} $$
|
|
|
|
$$ S = \{(-1, 1), (1, 1), (2, 2), (4, 2)\} $$
|
|
|
|
$$ R \cup S = \{(-1, 1), (1, 1), (2, 2), (4, 2)\} = S $$
|
|
|
|
$$ R \cap S = \{(-1, 1), (1, 1), (2, 2)\} = R $$
|
|
|
|
21. Define relations $R$ and $S$ on $\mathbb{R}$ as follows:
|
|
|
|
$$ R = \{(x, y) \in \mathbb{R} \times \mathbb{R} | x < y\} \text{ and } S = \{(x, y) \in \mathbb{R} \times \mathbb{R} | x = y\}$$
|
|
|
|
That is, $R$ is the "less than" relation and $S$ is the "equals" relation on
|
|
$\mathbb{R}$. Graph $R$, $S$, $R \cup S$, and $R \cap S$ in the Cartesian plane.
|
|
|
|
Think on this and then see appendix b (Page 975).
|
|
|
|
22. Define relations $R$ and $S$ on $\mathbb{R}$ as follows:
|
|
|
|
$$ R = \{(x, y) \in \mathbb{R} \times \mathbb{R} | x^2 + y^2 = 4\} \text{ and } S = \{(x, y) \in \mathbb{R} \times \mathbb{R} | x = y\} $$
|
|
|
|
Graph $R$, $S$, $R \cup S$, and $R \cap S$ in the Cartesian plane.
|
|
|
|
23. Define relations $R$ and $S$ on $\mathbb{R}$ as follows:
|
|
|
|
$$ R = \{(x, y) \in \mathbb{R} \times \mathbb{R} | y = |x|\} \text{ and } S = \{(x, y) \in \mathbb{R} \times \mathbb{R} | y = 1\} $$
|
|
|
|
Graph $R$, $S$, $R \cup S$, and $R \cap S$ in the Cartesian plane.
|
|
|
|
$R$ is a circle about the origin (with intersections along the axis along
|
|
$(-2, 0), (0, 2), (2, 0), (-2, 0)$). $S$ is a straight diagonal line ascending
|
|
from the left to the right, intersecting the origin $(0, 0)$.
|
|
|
|
$R \cup S$ is just the two graphs drawn together.
|
|
|
|
$R \cap S$ is only the two points along which the two graphs intersect.
|
|
|
|
(Done by hand.)
|
|
|
|
24. In Example 8.1.7 consider the query SELECT Patient_ID#, Name FROM S WHERE
|
|
Primary_Diagnosis = X. The response query is the projection onto the first
|
|
two coordinates of the intersection of the database with the set
|
|
$A_1 \times A_2 \times A_3 \times \{X\}$.
|
|
|
|
a. Find the result of the query SELECT Patient_ID#, Name FROM S WHERE
|
|
Primary_Diagnosis = pneumonia.
|
|
|
|
(574329, Tak Kurosawa),
|
|
|
|
(011985, John Schmidt)
|
|
|
|
b. Find the result of the query SELECT Patient_ID#, Name FROM S WHERE
|
|
Primary_Diagnosis = appendicitis.
|
|
|
|
(466581, Mary Lazars),
|
|
|
|
(778400, Jamal Baskers)
|
|
|
|
---
|
|
|
|
Page 526
|
|
|
|
**Exercise Set 8.2**
|
|
|
|
In 1-8, a number of relations are defined on the set $A = \{0, 1, 2, 3\}$. For
|
|
each relation:
|
|
|
|
a. Draw the directed graph.
|
|
|
|
b. Determine whether the relation is reflexive.
|
|
|
|
c. Determine whether the relation is symmetric.
|
|
|
|
d. Determine whether the relation is transitive.
|
|
|
|
Give a counterexample in each case in which the relation does not satisfy one of
|
|
the properties.
|
|
|
|
1. $R_1 = \{(0, 0), (0, 1), (0, 3), (1, 1), (1, 0), (2, 3), (3, 3)\}$
|
|
|
|
a. Draw the directed graph.
|
|
|
|
(Done by hand.)
|
|
|
|
b. Determine whether the relation is reflexive.
|
|
|
|
No, $2 \cancel{R_1} 2$.
|
|
|
|
c. Determine whether the relation is symmetric.
|
|
|
|
No, $0 R_1 3$, but $3 \cancel{R_1} 0$.
|
|
|
|
d. Determine whether the relation is transitive.
|
|
|
|
No, $1 R_1 0$ and $0 R_1 3$, but $1 \cancel{R_1} 3$
|
|
|
|
2. $R_2 = \{(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3)\}$
|
|
|
|
a. Draw the directed graph.
|
|
|
|
(Done by hand.)
|
|
|
|
b. Determine whether the relation is reflexive.
|
|
|
|
No, since $3 \cancel{R_2} 3$.
|
|
|
|
c. Determine whether the relation is symmetric.
|
|
|
|
No, $0 R_2 1$, but $1 \cancel{R_2} 0$.
|
|
|
|
d. Determine whether the relation is transitive.
|
|
|
|
No, $0 R_2 1$ and $1 R_2 2$, but $0 \cancel{R_2} 2$.
|
|
|
|
3. $R_3 = \{(2, 3), (3, 2)\}$
|
|
|
|
a. Draw the directed graph.
|
|
|
|
(Done by hand.)
|
|
|
|
b. Determine whether the relation is reflexive.
|
|
|
|
No, $2 \cancel{R_3} 2$.
|
|
|
|
c. Determine whether the relation is symmetric.
|
|
|
|
Yes, $2 R_3 3$ and $3 R_3 2$.
|
|
|
|
d. Determine whether the relation is transitive.
|
|
|
|
No, $2 R_3 3$ and $3 R_3 2$, but $2 \cancel{R_3} 2$.
|
|
|
|
4. $R_4 = \{(1, 2), (2, 1), (1, 3), (3, 1)\}$
|
|
|
|
a. Draw the directed graph.
|
|
|
|
(Done by hand.)
|
|
|
|
b. Determine whether the relation is reflexive.
|
|
|
|
No, $1 \cancel{R_4} 1$.
|
|
|
|
c. Determine whether the relation is symmetric.
|
|
|
|
Yes, $1 R_4 2$ and $2 R_4 1$ and $1 R_4 3$ and $3 R_4 1$.
|
|
|
|
d. Determine whether the relation is transitive.
|
|
|
|
No, $1 R_4 2$ and $2 R_4 1$, but $1 \cancel{R_4} 1$.
|
|
|
|
5. $R_5 = \{(0, 0), (0, 1), (0, 2), (1, 2)\}$
|
|
|
|
a. Draw the directed graph.
|
|
|
|
(Done by hand.)
|
|
|
|
b. Determine whether the relation is reflexive.
|
|
|
|
No, $1 \cancel{R_5} 1$.
|
|
|
|
c. Determine whether the relation is symmetric.
|
|
|
|
No, $0 R_5 1$, but $1 \cancel{R_5} 0$.
|
|
|
|
d. Determine whether the relation is transitive.
|
|
|
|
Yes, $0 R_5 1$ and $1 R_5 2$, and $0 R_5 2$.
|
|
|
|
6. $R_6 = \{(0, 1), (0, 2)\}$
|
|
|
|
a. Draw the directed graph.
|
|
|
|
(Done by hand.)
|
|
|
|
b. Determine whether the relation is reflexive.
|
|
|
|
No, $0 \cancel{R_6} 0$.
|
|
|
|
c. Determine whether the relation is symmetric.
|
|
|
|
No, $0 R_6 1$, but $1 \cancel{R_6} 0$.
|
|
|
|
d. Determine whether the relation is transitive.
|
|
|
|
Yes, vacuously.
|
|
|
|
7. $R_7 = \{(0, 3), (2, 3)\}$
|
|
|
|
a. Draw the directed graph.
|
|
|
|
(Done by hand.)
|
|
|
|
b. Determine whether the relation is reflexive.
|
|
|
|
No, $0 \cancel{R_7} 0$.
|
|
|
|
c. Determine whether the relation is symmetric.
|
|
|
|
No, $0 R_7 3$, but $3 \cancel{R_7} 0$.
|
|
|
|
d. Determine whether the relation is transitive.
|
|
|
|
Yes, vacuously.
|
|
|
|
8. $R_8 = \{(0, 0), (1, 1)\}$
|
|
|
|
a. Draw the directed graph.
|
|
|
|
(Done by hand.)
|
|
|
|
b. Determine whether the relation is reflexive.
|
|
|
|
Yes, both $0 R_8 0$ and $1 R_8 1$.
|
|
|
|
c. Determine whether the relation is symmetric.
|
|
|
|
Yes, since $0 R_8 0$ and $0 R_8 0$, and also $1 R_8 1$ and $1 R_8 1$.
|
|
|
|
d. Determine whether the relation is transitive.
|
|
|
|
Yes, vacuously.
|
|
|
|
In 9-33, determine whether the given relation is reflexive, symmetric,
|
|
transitive, or none of these. Justify your answers.
|
|
|
|
9. $R$ is the "greater than or equal to" relation on the set of real numbers:
|
|
For every $x, y \in \mathbb{R}$, $x R y \Leftrightarrow x \geq y$.
|
|
|
|
a. Is $R$ reflexive?
|
|
|
|
Yes, since $\forall x \in \mathbb{R}, x = x$, it follows that
|
|
$\forall x \in \mathbb{R}, x \geq x$.
|
|
|
|
b. Is $R$ symmetric?
|
|
|
|
No, since $\forall x, y \in \mathbb{R}, x \geq y \to y \geq x$ cannot be true.
|
|
Consider the example that $x = 5$ and $y = 4$, then $x \geq y$, but
|
|
$y \cancel{\geq} x$.
|
|
|
|
c. Is $R$ transitive?
|
|
|
|
Yes, since
|
|
$\forall x, y, z \in \mathbb{R}, (x \geq y \wedge y \geq z) \to x \geq z$ is
|
|
true by the transitive law of greatness (See appendix A, T18).
|
|
|
|
10. $C$ is the circle relation on the set of real numbers: For every
|
|
$x, y \in \mathbb{R}, x C y \Leftrightarrow x^2 + y^2 = 1$.
|
|
|
|
a. Is $C$ reflexive?
|
|
|
|
No, $C$ is not reflexive. The statement claims that
|
|
$\forall x \in \mathbb{R}, x C x \Leftrightarrow x^2 + x^2 = 1$, but consider
|
|
$x = 0$, then $0^2 + 0^2 = 1$, but $0 \neq 1$, this is a contradiction.
|
|
|
|
b. Is $C$ symmetric?
|
|
|
|
Yes, $C$ is symmetric. The statement claims that
|
|
$x, y \in \mathbb{R}, (x^2 + y^2 = 1) \to (y^2 + x^2 = 1)$. This is true by the
|
|
commutative laws of addition.
|
|
|
|
c. Is $C$ transitive?
|
|
|
|
No, $C$ is not transitive. The statement claims that
|
|
$x, y, z \in \mathbb{R}, [(x^2 + y^2 = 1) \wedge (y^2 + z^2 = 1)] \to x^2 + z^2 = 1$.
|
|
Consider $x = 1$, $y = 0$, and $z = 1$, then $x^2 + y^2 = (1)^2 + (0)^2 = 1$ and
|
|
$y^2 + z^2 = (0)^2 + (1)^2 = 1$, but $x^2 + z^2 = (1)^2 + (1)^2 = 2 \neq 1$.
|
|
|
|
11. $D$ is the relation defined on $\mathbb{R}$ as follows: For every
|
|
$x, y \in \mathbb{R}, x D y \Leftrightarrow xy \geq 0$.
|
|
|
|
a. Is $D$ reflexive?
|
|
|
|
Yes, $D$ is reflexive. $\forall x \in \mathbb{R} x \cdot x \geq 0$ is a true
|
|
statement, as even if $x$ is negative, any negative number times itself will
|
|
always be positive, and so $x \geq 0$ is true. If $x = 0$, then $x \geq 0$ is a
|
|
true statement. If $x$ is positive, then any positive number times itself will
|
|
be positive, and so $x \geq 0$ is true.
|
|
|
|
b. Is $D$ symmetric?
|
|
|
|
Yes, $D$ is symmetric,
|
|
$\forall x, y \in \mathbb{R}, (xy \geq 0) \to (yx \geq 0)$ is true by the
|
|
commutative laws of multiplication since $xy = yx$.
|
|
|
|
c. Is $D$ transitive?
|
|
|
|
No, $D$ is not transitive. The statement claims
|
|
$\forall x, y, z \in \mathbb{R}, [(xy \geq 0) \wedge (yz \geq 0)] \to (xz \geq 0)$.
|
|
This is not true, consider $x = 1$, $y = 0$, and $z = -1$, then
|
|
$xy = (1)(0) = 0 \geq 0$, and $yz = (0)(-1) = 0 \geq 0$, but
|
|
$xz = (1)(-1) = -1 \cancel{\geq} 0$.
|
|
|
|
12. $E$ is the congruence modulo $4$ relation on $\mathbb{Z}$: For every
|
|
$m, n \in \mathbb{Z}, m E n \Leftrightarrow 4 | (m - n)$.
|
|
|
|
a. Is $E$ reflexive?
|
|
|
|
Yes, $E$ is reflexive. The statement claims
|
|
$\forall m \in \mathbb{Z}, 4 | (m - m)$. Since any integer subtracted from
|
|
itself is $0$, this means that:
|
|
|
|
$$ 4 | (m - m) = 4 | 0 $$
|
|
|
|
Which is true since $4 = 4 \cdot 0$.
|
|
|
|
b. Is $E$ symmetric?
|
|
|
|
Yes, $E$ is symmetric. The statement claims
|
|
$\forall m, n \in \mathbb{Z}, [4 | (m - n)] \to [4 | (n - m)]$.
|
|
|
|
Since $4 | (m - n)$, this means that $m - n = 4k$ for some integer $k$. It
|
|
follows then that:
|
|
|
|
$$ n - m = -1(m - n) $$
|
|
|
|
$$ = -1(4k) $$
|
|
|
|
$$ = 4(-k) $$
|
|
|
|
Now, $-k$ is an integer by the multiplication of integers. It follows then that
|
|
$4 | (n - m)$. This is what was to be shown.
|
|
|
|
c. Is $E$ transitive?
|
|
|
|
Yes, $E$ is transitive. The statement claims that
|
|
$\forall m, n, p \in \mathbb{Z}, [(4 | (m - n)) \wedge (4 | (n - p))] \to (4 | (m - p))$.
|
|
|
|
Since $4 | (m - n)$ and $4 | (n - p)$, it can be said that $m - n = 4r$ and
|
|
$n - p = 4s$ for some integers $r$ and $s$. It follows by addition of these two
|
|
terms, and substitution, that:
|
|
|
|
$$ (m - n) + (n - p) = 4r + 4s $$
|
|
|
|
and also that:
|
|
|
|
$$ (m - n) + (n - p) = m - p $$
|
|
|
|
Then, setting the substitution equal to the evaluation/simplification:
|
|
|
|
$$ 4r + 4s = m - p $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ 4(r + s) = m - p $$
|
|
|
|
Now, $r + s$ is an integer by the sum of integers. It follows that
|
|
$4 | (m - p)$. This is what was to be shown.
|
|
|
|
13. $F$ is the congruence modulo $5$ relation on $\mathbb{Z}$: For every
|
|
$m, n \in \mathbb{Z}, m F n \Leftrightarrow 5 | (m - n)$.
|
|
|
|
a. Is $F$ reflexive?
|
|
|
|
Yes, $F$ is reflexive. The statement claims that
|
|
$\forall m \in \mathbb{Z}, 5 | (m - m)$. This is true since $m - m = 0$, and
|
|
$5 | 0$ is true since $5 = 5 \cdot 0$.
|
|
|
|
b. Is $F$ symmetric?
|
|
|
|
Yes, $F$ is symmetric. The statement claims that
|
|
$\forall m, n \in \mathbb{Z}, (5 | (m - n)) \to (5 | (n - m))$.
|
|
|
|
Since $5 | m - n$, it can be said that $m - n = 5k$ for some integer $k$. Then,
|
|
consider:
|
|
|
|
$$ m - n = -1(n - m) $$
|
|
|
|
By substitution then:
|
|
|
|
$$ 5k = -1(5k) $$
|
|
|
|
$$ 5k = 5(-k) $$
|
|
|
|
Now, $-k$ is an integer by the multiplication of integers. It follows that
|
|
$5 | (n - m)$. This is what was to be shown.
|
|
|
|
c. Is $F$ transitive?
|
|
|
|
Yes, $F$ is transitive. The statement claims that
|
|
$\forall m, n, p \in \mathbb{Z}, [(5 | (m - n)) \wedge (5 | (n - p))] \to [5 | (m - p)]$.
|
|
|
|
Since $5 | (m - n)$ and $5 | (n - p)$, it can be said that $m - n = 5r$ and
|
|
$n - p = 5s$ for some integers $r$ and $s$. Adding $m - n$ and $n - p$ gives
|
|
$m - p$:
|
|
|
|
$$ (m - n) + (n - p) = m - p $$
|
|
|
|
Then, by substitution:
|
|
|
|
$$ 5r + 5s = m - p $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ 5(r + s) = m - p $$
|
|
|
|
Now, $r + s$ is an integer by the sum of integers. It follows that
|
|
$5 | (m - p)$. This is what was to be shown.
|
|
|
|
14. $O$ is the relation defined on $\mathbb{Z}$ as follows: For every
|
|
$m, n \in \mathbb{Z}, m O n \Leftrightarrow m - n \text{ is odd}$.
|
|
|
|
a. Is $O$ reflexive?
|
|
|
|
No, $O$ is not reflexive. The statement claims that
|
|
$\forall m \in \mathbb{Z}, m - m \text{ is odd}$. Since $m - m = 0$, and $0$ is
|
|
even (since $0 = 2(0)$), by the definition of even, $m - m$ cannot be odd.
|
|
Therefore $O$ is not reflexive.
|
|
|
|
b. Is $O$ symmetric?
|
|
|
|
Yes, $O$ is symmetric. The statement claims that
|
|
$\forall m, n \in \mathbb{Z}, (m - n \text{ is odd}) \to (n - m \text{ is odd})$.
|
|
|
|
Since $m - n$ is odd, it can be said that $m - n = 2k + 1$ for some integer $k$.
|
|
Consider that:
|
|
|
|
$$ m - n = -1(n - m) $$
|
|
|
|
Then, by substitution:
|
|
|
|
$$ 2k + 1 = -1(n - m) $$
|
|
|
|
By algebra:
|
|
|
|
$$ -1(2k + 1) = n - m $$
|
|
|
|
$$ -2k - 1 = n - m $$
|
|
|
|
$$ 2(-k - 1) + 1 = n - m $$
|
|
|
|
Now, $-k - 1$ is an integer by the multiplication and sum of integers. Therefore
|
|
$n - m$ is odd. This is what was to be shown.
|
|
|
|
c. Is $O$ transitive?
|
|
|
|
No, $O$ is not transitive. The statement claims that
|
|
$\forall m, n, p \in \mathbb{Z} [(m - n \text{ is odd}) \wedge (n - p \text{ is odd})] \to [m - p \text{ is odd}]$.
|
|
This is not true for all integers. Consider $m = 2$, $n = 1$, and $p = 0$. Then
|
|
$m - n = 2 - 1 = 1 \text{ is odd}$, and $n - p = 1 - 0 = 1 \text{ is odd}$, but
|
|
$m - p = 2 - 0 = 2 \text{ is even}$. Therefore $0$ is not transitive.
|
|
|
|
15. $D$ is the "divides" relation on $\mathbb{Z}^+$: For all positive integers
|
|
$m$ and $n$, $m D n \Leftrightarrow m | n$.
|
|
|
|
a. Is $D$ reflexive?
|
|
|
|
Yes, $D$ is reflexive. The statement claims $\forall m \in \mathbb{Z}^+, m | m$.
|
|
This is true since any integer divides itself by the definition of divisibility.
|
|
|
|
b. Is $D$ symmetric?
|
|
|
|
No, $D$ is not symmetric. The statement claims
|
|
$\forall m, n \in \mathbb{Z}^+, (m | n) \to (n | m)$, but this is not true for
|
|
all positive integers. Consider $m = 2$ and $n = 4$, then $2 | 4$ is true since
|
|
$2 = 2 \cdot 2 = 4$, but $4 \cancel{|} 2$ since $4 \neq 4k = 2$ for some integer
|
|
$k$.
|
|
|
|
c. Is $D$ transitive?
|
|
|
|
Yes, $D$ is transitive. The statement claims
|
|
$\forall m, n, p \in \mathbb{Z}^+, [(m | n) \wedge (n | p)] \to [m | p]$. This
|
|
is true by the transitivity of divisibility (see Theorem 4.4.3).
|
|
|
|
16. $A$ is the "absolute value" relation on $\mathbb{R}$: For all real numbers
|
|
$x$ and $y$, $x A y \Leftrightarrow |x| = |y|$.
|
|
|
|
a. Is $A$ reflexive?
|
|
|
|
Yes, $A$ is reflexive. The statement claims
|
|
$\forall x \in \mathbb{R}, |x| = |x|$. This is trivially true.
|
|
|
|
b. Is $A$ symmetric?
|
|
|
|
Yes, $A$ is symmetric. The statement claims that
|
|
$\forall x, y \in \mathbb{R}, (|x| = |y|) \to (|y| = |x|)$. This is true by the
|
|
definition of equality.
|
|
|
|
c. Is $A$ transitive?
|
|
|
|
Yes, $A$ is transitive. The statement claims that
|
|
$\forall x, y, z \in \mathbb{R}, [(|x| = |y|) \wedge (|y| = |z|)] \to |x| = |z|$
|
|
|
|
This is true by the transitivity of equality (since $|x| = |y| = |z|$).
|
|
|
|
17. Recall that a prime number is an integer that is greater than $1$ and has no
|
|
positive integer divisors other than $1$ and itself. (In particular, $1$ is
|
|
not prime.) A relation $P$ is defined on $\mathbb{Z}$ as follows: For every
|
|
$m, n \in \mathbb{Z}, m P n \Leftrightarrow \exists \text{ a prime number } p \text{ such that } p | m \text{ and } p | n$.
|
|
|
|
a. Is $P$ reflexive?
|
|
|
|
No, $P$ is not reflexive. The statement claims
|
|
$\forall m \in \mathbb{Z}, \exists \text{ a prime number } p \text{ such that } p | m$.
|
|
Consider $m = 1$ (note that $1 \in \mathbb{Z}$), then there is no such prime
|
|
number $p$ that divides $m$.
|
|
|
|
b. Is $P$ symmetric?
|
|
|
|
Yes, $P$ is symmetric. The statement claims
|
|
$\forall m, n \in \mathbb{Z}, \exists \text{ some prime number } p \text{ such that } p | m \wedge p | n \to p | n \wedge p | m$.
|
|
|
|
Since there is a prime number $p$ that divides $m$ and $n$, it is trivially true
|
|
that $p$ divides $n$ and $m$.
|
|
|
|
c. Is $P$ transitive?
|
|
|
|
No, $P$ is not transitive. The statement claims that:
|
|
|
|
$$ \forall m, n, o \in \mathbb{Z}, [\exists \text{ some prime } p_1, p_1 | m \wedge p_1 | n] \wedge [\exists \text{ some prime } p_2, p_2 | n \wedge p_2 | o] \to [\exists \text{ some prime } p_3, p_3 | m \wedge p_3 | o] $$
|
|
|
|
But this is not true for all integers $m$, $n$, and $o$.
|
|
|
|
Consider $m = 6$, $n = 15$, $o = 35$.
|
|
|
|
Then there exists the prime number $p_1 = 3$ such that $3 | m$ since $3 | 6$
|
|
since $6 = 3 \cdot 2$. Additionally, $3 | n$ since $3 | 15$ since
|
|
$15 = 3 \cdot 5$, so the first term of the supposition is true.
|
|
|
|
Next, there exists the prime number $p_2 = 5$ such that $5 | n$ since $5 | 15$
|
|
since $15 = 5 \cdot 3$. Additionally $5 | o$ since $5 | 35$ since
|
|
$35 = 5 \cdot 7$, so the second term of the supposition is true.
|
|
|
|
Then, the conclusion claims that there exists some prime $p_3$ such $p_3 | m$
|
|
and $p_3 | o$, but the only prime numbers that divide $m$ are $3$ and $2$ since
|
|
$m = 6$, and the only prime numbers that divide $o$ are $7$ and $5$ since
|
|
$o = 35$. None of these primes are equal to each other, and so $p_3$ does not
|
|
exist. Therefore $P$ is not transitive.
|
|
|
|
18. Define a relation $Q$ on $\mathbb{R}$ as follows: For all real numbers $x$
|
|
and $y$, $x Q y \Leftrightarrow x - y$ is rational.
|
|
|
|
_Hint:_ $Q$ is reflexive, symmetric, and transitive.
|
|
|
|
a. Is $Q$ reflexive?
|
|
|
|
Yes, $Q$ is reflexive. The statement claims that
|
|
$\forall x \in \mathbb{R}, x - x \text{ is rational}$. This is true since
|
|
$x - x = 0$, and $0$ is rational since $0 = \dfrac{0}{1}$.
|
|
|
|
b. Is $Q$ symmetric?
|
|
|
|
Yes, $Q$ is symmetric. The statement claims that
|
|
$\forall x, y \in \mathbb{R}, (x - y \text{ is rational }) \to (y - x \text{ is rational})$.
|
|
|
|
Since $x - y$ is rational, it can be said that $x - y = \dfrac{a}{b}$, where $a$
|
|
is some integer and $b$ is some integer with $b \neq 0$. Now, consider that:
|
|
|
|
$$ x - y = -1(y - x) $$
|
|
|
|
$$ -1(x - y) = y - x $$
|
|
|
|
Then, by substitution:
|
|
|
|
$$ -1\left(\frac{a}{b}\right) = y - x $$
|
|
|
|
Now, $-1\left(\dfrac{a}{b}\right)$ is a rational number (since $-1$ multiplied
|
|
by a rational number is a rational number). Therefore $y - x$ is rational. This
|
|
is what was to be shown.
|
|
|
|
c. Is $Q$ transitive?
|
|
|
|
Yes, $Q$ is transitive. The statement claims that
|
|
$\forall x, y, z \in \mathbb{R}, [(x - y \text{ is rational}) \wedge (y - z \text{ is rational})] \to x - z \text{ is rational}$.
|
|
|
|
Since $x - y$ is rational and $y - z$ is rational, it can be said that
|
|
$x - y = \dfrac{a}{b}$ and $y - z = \dfrac{c}{d}$, where
|
|
$a, b, c, d \in \mathbb{Z}$ with $b \neq 0$ and $d \neq 0$.
|
|
|
|
Then, consider the addition of $x - y$ and $y - z$:
|
|
|
|
$$ (x - y) + (y - z) = x - z $$
|
|
|
|
Then, by substitution:
|
|
|
|
$$ x - z = \frac{a}{b} + \frac{c}{d} $$
|
|
|
|
$$ = \frac{ad + cb}{bd}$$
|
|
|
|
Now, $ad + cb$ is an integer by the product and sum of integers, and $bd$ is an
|
|
integer by the product of integers and $bd \neq 0$ (since $b \neq 0$ and
|
|
$d \neq 0$). Thus $\dfrac{ad + cb}{bd}$ is a rational number, and therefore
|
|
$x - z$ is rational. This is what was to be shown.
|
|
|
|
19. Define a relation $I$ on $\mathbb{R}$ as follows: For all real numbers $x$
|
|
and $y$, $x I y \Leftrightarrow x - y$ is irrational.
|
|
|
|
a. Is $I$ reflexive?
|
|
|
|
No, $I$ is not reflexive. The statement claims that
|
|
$\forall x \in \mathbb{R}, x - x \text{ is irrational}$. Since $x - x = 0$, and
|
|
$0 = \dfrac{0}{1}$, it follows that $x - x$ is rational. Therefore $I$ is not
|
|
reflexive.
|
|
|
|
b. Is $I$ symmetric?
|
|
|
|
Yes, $I$ is symmetric. The statement claims
|
|
$\forall x, y \in \mathbb{R}, (x - y \text{ is irrational}) \to (y - x \text{ is irrational})$.
|
|
|
|
Consider that:
|
|
|
|
$$ x - y = -1(y - x) $$
|
|
|
|
$$ -1(x - y) = y - x $$
|
|
|
|
Now, the product of $-1$ and an irrational number ($x - y$) is irrational. It
|
|
follows that $y - x$ is irrational. This is what was to be shown.
|
|
|
|
c. Is $I$ transitive?
|
|
|
|
The statement claims that
|
|
$\forall x, y, z \in \mathbb{R}, [(x - y \text{ is irrational}) \wedge (y - z \text{ is irrational})] \to x - z \text{ is irrational}$.
|
|
But this is not true for all integers $x$, $y$, and $z$.
|
|
|
|
Consider $x = \sqrt{2}$, $y = 0$, and $z = \sqrt{2}$.
|
|
|
|
Then $x - y = \sqrt{2} - 0 = \sqrt{2}$, which is irrational. Additionally,
|
|
$y - z = 0 - \sqrt{2} = -\sqrt{2}$, which is irrational. Thus the supposition is
|
|
true.
|
|
|
|
Then $x - z = \sqrt{2} - \sqrt{2} = 0$, which is rational (since
|
|
$0 = \dfrac{0}{1}$). Therefore $I$ is not transitive.
|
|
|
|
20. Let $X = \{a, b, c\}$ and $\mathscr{P}(X)$ be the power set of $X$ (the set
|
|
of all subsets of $X$). A relation $\mathbf{E}$ is defined on
|
|
$\mathscr{P}(X)$ as follows: For every
|
|
$A, B \in \mathscr{P}(X), A \mathbf{E} B \Leftrightarrow \text{ the number of elements in } A \text{ equals the number of elements in } B$.
|
|
|
|
a. Is $E$ reflexive?
|
|
|
|
Yes, $E$ is reflexive. The statement claims that
|
|
$\forall A \in \mathscr{P}(X), \text{ the number of elements in } A \text{ equals the number of elements in } A$.
|
|
|
|
This is trivially true.
|
|
|
|
b. Is $E$ symmetric?
|
|
|
|
Yes, $E$ is symmetric. The statement claims that
|
|
$\forall A, B \in \mathscr{P}(X), (\text{the number of elements in } A \text{ equals the number of elements in } B) \to (\text{the number of elements in } B \text{ equals the number of elements in } A)$.
|
|
|
|
This is trivially true (by the commutative laws of equality).
|
|
|
|
c. Is $E$ transitive?
|
|
|
|
Yes, $E$ is transitive. The statement claims that
|
|
$\forall A, B, C \in \mathscr{P}(X), [(\text{ the
|
|
number of elements in } A \text{ equals the number of elements in } B) \wedge
|
|
(\text{ the number of elements in } B \text{ equals the number of elements in }
|
|
C)] \to \text{the number of elements in } A \text{ equals the number of elements
|
|
in } C$.
|
|
|
|
This is trivially true (by the transitivity of equality).
|
|
|
|
21. Let $X = \{a, b, c\}$ and $\mathscr{P}(X)$ be the power set of $X$. A
|
|
relation $\mathbf{L}$ is defined on $\mathscr{P}(X)$ as follows: For every
|
|
$A, B \in \mathscr{P}(X), A \mathbf{L} B \Leftrightarrow \text{ the number of elements in } A \text{ is less than the number of elements in } B$.
|
|
|
|
a. Is $L$ reflexive?
|
|
|
|
No, $L$ is not reflexive. The statement claims
|
|
$\forall A \in \mathscr{P}(X), \text{ the number of elements in } A \text{ is less than the number of elements in } A$.
|
|
|
|
This cannot be true, since the number of elements in $A$ will always equal the
|
|
number of elements in $A$.
|
|
|
|
b. Is $L$ symmetric?
|
|
|
|
No, $L$ is not symmetric. The statement claims that
|
|
$\forall A, B \in \mathscr{P}(X), (\text{the number of elements in } A \text{ is less than the number of elements in } B) \to (\text{the number of elements in } B \text{ is less than the number of elements in } A)$.
|
|
|
|
Let $x= \text{ the number of elements in } A$ and
|
|
$y = \text{ the number of elements in } B$. Then, by the supposition, $x < y$.
|
|
By the definition of inequality, this means that $y \cancel{<} x$. Therefore $L$
|
|
is not symmetric.
|
|
|
|
c. Is $L$ transitive?
|
|
|
|
Yes, $L$ is transitive. The statement claims that
|
|
$\forall A, B, C \in \mathscr{P}(X), [(\text{the number of elements in } A \text{ is less than the number of elements in } B) \wedge (\text{the number of elements in } B \text{ is less than the number of elements in } C)] \to \text{ the number of elements in } A \text{ is less than the number of elements in } C$.
|
|
|
|
Let $x = \text{ the number of elements in } A$,
|
|
$y = \text{ the number of elements in } B$, and
|
|
$z = \text{ the number of elements in } C$.
|
|
|
|
Then, by the supposition, $x < y$ and $y < z$. Since $x < y < z$ (by the
|
|
transitivity of inequality), it follows that $x < z$. This is what was to be
|
|
shown. Therefore $L$ is transitive.
|
|
|
|
22. Let $X = \{a, b, c\}$ and $\mathscr{P}(X)$ be the power set of $X$. A
|
|
relation $\mathbf{N}$ is defined on $\mathscr{P}(X)$ as follows: For every
|
|
$A, B \in \mathscr{P}(X), A \mathbf{N} B \Leftrightarrow \text{ the number of elements in } A \text{ is not equal to the number of elements in } B$.
|
|
|
|
a. Is $\mathbf{N}$ reflexive?
|
|
|
|
No, $\mathbf{N}$ is not reflexive. The statement claims
|
|
$\forall A \in \mathscr{P}(X), \text{ the number of elements in } A \text{ is not equal to the number of elements in } A$.
|
|
|
|
This is trivially false.
|
|
|
|
b. Is $\mathbf{N}$ symmetric?
|
|
|
|
Yes, $\mathbf{N}$ is symmetric. The statement claims
|
|
$\forall A, B \in \mathscr{P}(X), (\text{the number of elements in } A \text{ is not equal to the number of elements in } B) \to (\text{ the number of elements in } B \text{ is not equal to the number of elements in } A)$.
|
|
|
|
This is true.
|
|
|
|
Let $x = \text{ the number of elements in } A$,
|
|
$y = \text{ the number of elements in } B$. Then, by the supposition,
|
|
$x \neq y$. It follows by the definition of inequality that $y \neq x$.
|
|
|
|
Therefore $\mathbf{N}$ is symmetric.
|
|
|
|
c. Is $\mathbf{N}$ transitive?
|
|
|
|
No, $\mathbf{N}$ is not transitive. The statement claims
|
|
$\forall A, B, C \in \mathscr{P}(X), [(\text{the number of elements in } A \text{ is not equal to the number of elements in } B) \wedge (\text{the number of elements in } B \text{ is not equal to the number of elements in } C)] \to \text{the number of elements in } A \text{ is not equal to the number of elements in } C$.
|
|
But this is not true for all subsets $A$, $B$, and $C$.
|
|
|
|
Consider $A = \{a\}$, $B = \{a, b\}$, and $C = \{c\}$.
|
|
|
|
Then, by the supposition, the number of elements in $A$ does not equal the
|
|
number of elements in $B$, and the number of elements in $B$ does not equal the
|
|
number of elements in $C$, but the number of elements in $A$ is equal to the
|
|
number of elements in $C$.
|
|
|
|
Therefore, $\mathbf{N}$ is not transitive.
|
|
|
|
23. Let $X$ be a nonempty set and $\mathscr{P}(X)$ the power set of $X$. Define
|
|
the "subset" relation $\mathbf{S}$ on $\mathscr{P}(X)$ as follows: For every
|
|
$A, B \in \mathscr{P}(X), A \mathbf{S} B \Leftrightarrow A \subseteq B$.
|
|
|
|
a. Is $\mathbf{S}$ reflexive?
|
|
|
|
Yes, $\mathbf{S}$ is reflexive. The statement claims
|
|
$\forall A \in \mathscr{P}(X), A \subseteq A$. By the definition of subset, this
|
|
is true.
|
|
|
|
b. Is $\mathbf{S}$ symmetric?
|
|
|
|
No, $\mathbf{S}$ is not symmetric. The statement claims
|
|
$\forall A, B \in \mathscr{P}(X), (A \subseteq B) \to (B \subseteq A)$.
|
|
|
|
Consider $X = \{1, 2, 3\}$, $A = \{1\}$, $B = \{1, 2\}$. Then, by the
|
|
supposition $A, B \in \mathscr{P}(X)$, and $A \subseteq B$, but
|
|
$B \nsubseteq A$. Therefore $\mathbf{S}$ is not symmetric.
|
|
|
|
c. Is $\mathbf{S}$ transitive?
|
|
|
|
Yes, $\mathbf{S}$ is transitive. The statement claims that
|
|
$\forall A, B, C \in \mathscr{P}(X), [(A \subseteq B) \wedge (B \subseteq C)] \to [A \subseteq C]$.
|
|
|
|
By the supposition $A \subseteq B$ and $B \subseteq C$, it follows by the
|
|
transitivity property of subset that $A \subseteq B \subseteq C$, and thus
|
|
$A \subseteq C$. Therefore $\mathbf{S}$ is transitive.
|
|
|
|
24. Let $X$ be a nonempty set and $\mathscr{P}(X)$ the power set of $X$. Define
|
|
the "not equal to" relation $\mathbf{U}$ on $\mathscr{P}(X)$ as follows: For
|
|
every $A, B \in \mathscr{P}(X), A \mathbf{U} B \Leftrightarrow A \neq B$.
|
|
|
|
a. Is $\mathbf{U}$ reflexive?
|
|
|
|
No, $\mathbf{U}$ is not reflexive. The statement claims
|
|
$\forall A \in \mathscr{P}(X), A \neq A$. This is trivially false.
|
|
|
|
b. Is $\mathbf{U}$ symmetric?
|
|
|
|
Yes, $\mathbf{U}$ is symmetric. The statement claims
|
|
$\forall A, B \in \mathscr{P}, (A \neq B) \to (B \neq A)$. This is true by the
|
|
definition of inequality.
|
|
|
|
c. Is $\mathbf{U}$ transitive?
|
|
|
|
No, $\mathbf{U}$ is not transitive. The statement claims
|
|
$\forall A, B, C \in \mathscr{P}, [(A \neq B) \wedge (B \neq C)] \to [A \neq C]$.
|
|
|
|
Let $X = \{1, 2, 3\}$, $A = \{1\}$, $B = \{2\}$, and $C = \{1\}$. Then, by the
|
|
supposition, $A, B, C \in \mathscr{P}(X)$, $A \neq B$ and $B \neq C$, but
|
|
$A = C$.
|
|
|
|
Therefore $\mathbf{U}$ is not transitive.
|
|
|
|
25. Let $A$ be the set of all strings of _a_'s and _b_'s of length $4$. Define a
|
|
relation $R$ on $A$ as follows: For every
|
|
$s, t \in A, s R t \Leftrightarrow s \text{ has the same first two characters as } t$.
|
|
|
|
a. Is $R$ reflexive?
|
|
|
|
Yes, $R$ is reflexive. The statement claims
|
|
$\forall s \in A, s \text{ has the same first two characters as } s$. This is
|
|
trivially true.
|
|
|
|
b. Is $R$ symmetric?
|
|
|
|
Yes, $R$ is symmetric. The statement claims
|
|
$\forall s, t \in A, (s \text{ has the same first two characters as } t) \to (t \text{ has the same first two characters as} s)$.
|
|
|
|
This is trivially true.
|
|
|
|
c. Is $R$ transitive?
|
|
|
|
Yes, $R$ is transitive. The statement claims
|
|
$\forall s, t, u \in A, [(s \text{ has the same first two characters as } t) \wedge (t \text{ has the same first two characters as } u)] \to s \text{ has the same first two characters as } u$.
|
|
|
|
This is true by the transitivity of equality, since $s$ and $t$ have the same
|
|
first two characters, and $t$ and $u$ have the same first two characters, it
|
|
follows that $s$ and $u$ have the same first two characters. Therefore $R$ is
|
|
transitive.
|
|
|
|
26. Let $A$ be the set of all strings of 0's, 1's, and 2's that have length 4
|
|
and for which the sum of the characters in the string is less than or equal
|
|
to 2. Define a relation $R$ on $A$ as follows: For every
|
|
$s, t \in A, s R t \Leftrightarrow \text{ the sum of the characters of } s \text{ equals the sum of the characters of } t$.
|
|
|
|
a. Is $R$ reflexive?
|
|
|
|
Yes, $R$ is reflexive. The statement claims
|
|
$\forall s \in A, \text{ the sum of the characters of } s \text{ equals the sum of the characters of } s$.
|
|
This is trivially true.
|
|
|
|
b. Is $R$ symmetric?
|
|
|
|
Yes, $R$ is symmetric. The statement claims
|
|
$\forall s, t \in A, (\text{ the sum of the characters of} s \text{ equals the sum of the characters of } t) \to (\text{ the sum of the characters of } t \text{ equals the sum of the characters of } s)$.
|
|
|
|
Let $x = \text{ the sum of the characters of } s$ and
|
|
$y = \text{ the sum of the characters of } t$. Then, by the supposition,
|
|
$x = y$. It follows by symmetry of equality that $y = x$. This is what was to be
|
|
shown. Therefore $R$ is symmetric.
|
|
|
|
c. Is $R$ transitive?
|
|
|
|
Yes, $R$ is transitive. The statement claims
|
|
$\forall s, t, u \in A, [(\text{ the sum of the characters of } s \text{ equals the sum of the characters of } t) \wedge (\text{ the sum of the characters of } t \text{ equals the sum of the characters of } u)] \to \text{ the sum of the characters of } s \text{ equals the sum of the characters of } u$.
|
|
|
|
Let $x = \text{ the sum of the characters of } s$,
|
|
$y = \text{ the sum of the characters of } t$, and
|
|
$z = \text{ the sum of the characters of } u$.
|
|
|
|
By the supposition $x = y$ and $y = z$. By the transitivity of equality,
|
|
$x = y = z$, and it follows that $x = z$. This is what was to be shown.
|
|
Therefore $R$ is transitive.
|
|
|
|
27. Let $A$ be the set of all English statements. A relation $\mathbf{I}$ is
|
|
defined on $A$ as follows: For every $p, q \in A$,
|
|
|
|
$$ p \mathbf{I} q \Leftrightarrow p \to q \text{ is true} $$
|
|
|
|
a. Is $\mathbf{I}$ reflexive?
|
|
|
|
Yes $\mathbf{I}$ is reflexive. The statement claims
|
|
$\forall p \in A, p \to p \text{ is true}$. This is true by the law of identity
|
|
(tautology).
|
|
|
|
b. Is $\mathbf{I}$ symmetric?
|
|
|
|
No, $\mathbf{I}$ is not symmetric. The statement claims
|
|
$\forall p, q \in A, (p \to q) \to (q \to p)$.
|
|
|
|
Consider $p$ is the statement "All pigs can fly", and $q$ is the statement "The
|
|
sky is blue". Then, by the supposition $p, q \in A$, and $p \to q$ is vacuously
|
|
true. But, $q \to p$ is false, since $q$ is true and $p$ is false.
|
|
|
|
Therefore $\mathbf{I}$ is not symmetric.
|
|
|
|
c. Is $\mathbf{I}$ transitive?
|
|
|
|
Yes, $\mathbf{I}$ is transitive. The statement claims
|
|
$\forall p, q, r \in A, [(p \to q) \wedge (q \to r)] \to (p \to r)$.
|
|
|
|
This is true, since $p \to q$ and $q \to r$ is true, it follows that
|
|
$p \to q \to r$, and that $p \to r$ is true.
|
|
|
|
28. Let $A = \mathbb{R} \times \mathbb{R}$. A relation $\mathbf{F}$ is defined
|
|
on $A$ as follows: For every $(x_1, y_1)$ and $(x_2, y_2)$ in $A$,
|
|
|
|
$$ (x_1, y_2) \mathbf{F} (x_2, y_2) \Leftrightarrow x_1 = x_2 $$
|
|
|
|
a. Is $\mathbf{F}$ reflexive?
|
|
|
|
Yes, $\mathbf{F}$ is reflexive. The statement claims
|
|
$\forall (x_1, y_1) \in A, x_1 = x_1$. This is trivially true.
|
|
|
|
b. Is $\mathbf{F}$ symmetric?
|
|
|
|
Yes, $\mathbf{F}$ is symmetric. The statement claims
|
|
$\forall (x_1, y_1), (x_2, y_2) \in A, (x_1 = x_2) \to (x_2 = x_1)$.
|
|
|
|
This is true by the symmetry of equality.
|
|
|
|
c. Is $\mathbf{F}$ transitive?
|
|
|
|
The statement claims
|
|
$\forall (x_1, y_1), (x_2, y_2), (x_3, y_3) \in A, [(x_1 = x_2) \wedge (x_2 = x_3)] \to x_1 = x_3$.
|
|
|
|
This is true by the transitivity of equality.
|
|
|
|
29. Let $A = \mathbb{R} \times \mathbb{R}$. A relation $\mathbf{S}$ is defined
|
|
on $A$ as follows: For every $(x_1, y_1)$ and $(x_2, y_2)$ in $A$,
|
|
|
|
$$ (x_1, y_2) \mathbf{S} (x_2, y_2) \Leftrightarrow y_1 = y_2 $$
|
|
|
|
a. Is $\mathbf{S}$ reflexive?
|
|
|
|
Yes, $\mathbf{S}$ is reflexive. The statement claims
|
|
$\forall (x_1, y_1) \in A, y_1 = y_1$. This is trivially true.
|
|
|
|
b. Is $\mathbf{S}$ symmetric?
|
|
|
|
Yes, $\mathbf{S}$ is symmetric. The statement claims
|
|
$\forall (x_1, y_1), (x_2, y_2) \in A, (y_1 = y_2) \to (y_2 = y_1)$.
|
|
|
|
This is true by the symmetry of equality.
|
|
|
|
c. Is $\mathbf{S}$ transitive?
|
|
|
|
Yes, $\mathbf{S}$ is transitive. The statement claims
|
|
$\forall (x_1, y_1), (x_2, y_2), (x_3, y_3) \in A, [(y_1 = y_2) \wedge (y_2 = y_3)] \to y_1 = y_3$.
|
|
|
|
This is true by the transitivity of equality.
|
|
|
|
30. Let $A$ be the "punctured plane"; that is, $A$ is the set of all points in
|
|
the Cartesian plane except the origin $(0, 0)$. A relation $R$ is defined on
|
|
$A$ as follows: For every $p_1$ and $p_2$ in $A$,
|
|
$p_1 R p_2 \Leftrightarrow p_1 \text{ and } p_2 \text{ lie on the same half line emanating from the origin}$.
|
|
|
|
a. Is $$ reflexive?
|
|
|
|
b. Is $$ symmetric?
|
|
|
|
c. Is $$ transitive?
|
|
|
|
31. Let $A$ be the set of people living in the world today. A relation $R$ is
|
|
defined on $A$ as follows: For all people $p$ and $q$ in $A$,
|
|
|
|
$$ p R q \Leftrightarrow p \text{ lives within 100 miles of } q $$
|
|
|
|
a. Is $$ reflexive?
|
|
|
|
Omitted.
|
|
|
|
b. Is $$ symmetric?
|
|
|
|
Omitted.
|
|
|
|
c. Is $$ transitive?
|
|
|
|
Omitted.
|
|
|
|
32. Let $A$ be the set of all lines in the plane. A relation $R$ is defined on
|
|
$A$ as follows: For every $l_1$ and $l_2$ in $A$,
|
|
$l_1 R l_2 \Leftrightarrow l_1 \text{ is parallel to } l_2$. (Assume that a
|
|
line is parallel to itself.)
|
|
|
|
a. Is $$ reflexive?
|
|
|
|
Omitted.
|
|
|
|
b. Is $$ symmetric?
|
|
|
|
Omitted.
|
|
|
|
c. Is $$ transitive?
|
|
|
|
Omitted.
|
|
|
|
33. Let $A$ be the set of all lines in the plane. A relation $R$ is defined on
|
|
$A$ as follows: For every $l_1$ and $l_2$ in $A$,
|
|
|
|
$$ l_1 R l_2 \Leftrightarrow l_1 \text{ is perpendicular to } l_2 $$
|
|
|
|
a. Is $$ reflexive?
|
|
|
|
Omitted.
|
|
|
|
b. Is $$ symmetric?
|
|
|
|
Omitted.
|
|
|
|
c. Is $$ transitive?
|
|
|
|
Omitted.
|
|
|
|
In 34-36, assume that $R$ is a relation on a set $A$. Prove or disprove each
|
|
statement.
|
|
|
|
34. If $R$ is reflexive, then $R^{-1}$ is reflexive.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ is any relation on a set $A$, such that $R$ is reflexive.
|
|
|
|
By the definition of reflexive, this means that $\forall x \in A, (x, x) \in R$,
|
|
or $\forall x \in A, x R x$. Then, by definition of an inverse relation, it
|
|
follows that $(x, x) \in R^{-1}$, or $x R^{-1} x$.
|
|
|
|
Therefore, $R^{-1}$ is reflexive.
|
|
|
|
Q.E.D.
|
|
|
|
35. If $R$ is symmetric, then $R^{-1}$ is symmetric.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ is any relation on a set $A$, such that $R$ is symmetric.
|
|
|
|
By the definition of symmetric, this means that
|
|
$\forall (x, y) \in A, (x, y) \in R \to (y, x) \in R$. Since $(y, x) \in R$, it
|
|
follows, by definition of inverse relation, that $(x, y) \in R^{-1}$.
|
|
Furthermore, since $(x, y) \in R$, it follows that $(y, x) \in R^{-1}$.
|
|
|
|
Therefore $R^{-1}$ is symmetric.
|
|
|
|
Q.E.D.
|
|
|
|
36. If $R$ is transitive, then $R^{-1}$ is transitive.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ is any relation on a set $A$ such that $R$ is transitive.
|
|
|
|
By the definition of transitive, this means that
|
|
$\forall x, y, z \in A, [(x, y) \in R \wedge (y, z) \in R] \to (x, z) \in R$.
|
|
|
|
Since $(x, y), (y, z), (x, z) \in R$, it follows by the definition of inverse
|
|
that $(y, x), (z, y), (z, x) \in R^{-1}$. This means that
|
|
$\forall x, y, z \in A, [(z, y) \in R^{-1} \wedge (y, x) \in R^{-1}] \to (z, x) \in R^{-1}$.
|
|
|
|
Therefore $R^{-1}$ is transitive.
|
|
|
|
Q.E.D.
|
|
|
|
In 37-42, assume that $R$ and $S$ are relations on a set $A$. Prove or disprove
|
|
each statement.
|
|
|
|
37. If $R$ and $S$ are reflexive, is $R \cap S$ reflexive? Why?
|
|
|
|
$R \cap S$ is reflexive.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ and $S$ are any relations on some set $A$ such that $R$ and $S$ are
|
|
reflexive.
|
|
|
|
By the definition of reflexive, this means that $\forall x \in A, (x, x) \in R$,
|
|
and $\forall x \in A, (x, x) \in S$.
|
|
|
|
Since $(x, x) \in R$ and $(x, x) \in S$, it follows (by the definition of
|
|
intersection), that $(x, x) \in R \cap S$.
|
|
|
|
Therefore $R \cap S$ is reflexive.
|
|
|
|
Q.E.D.
|
|
|
|
38. If $R$ and $S$ are symmetric, is $R \cap S$ symmetric? Why?
|
|
|
|
$R \cap S$ is symmetric.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ and $S$ are any relations on a set $A$ such that $R$ and $S$ are
|
|
symmetric.
|
|
|
|
By the definition of symmetric, this means that
|
|
$\forall x, y \in A, (x, y) \in R \to (y, x) \in R$. Similarly,
|
|
$\forall x, y \in A, (x, y) \in S \to (y, x) \in S$.
|
|
|
|
Since $(x, y) \in R$, $(y, x) \in R$, $(x, y) \in S$, $(y, x) \in S$, it follows
|
|
by the definition of intersection that $(x, y) \in R \cap S$ and
|
|
$(y, x) \in R \cap S$.
|
|
|
|
Therefore $R \cap S$ is symmetric.
|
|
|
|
Q.E.D.
|
|
|
|
39. If $R$ and $S$ are transitive, is $R \cap S$ transitive? Why?
|
|
|
|
$R \cap S$ is transitive.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ and $S$ are any relations on a set $A$ such that $R$ and $S$ are
|
|
transitive.
|
|
|
|
By the definition of transitive, this means that
|
|
$\forall x, y, z \in A, [(x, y) \in R \wedge (y, z) \in R] \to (x, z) \in R$.
|
|
Similarly,
|
|
$\forall x, y, z \in A, [(x, y) \in S \wedge (y, z) \in S] \to (x, z) \in S$.
|
|
|
|
Since $(x, y), (y, z), (x, z) \in R$ and $(x, y), (y, z), (x, z) \in S$, it
|
|
follows by the definition of intersection that
|
|
$(x, y), (y, z), (x, z) \in (R \cap S)$. Furthermore, this means that
|
|
$\forall x, y, z \in A, [(x, y) \in (R \cap S) \wedge (y, z) \in (R \cap S)] \to (x, z) \in (R \cap S)$.
|
|
|
|
Therefore, by the definition of transitive, $R \cap S$ is transitive.
|
|
|
|
Q.E.D.
|
|
|
|
40. If $R$ and $S$ are reflexive, is $R \cup S$ reflexive? Why?
|
|
|
|
$R \cup S$ is reflexive.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ and $S$ are any relations on some set $A$ such that $R$ and $S$ are
|
|
reflexive.
|
|
|
|
By the definition of reflexive, this means that $\forall x \in A, (x, x) \in R$,
|
|
and $\forall x \in A, (x, x) \in S$.
|
|
|
|
Since $(x, x) \in R$ and $(x, x) \in S$, it follows (by the definition of
|
|
union), that $(x, x) \in R \cup S$ (since in order to satisfy the definition of
|
|
union, $(x, x) \in R$ _or_ $(x, x) \in S$).
|
|
|
|
Therefore $R \cup S$ is reflexive.
|
|
|
|
Q.E.D.
|
|
|
|
41. If $R$ and $S$ are symmetric, is $R \cup S$ symmetric? Why?
|
|
|
|
$R \cup S$ is symmetric.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ and $S$ are any relations on a set $A$ such that $R$ and $S$ are
|
|
symmetric.
|
|
|
|
By the definition of symmetric, this means that
|
|
$\forall x, y \in A, (x, y) \in R \to (y, x) \in R$. Similarly,
|
|
$\forall x, y \in A, (x, y) \in S \to (y, x) \in S$.
|
|
|
|
Since $(x, y) \in R$, $(y, x) \in R$, $(x, y) \in S$, $(y, x) \in S$, it follows
|
|
by the definition of union that $(x, y) \in R \cup S$ and $(y, x) \in R \cup S$
|
|
(since in order to satisfy the definition of union, $(x, y) \in R$ and
|
|
$(y, x) \in R$ _or_ $(x, y ) \in S$ and $(y, x) \in S$).
|
|
|
|
Therefore $R \cup S$ is symmetric.
|
|
|
|
Q.E.D.
|
|
|
|
42. If $R$ and $S$ are transitive, is $R \cup S$ transitive? Why?
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Let $A = \{a, b, c, d\}$, $R = {(a, b), (b, c), (a, c)}$, and
|
|
$S = \{(b, c), (c, d), (b, d)\}$. Note that $R$ and $S$ are transitive. However,
|
|
when we take the union, $R \cup S$:
|
|
|
|
$$ (R \cup S) = \{(a, b), (b, c), (a, c), (c, d), (b, d)\} $$
|
|
|
|
Note that $(a, b), (b, d) \in (R \cup S)$, but $(a, d) \notin (R \cup S)$. By
|
|
the definition of transitive, it follows that $R \cup S$ is not transitive.
|
|
|
|
Q.E.D.
|
|
|
|
In 43-50, the following definitions are used: A relation on a set $A$ is defined
|
|
to be
|
|
|
|
irreflexive if, and only if, for every $x \in A, x \cancel{R} x$;
|
|
|
|
asymmetric if, and only if, for every $x, y \in A$ if $x R y$ then
|
|
$y \cancel{R} x$;
|
|
|
|
intransitive if, and only if, for every $x, y, z \in A$, if $x R y$ and $y R z$
|
|
then $x \cancel{R} z$.
|
|
|
|
For each of the relations in the referenced exercise, determine whether the
|
|
relation is irreflexive, asymmetric, intransitive, or none of these.
|
|
|
|
43. Exercise 1
|
|
|
|
$R_1 = \{(0, 0), (0, 1), (0, 3), (1, 1), (1, 0), (2, 3), (3, 3)\}$
|
|
|
|
a. Irreflexive?:
|
|
|
|
No, since $0 R_1 0$, $R_1$ is not irreflexive.
|
|
|
|
b. Asymmetric?:
|
|
|
|
No, since $(0, 1) \in R_1$ and $(1, 0) \in R_1$, $R_1$ is not asymmetric.
|
|
|
|
c. Intransitive?:
|
|
|
|
No, since $(0, 1), (1, 0), (0, 0) \in R_1$, $R_1$ is not intransitive.
|
|
|
|
44. Exercise 2
|
|
|
|
$R_2 = \{(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3)\}$
|
|
|
|
a. Irreflexive?:
|
|
|
|
No, since $0 R_2 0$, $R_2$ is not irreflexive.
|
|
|
|
b. Asymmetric?:
|
|
|
|
No, since $(0, 0) \in R_2$ and $(0, 0) \in R_2$, $R_2$ is not asymmetric.
|
|
|
|
c. Intransitive?:
|
|
|
|
No, since $(1, 1), (1, 2), (2, 2) \in R_2$, $R_2$ is not intransitive.
|
|
|
|
45. Exercise 3
|
|
|
|
$R_3 = \{(2, 3), (3, 2)\}$
|
|
|
|
a. Irreflexive?:
|
|
|
|
Yes, $R_3$ is irreflexive.
|
|
|
|
b. Asymmetric?:
|
|
|
|
No, since $(2, 3), (3, 2) \in R_3$, $R_3$ is not asymmetric.
|
|
|
|
c. Intransitive?:
|
|
|
|
Yes, since $(2, 3), (3, 2) \in R_3$, but $(2, 2) \notin R_3$, $R_3$ is
|
|
intransitive.
|
|
|
|
46. Exercise 4
|
|
|
|
$R_4 = \{(1, 2), (2, 1), (1, 3), (3, 1)\}$
|
|
|
|
a. Irreflexive?:
|
|
|
|
Yes, $R_4$ is irreflexive.
|
|
|
|
b. Asymmetric?:
|
|
|
|
No, since $(1, 2), (2, 1) \in R_4$, $R_4$ is not asymmetric.
|
|
|
|
c. Intransitive?:
|
|
|
|
Yes, $R_4$ is intransitive.
|
|
|
|
$$ (1, 2), (2, 1) \in R_4, \text{ but } (1, 1) \notin R_4 $$
|
|
|
|
$$ (2, 1), (1, 3) \in R_4, \text{ but } (2, 3) \notin R_4 $$
|
|
|
|
$$ (1, 3), (3, 1) \in R_4 , \text{ but } (1, 1) \notin R_4 $$
|
|
|
|
etc. (note that a more rigorous proof would check all examples.)
|
|
|
|
47. Exercise 5
|
|
|
|
$R_5 = \{(0, 0), (0, 1), (0, 2), (1, 2)\}$
|
|
|
|
a. Irreflexive?:
|
|
|
|
No, since $(0, 0) \in R_5$
|
|
|
|
b. Asymmetric?:
|
|
|
|
No, since $(0, 0) \in R_5$ and $(0, 0) \in R_5$.
|
|
|
|
c. Intransitive?:
|
|
|
|
No, since $(0, 1), (1, 2), (0, 2) \in R_5$.
|
|
|
|
48. Exercise 6
|
|
|
|
$R_6 = \{(0, 1), (0, 2)\}$
|
|
|
|
a. Irreflexive?:
|
|
|
|
Yes.
|
|
|
|
b. Asymmetric?:
|
|
|
|
Yes.
|
|
|
|
c. Intransitive?:
|
|
|
|
Yes, since there is no $(1, x)$ for some element $x$, nor is there $(2, y)$ for
|
|
some element $y$, the supposition is always false, and is therefore the if/then
|
|
proposition is vacuously true.
|
|
|
|
49. Exercise 7
|
|
|
|
$R_7 = \{(0, 3), (2, 3)\}$
|
|
|
|
a. Irreflexive?:
|
|
|
|
Yes.
|
|
|
|
b. Asymmetric?:
|
|
|
|
Yes.
|
|
|
|
c. Intransitive?:
|
|
|
|
Yes (see Exercise 48 for vacuous truth explanation, which applies here as well.)
|
|
|
|
50. Exercise 8
|
|
|
|
$R_8 = \{(0, 0), (1, 1)\}$
|
|
|
|
a. Irreflexive?:
|
|
|
|
No, since $(0, 0) \in R_8$.
|
|
|
|
b. Asymmetric?:
|
|
|
|
No, since $(0, 0) \in R_8$.
|
|
|
|
c. Intransitive?:
|
|
|
|
No, since $(0, 0), (0, 0), (0, 0) \in R_8$.
|
|
|
|
In 51-53, $R$, $S$, and $T$ are relations defined on $A = \{0, 1, 2, 3\}$.
|
|
|
|
51. Let $R = \{(0, 1), (0, 2), (1, 1), (1, 3), (2, 2), (3, 0)\}$.
|
|
|
|
Find $R^t$, the transitive closure of $R$.
|
|
|
|
First, by definition of the transitive closure, $R \subseteq R^t$, so (building
|
|
$R^t$, _i.e._ not finished):
|
|
|
|
$$ R^t = \{(0, 1), (0, 2) (1, 1), (1, 3), (2, 2), (3, 0)\} $$
|
|
|
|
Since $R^t$ must be transitive, every ordered pair triple must have a transitive
|
|
"third":
|
|
|
|
$$ (0, 1), (1, 1) \to (0, 1) $$
|
|
|
|
$$ (0, 1), (1, 3) \to (0, 3) $$
|
|
|
|
$$ (0, 2), (2, 2) \to (0, 2) $$
|
|
|
|
$$ (1, 1), (1, 3) \to (1, 3) $$
|
|
|
|
$$ (1, 3), (3, 0) \to (1, 0) $$
|
|
|
|
$$ (3, 0), (0, 1) \to (3, 1) $$
|
|
|
|
$$ (3, 0), (0, 2) \to (3, 2) $$
|
|
|
|
Now, add all missing ordered pairs to $R^t$:
|
|
|
|
$$ R^t = \{(0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 3), (2, 2), (3, 0), (3, 1), (3, 2)\} $$
|
|
|
|
Now, check to be sure all ordered triples yields:
|
|
|
|
$$ (3, 1), (1, 3) \to (3, 3) $$
|
|
|
|
$$ \boxed{R^t = \{(0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 3), (2, 2), (3, 0), (3, 1), (3, 2), (3, 3)\}} $$
|
|
|
|
52. Let $S = \{(0, 0), (0, 3), (1, 0), (1, 2), (2, 0), (3, 2)\}$.
|
|
|
|
Find $S^t$, the transitive closure of $S$.
|
|
|
|
$$ S^t = \{(0, 0), (0, 3), (1, 0), (1, 2), (2, 0), (3, 2)\} $$
|
|
|
|
Then:
|
|
|
|
$$ (0, 0), (0, 3) \to (0, 3) $$
|
|
|
|
$$ (0, 3), (3, 2) \to (0, 2) $$
|
|
|
|
$$ (1, 0), (0, 0) \to (1, 0) $$
|
|
|
|
$$ (1, 0), (0, 3) \to (1, 3) $$
|
|
|
|
$$ (1, 2), (2, 0) \to (1, 0) $$
|
|
|
|
$$ (2, 0), (0, 0) \to (2, 0) $$
|
|
|
|
$$ (2, 0), (0, 3) \to (2, 3) $$
|
|
|
|
$$ (3, 2), (2, 0) \to (3, 0) $$
|
|
|
|
New:
|
|
|
|
$$ S^t = \{(0, 0), (0, 2), (0, 3), (1, 0), (1, 2), (1, 3), (2, 0), (2, 3), (3, 0), (3, 2)\} $$
|
|
|
|
Check again:
|
|
|
|
$$ (2, 0), (0, 2) \to (2, 2) $$
|
|
|
|
$$ (3, 0), (0, 3) \to (3, 3) $$
|
|
|
|
Finally:
|
|
|
|
$$ S^t = \{(0, 0), (0, 2), (0, 3), (1, 0), (1, 2), (1, 3), (2, 0), (2, 2), (2, 3), (3, 0), (3, 2), (3, 3)\} $$
|
|
|
|
53. Let $T = \{(0, 2), (1, 0), (2, 3), (3, 1)\}$.
|
|
|
|
Find $T^t$, the transitive closure of $T$.
|
|
|
|
$$ $T^t = \{(0, 2), (1, 0), (2, 3), (3, 1)\}. $$
|
|
|
|
$$ (0, 2), (2, 3) \to (0, 3) $$
|
|
|
|
$$ (1, 0), (0, 2) \to (1, 2) $$
|
|
|
|
$$ (2, 3), (3, 1) \to (2, 1) $$
|
|
|
|
$$ (3, 1), (1, 0) \to (3, 0) $$
|
|
|
|
Now:
|
|
|
|
$$ $T^t = \{(0, 2), (0, 3), (1, 0), (1, 2), (2, 1), (2, 3), (3, 0), (3, 1)\}. $$
|
|
|
|
Furthermore:
|
|
|
|
$$ (0, 2), (2, 1) \to (0, 1) $$
|
|
|
|
$$ (0, 3), (3, 0) \to (0, 0) $$
|
|
|
|
$$ (1, 0), (0, 3) \to (1, 3) $$
|
|
|
|
$$ (1, 2), (2, 1) \to (1, 1) $$
|
|
|
|
$$ (2, 3), (3, 0) \to (2, 0) $$
|
|
|
|
Now:
|
|
|
|
$$ $T^t = \{(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2,
|
|
0), (2, 1), (2, 3), (3, 0), (3, 1)\}. $$
|
|
|
|
And:
|
|
|
|
$$ (2, 1), (1, 2) \to (2, 2) $$
|
|
|
|
$$ (3, 1), (1, 2) \to (3, 2) $$
|
|
|
|
$$ (3, 1), (1, 3) \to (3, 3) $$
|
|
|
|
So:
|
|
|
|
$$ $T^t = \{(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2,
|
|
0), (2, 1), (2, 2), (2, 3), (3, 0), (3, 1), (3, 2), (3, 3)\}. $$
|
|
|
|
54. Write a computer algorithm to test whether a relation $R$ defined on a
|
|
finite set $A$ is reflexive, where
|
|
|
|
$$ A = \{a[1], a[2], \dots, a[n]\} $$
|
|
|
|
Omitted.
|
|
|
|
55. Write a computer algorithm to test whether a relation $R$ defined on a
|
|
finite set $A$ is symmetric, where
|
|
|
|
$$ A = \{a[1], a[2], \dots, a[n]\} $$
|
|
|
|
Omitted.
|
|
|
|
56. Write a computer algorithm to test whether a relation $R$ defined on a
|
|
finite set $A$ is transitive, where
|
|
|
|
$$ A = \{a[1], a[2], \dots, a[n]\} $$
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
Page 543
|
|
|
|
**Exercise Set 8.3**
|
|
|
|
1. Suppose that $S = \{a, b, c, d, e\}$ and $R$ is a relation on $S$ such that
|
|
$a R b$, $b R c$, and $d R e$. List all of the following that must be true if
|
|
$R$ is (a) reflexive (but not symmetric or transitive), (b) symmetric (but
|
|
not reflexive or ransitive), \(c\) transitive (but not reflexive or
|
|
symmetric), and (d) an equivalence relation.
|
|
|
|
$$ c R b \quad c R c \quad a R c \quad b R a $$
|
|
|
|
$$ a R d \quad e R a \quad e R d \quad c R a $$
|
|
|
|
a. reflexive
|
|
|
|
$c R c$
|
|
|
|
b. symmetric
|
|
|
|
$b R a$, $c R b$, $e R d$
|
|
|
|
c. transitive
|
|
|
|
$a R c$
|
|
|
|
d. equivalence relation
|
|
|
|
$c R c$, $b R a$, $c R b$, $e R d$, $a R c$, $c R a$
|
|
|
|
2. Each of the following partitions of $\{0, 1, 2, 3, 4\}$ induces a relation
|
|
$R$ on $\{0, 1, 2, 3, 4\}$. In each case, find the ordered pairs in $R$.
|
|
|
|
a. $\{0, 2\}, \{1\}, \{3, 4\}$
|
|
|
|
$$ R = \{(0, 0), (0, 2), (2, 0), (2, 2), (1, 1), (3, 3), (3, 4), (4, 3) , (4, 4)\} $$
|
|
|
|
b. $\{0\}, \{1, 3, 4\}, \{2\}$
|
|
|
|
$$ R = \{(0, 0), (1, 1), (1, 3), (1, 4), (2, 2), (3, 1), (3, 3), (3, 4), (4, 1), (4, 3), (4, 4)\} $$
|
|
|
|
c. $\{0\}$, $\{1, 2, 3, 4\}$
|
|
|
|
$$ R = \{(0, 0), (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)\} $$
|
|
|
|
In each of 3-6, the relation $R$ is an equivalence relation on $A$. As in
|
|
example 8.3.5, first find the specified equivalence classes. Then state the
|
|
number of distinct equivalence classes for $R$ and list them.
|
|
|
|
3.
|
|
|
|
$$ A = \{0, 1, 2, 3, 4\} $$
|
|
|
|
$$ R = \{(0, 0), (0, 4), (1, 1), (1, 3), (2, 2), (3, 1), (3, 3), (4, 0), (4, 4)\} $$
|
|
|
|
equivalence classes: $[0], [1], [2], [3]$
|
|
|
|
$$ [0] = \{x \in A | x R 0\} = \{0, 4\} $$
|
|
|
|
$$ [1] = \{x \in A | x R 1\} = \{1, 3\} $$
|
|
|
|
$$ [2] = \{x \in A | x R 2\} = \{2\} $$
|
|
|
|
$$ [3] = \{x \in A | x R 3\} = \{1, 3\} $$
|
|
|
|
The distinct number of classes is $3$. List:
|
|
|
|
$$ [0] = \{0, 4\}, [1] = \{1, 3\} = [3], [2] = \{2\} $$
|
|
|
|
4.
|
|
|
|
$$ A = \{a, b, c, d\} $$
|
|
|
|
$$ R = \{(a, a), (b, b), (b, d), (c, c), (d, b), (d, d)\} $$
|
|
|
|
equivalence classes: $[a], [b], [c], [d]$
|
|
|
|
$$ [a] = \{x \in A | x R a\} = \{a\} $$
|
|
|
|
$$ [b] = \{x \in A | x R b\} = \{b, d\} $$
|
|
|
|
$$ [c] = \{x \in A | x R c\} = \{c\} $$
|
|
|
|
$$ [d] = \{x \in A | x R d\} = \{b, d\} $$
|
|
|
|
The number of distinct classes is 3. List:
|
|
|
|
$$ [a] = \{a\}, [b] = \{b, d\} = [d], [c] = \{c\} $$
|
|
|
|
5.
|
|
|
|
$$ A = \{1, 2, 3, 4, \dots, 20\} $$
|
|
|
|
$R$ is defined on $A$ as follows:
|
|
|
|
$$ \text{For all } x, y \in A, x R y \Leftrightarrow 4 | (x - y) $$
|
|
|
|
equivalence classes: $[1], [2], [3], [4], [5]$
|
|
|
|
$$ [1] = \{1, 5, 9, 13, 17\} $$
|
|
|
|
$$ [2] = \{2, 6, 10, 14, 18\} $$
|
|
|
|
$$ [3] = \{3, 7, 11, 15, 19\} $$
|
|
|
|
$$ [4] = \{4, 8, 12, 16, 20\} $$
|
|
|
|
$$ [5] = \{1, 5, 9, 13, 17\} $$
|
|
|
|
There are 4 distinct classes:
|
|
|
|
$$ [1] = \{1, 5, 9, 13, 17\} = [5], [2] = \{2, 6, 10, 14, 18\}, [3] = \{3, 7, 11, 15, 19\}, [4] = \{4, 8, 12, 16, 20\} $$
|
|
|
|
6.
|
|
|
|
$$ A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4, 5\} $$
|
|
|
|
$R$ is defined on $A$ as follows:
|
|
|
|
$$ \text{For all } x, y \in A, x R y \Leftrightarrow 3 | (x - y) $$
|
|
|
|
equivalence classes: $[0], [1], [2], [3]$
|
|
|
|
$$ [0] = \{-3, 0, 3\} $$
|
|
|
|
$$ [1] = \{-2, 1, 4\} $$
|
|
|
|
$$ [2] = \{-4, -1, 2, 5\} $$
|
|
|
|
$$ [3] = \{-3, 0, 3\} $$
|
|
|
|
There are 3 distinct equivalence classes:
|
|
|
|
$$ [0] = \{-3, 0, 3\} = [3], [1] = \{-2, 1, 4\}, [2] = \{-4, -1, 2, 5\} $$
|
|
|
|
In each of 7-14, the relation $R$ is an equivalence relation on the set $A$.
|
|
Find the distinct equivalence classes of $R$.
|
|
|
|
7. $A = \{(1, 3), (2, 4), (-4, -8), (3, 9), (1, 5), (3, 6)\}$. $R$ is defined on
|
|
$A$ as follows: For every $(a, b), (c, d) \in A$,
|
|
|
|
$$ (a, b) R (c, d) \Leftrightarrow ad = bc $$
|
|
|
|
$$ \{(1, 3), (3, 9)\}, \{(2, 4), (-4, -8), (3, 6)\}, \{(1, 5)\} $$
|
|
|
|
8. $X = \{a, b, c\}$ and $A = \mathscr{P}(X)$. $R$ is defined on $A$ as follows:
|
|
For all sets $u$ and $v$ in $\mathscr{P}(X)$,
|
|
|
|
$$ u R v \Leftrightarrow N(u) = N(v) $$
|
|
|
|
(That is, the number of elements in $u$ equals the number of elements in $v$.)
|
|
|
|
$$ \mathscr{P}(X) = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}\} $$
|
|
|
|
$$ \{\emptyset\}, \{\{a\}, \{b\}, \{c\}\}, \{\{a, b\}, \{a, c\}, \{b, c\}\}, \{\{a, b, c\}\} $$
|
|
|
|
9. $X = \{-1, 0, 1\}$ and $A = \mathscr{P}(X)$. $R$ is defined on
|
|
$\mathscr{P}(X)$ as follows: For all sets $s$ and $t$ in $\mathscr{P}(X)$,
|
|
|
|
$$ s R t \Leftrightarrow \text{ the sum of the elements in } s \text{ equals the sum of the elements in } t $$
|
|
|
|
$$ \mathscr{P}(X) = \{\emptyset, \{-1\}, \{0\}, \{1\}, \{-1, 0\}, \{-1, 1\}, \{0, 1\}, \{-1, 0, 1\}\} $$
|
|
|
|
$$ \{\{\emptyset\}, \{-1\}, \{-1, 0\}\}, \{\{0\}, \{-1, 1\}, \{-1, 0, 1\}\}, \{\{1\}, \{0, 1\}\} $$
|
|
|
|
10. $A = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$. $R$ is defined on $A$ as
|
|
follows: For all $m, n \in \mathbb{Z}$,
|
|
|
|
$$ m R n \Leftrightarrow 3 |(m^2 - n^2) $$
|
|
|
|
$$ \{-5, -4, -2, -1, 1, 2, 4, 5\}, \{-3, 0, 3\} $$
|
|
|
|
11. $A = \{-4, -3, -2< -1, 0, 1, 2, 3, 4\}$. $R$ is defined on $A$ as follows:
|
|
For every $(m, n) \in A$,
|
|
|
|
$$ m R n \Leftrightarrow 4 | (m^2 - n^2) $$
|
|
|
|
$$ [0] = \{x \in A | 4 | (x^2 - 0^2)\} = \{x \in A | 4 | x^2\} $$
|
|
|
|
$$ = \{-4, -2, 0, 2, 4\} $$
|
|
|
|
$$ [1] = \{x \in A | 4 | (x^2 - 1^2) = \{x \in A | 4 | (x^2 - 1)\}\} $$
|
|
|
|
$$ = \{-3, -1, 1, 3\} $$
|
|
|
|
12. $A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$. $R$ is defined on $A$ as follows:
|
|
For all $(m, n) \in A$,
|
|
|
|
$$ m R n \Leftrightarrow 5 | (m^2 - n^2) $$
|
|
|
|
$$ [0] = \{x \in A | 5 | (x^2 - 0^2)\} = \{x \in A | 5 | x^2\} $$
|
|
|
|
$$ = \{0\} $$
|
|
|
|
$$ [1] = \{x \in A | 5 | (x^2 - 1^2)\} = \{x \in A | 5 | (x^2 - 1) \} $$
|
|
|
|
$$ = \{-4, -1, 1, 4\} $$
|
|
|
|
$$ [2] = \{x \in A | 5 | (x^2 - 2^2)\} = \{x \in A | 5 | (x^2 - 4)\} $$
|
|
|
|
$$ = \{-3, -2, 2, 3\} $$
|
|
|
|
13. $A$ is the set of all strings of length 4 in _a_'s and _b_'s. $R$ is defined
|
|
on $A$ as follows: For all strings $s$ and $t$ in $A$,
|
|
|
|
$$ s R t \Leftrightarrow s \text{ has the same first two characters as } t $$
|
|
|
|
$$ A = \{aaaa, aaab, aabb, aaba, abbb, abba, abaa, abab, bbbb, bbba, bbaa, bbab, baaa, baab, babb, baba\} $$
|
|
|
|
$$ \{aaaa, aaab, aabb, aaba\}, \{abbb, abba, abaa, abab\}, \{bbbb, bbba, bbaa, bbab\}, \{baaa, baab, babb, baba\} $$
|
|
|
|
14. $A$ is the set of all strings of 0's, 1's, and 2's that have length 4 and
|
|
for which the sum of the characters in the string is less than or equal
|
|
to 2. $R$ is defined on $A$ as follows: For every $s, t \in A$,
|
|
|
|
$$ s R t \Leftrightarrow \text{ the sum of the characters of } s \text{ equals the sum of the characters of } t $$
|
|
|
|
$$ \{0000\}, \{0001, 0010, 0100, 1000\}, \{0011, 0101, 1001, 1010, 1100, 0002, 0020, 0200, 2000\}$$
|
|
|
|
15. Determine which of the following congruence relations are true and which are
|
|
false.
|
|
|
|
$$ m \equiv n (\mod d) \Leftrightarrow d | (m - n) $$
|
|
|
|
a. $17 \equiv 2 (\mod 5)$
|
|
|
|
$$ 5 | 17 - 2 $$
|
|
|
|
$$ 5 | 15 $$
|
|
|
|
Yes, this congruence relation is true, because $15 = 5 \cdot 3$, therefore
|
|
$5 | 15$.
|
|
|
|
b. $4 \equiv -5 (\mod 7)$
|
|
|
|
$$ 7 | 4 - (-5) $$
|
|
|
|
$$ 7 | 9 $$
|
|
|
|
No, this congruence relation is not true, since $7 \cancel{|} 9$.
|
|
|
|
c. $-2 \equiv -8 (\mod 3)$
|
|
|
|
$$ 3 | -2 - (-8) $$
|
|
|
|
$$ 3 | 6 $$
|
|
|
|
Yes, this congruence relation is true, since $6 = 3 \dot 2$, therefore $3 | 6$.
|
|
|
|
d. $-6 \equiv 22 (\mod 2)$
|
|
|
|
$$ 2 | -6 - 22 $$
|
|
|
|
$$ 2 | -28 $$
|
|
|
|
Yes, this congruence relation is true, since $-28 = 2 \cdot -14$, therefore
|
|
$2 | -28$.
|
|
|
|
16.
|
|
|
|
a. Let $R$ be the relation of congruence modulo 3. Which of the following
|
|
equivalence classes are equal?
|
|
|
|
$$ [7], [-4], [-6], [17], [4], [27], [19] $$
|
|
|
|
$$ 7 \mod 3 = 1, -4 \mod 3 = 2, -6 \mod 3 = 0, 17 \mod 3 = 2, 4 \mod 3 = 1, 27 \mod 3 = 0, 19 \mod 3 = 1 $$
|
|
|
|
$$ [7] = [4] = [19], [-4] = [17], [-6] = [27] $$
|
|
|
|
b. Let $R$ be the relation of congruence modulo 7. Which of the following
|
|
equivalence classes are equal?
|
|
|
|
$$ [35], [3], [-7], [12], [0], [-2], [17] $$
|
|
|
|
$$ 35 \mod 7 = 0, 3 \mod 7 = 3, -7 \mod 7 = 0, 12 \mod 7 = 5, 0 \mod 7 = 0, -2 \mod 7 = 5, 17 \mod 7 = 3 $$
|
|
|
|
$$ [35] = [-7] = [0], [12] = [-2], [3] = [17] $$
|
|
|
|
17.
|
|
|
|
a. Prove that for all integers $m$ and $n$, $m \equiv n (\mod 3)$ if, and only
|
|
if, $m \mod 3 = n \mod 3$.
|
|
|
|
**Proof:**
|
|
|
|
To prove that $m \equiv n (\mod 3) \Leftrightarrow m \mod 3 = n \mod 3$, it must
|
|
be shown that $m \equiv n (\mod 3) \to m \mod 3 = n \mod 3$, and it must also be
|
|
shown that $m \mod 3 = n \mod 3 \to m \equiv n (\mod 3)$.
|
|
|
|
_Proof ($m \equiv n (\mod 3)\to m \mod 3 = n \mod 3$):_
|
|
|
|
Suppose $m \in \mathbb{Z}$ and $n \in \mathbb{Z}$, such that
|
|
$m \equiv n (\mod 3)$.
|
|
|
|
It is to be shown that $m \mod 3 = n \mod 3$.
|
|
|
|
Since $m \equiv n (\mod 3)$, by the definition of congruence, this means that:
|
|
|
|
$$ 3 | (m - n) $$
|
|
|
|
By the definition of divisiblity:
|
|
|
|
$$ m - n = 3a $$
|
|
|
|
For some integer $a$.
|
|
|
|
Let $r = m \mod 3$.
|
|
|
|
Then, by the definition of modulo:
|
|
|
|
$$ m = 3b + r $$
|
|
|
|
for some integer $b$.
|
|
|
|
Since $m - n = 3a$, it follows by substitution that:
|
|
|
|
$$ m - n = (3b + r) - n = 3a $$
|
|
|
|
Equivalently (by algebra):
|
|
|
|
$$ (3b + r) - n = 3a $$
|
|
|
|
$$ -n = 3a - (3b + r) $$
|
|
|
|
$$ n = (3b + r) - 3a $$
|
|
|
|
$$ n = 3b + r - 3a $$
|
|
|
|
$$ n = 3b - 3a + r $$
|
|
|
|
$$ n = 3(b - a) + r $$
|
|
|
|
Now, $b - a$ is an integer (by the difference of integers), and $0 \leq r < 3$.
|
|
So, by definition of $\mod$, $n \mod 3 = r$, which equals $m \mod 3$.
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
_Proof ($m \mod 3 = n \mod 3 \to m \equiv n (\mod 3)$):_
|
|
|
|
Suppose $m \in \mathbb{Z}$ and $n \in \mathbb{Z}$ such that
|
|
$m \mod 3 = n \mod 3$.
|
|
|
|
It must be shown that $m \equiv n (\mod 3)$.
|
|
|
|
Let $r = m \mod 3 = n \mod 3$.
|
|
|
|
Then, by definition of $\mod$, $m = 3p + r$ and $n = 3q + r$ for some integers
|
|
$p$ and $q$.
|
|
|
|
By substitution:
|
|
|
|
$$ m - n = (3p + r) - (3q + r) $$
|
|
|
|
$$ = 3p + r - 3q - r $$
|
|
|
|
$$ = 3p - 3q $$
|
|
|
|
$$ = 3(p - q) $$
|
|
|
|
Now, $p - q$ is an integer (by the difference of integers). It follows by the
|
|
definition of divisibility, that $3 | (m - n)$. Therefore, by the definition of
|
|
congruence, $m \equiv n (\mod 3)$.
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $m \equiv n (\mod 3) \to m \mod 3 = n \mod 3$ and
|
|
it has also been shown that $m \mod 3 = n \mod 3 \to m \equiv n (\mod 3)$, it
|
|
can be concluded that $m \equiv n (\mod 3) \Leftrightarrow m \mod 3 = n \mod 3$.
|
|
|
|
b. Prove that for all integers $m$ and $n$ and any positive integer $d$,
|
|
$m \equiv n (\mod d)$ if, and only if, $m \mod d = n \mod d$.
|
|
|
|
**Proof:**
|
|
|
|
To prove that $m \equiv n (\mod d) \Leftrightarrow m \mod d = n \mod d$, it must
|
|
be shown that $m \equiv n (\mod d) \to m \mod d = n \mod d$, and it must also be
|
|
shown that $m \mod d = n \mod d \to m \equiv n (\mod d)$.
|
|
|
|
_Proof ($m \equiv n (\mod d)\to m \mod d = n \mod d$):_
|
|
|
|
Suppose $m \in \mathbb{Z}$, $n \in \mathbb{Z}$, and $d \in \mathbb{Z}^+$ such
|
|
that $m \equiv n (\mod d)$.
|
|
|
|
It is to be shown that $m \mod d = n \mod d$.
|
|
|
|
Since $m \equiv n (\mod d)$, by the definition of congruence, this means that:
|
|
|
|
$$ d | (m - n) $$
|
|
|
|
By the definition of divisiblity:
|
|
|
|
$$ m - n = da $$
|
|
|
|
For some integer $a$.
|
|
|
|
Let $r = m \mod d$.
|
|
|
|
Then, by the definition of modulo:
|
|
|
|
$$ m = db + r $$
|
|
|
|
for some integer $b$.
|
|
|
|
Since $m - n = da$, it follows by substitution that:
|
|
|
|
$$ m - n = (db + r) - n = da $$
|
|
|
|
Equivalently (by algebra):
|
|
|
|
$$ (db + r) - n = da $$
|
|
|
|
$$ -n = da - (db + r) $$
|
|
|
|
$$ n = (db + r) - da $$
|
|
|
|
$$ n = db + r - da $$
|
|
|
|
$$ n = db - da + r $$
|
|
|
|
$$ n = d(b - a) + r $$
|
|
|
|
Now, $b - a$ is an integer (by the difference of integers), and $0 \leq r < d$.
|
|
So, by definition of $\mod$, $n \mod d = r$, which equals $m \mod d$.
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
_Proof ($m \mod d = n \mod d \to m \equiv n (\mod d)$):_
|
|
|
|
Suppose $m \in \mathbb{Z}$, $n \in \mathbb{Z}$, $d \in \mathbb{Z}^+$ such that
|
|
$m \mod d = n \mod d$.
|
|
|
|
It must be shown that $m \equiv n (\mod d)$.
|
|
|
|
Let $r = m \mod d = n \mod d$.
|
|
|
|
Then, by definition of $\mod$, $m = dp + r$ and $n = dq + r$ for some integers
|
|
$p$ and $q$.
|
|
|
|
By substitution:
|
|
|
|
$$ m - n = (dp + r) - (dq + r) $$
|
|
|
|
$$ = dp + r - dq - r $$
|
|
|
|
$$ = dp - dq $$
|
|
|
|
$$ = d(p - q) $$
|
|
|
|
Now, $p - q$ is an integer (by the difference of integers). It follows by the
|
|
definition of divisibility, that $d | (m - n)$. Therefore, by the definition of
|
|
congruence, $m \equiv n (\mod d)$.
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $m \equiv n (\mod d) \to m \mod d = n \mod d$ and
|
|
it has also been shown that $m \mod d = n \mod d \to m \equiv n (\mod d)$, it
|
|
can be concluded that $m \equiv n (\mod d) \Leftrightarrow m \mod d = n \mod d$.
|
|
|
|
18.
|
|
|
|
a. Give an example of two sets that are distinct but not disjoint.
|
|
|
|
Consider $\{1, 2, 3\}$, $\{2\}$, then they are distinct since
|
|
$\{1, 2, 3\} \neq \{2\}$, but they are not disjoint since
|
|
$\{1, 2, 3\} \cap \{2\} = \{2\}$.
|
|
|
|
b. Find sets $A_1$ and $A_2$ and elements $x$, $y$, and $z$ such that $x$ and
|
|
$y$ are in $A_1$ and $y$ and $z$ are in $A_2$ but $x$ and $z$ are not both in
|
|
either of the sets $A_1$ or $A_2$.
|
|
|
|
$$ A_1 = \{x, y\} $$
|
|
|
|
$$ A_2 = \{y, z\} $$
|
|
|
|
In 19-31, (1) prove that the relation is an equivalence relation, and (2)
|
|
describe the distinct equivalence classes of each relation.
|
|
|
|
19. $A$ is the set of all students at your college.
|
|
|
|
a. $R$ is the relation defined on $A$ a follows: For every $x$ and $y$ in $A$,
|
|
|
|
$$ x R y \Leftrightarrow x \text{ has the same major (or double major) as } y $$
|
|
|
|
(Assume "undeclared" is a major.)
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $A$ is the set of all students at my college, and $R$ is a relation
|
|
defined on $A$ defined as follows:
|
|
|
|
$$ \forall x, y \in A, x R y \Leftrightarrow x \text{ has the same major (or
|
|
double major) as } y $$
|
|
|
|
It must be shown that $R$ is an equivalence relation.
|
|
|
|
To prove that $R$ is an equivalence relation, it must be shown that $R$ is
|
|
reflexive, symmetric, and transitive.
|
|
|
|
_Proof (that $R$ is reflexive):_
|
|
|
|
Let $x \in A$.
|
|
|
|
To prove that $R$ is reflexive, it must be shown that $x R x$. It is true that
|
|
$x$ has the same major as $x$. Therefore $R$ is reflexive. This is what was to
|
|
be shown.
|
|
|
|
_Proof (that $R$ is symmetric):_
|
|
|
|
Let $x, y \in A$.
|
|
|
|
To prove that $R$ is symmetric, it must be shown that
|
|
$(x, y) \in R \to (y, x) \in R$.
|
|
|
|
Suppose $(x, y) \in R$. Then, by definition of $R$, this means that $x$ has the
|
|
same major as $y$. By symmetric property of equality, this means that $y$ has
|
|
the same major as $x$. Therefore $(y, x) \in R$.
|
|
|
|
This is what was to be shown.
|
|
|
|
_Proof (that $R$ is transitive):_
|
|
|
|
Let $x, y, z \in A$.
|
|
|
|
To prove that $R$ is transitive, it must be shown that
|
|
$(x, y) \in R \wedge (y, z) \in R \to (x, z) \in R$.
|
|
|
|
Suppose $(x, y) \in R$ and $(y, z) \in R$. Then, by definition of $R$, this
|
|
means that $x$ has the same major as $y$, and $y$ has the same major as $z$. By
|
|
the transitive property of equality, it follows that $x$ has the same major as
|
|
$z$.
|
|
|
|
Therefore $(x, z) \in R$, and it can be concluded that $R$ is transitive.
|
|
|
|
This is what was to be shown.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $R$ is reflexive, symmetric, and transitive, it can
|
|
be concluded that $R$ is an equivalence relation.
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
There is one equivalence class for each major and double major at the college.
|
|
Each class consists of all students with that major (or double major).
|
|
|
|
b. $S$ is the relation defined on $A$ as follows: For every $x, y \in A$,
|
|
|
|
$$ x S y \Leftrightarrow x \text{ is the same age as } y $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $A$ is the set of all students at my college, with $S$ being a relation
|
|
defined on $A$ as follows:
|
|
|
|
$$ \forall x, y \in A, x S y \Leftrightarrow x \text{ is the same age as } y $$
|
|
|
|
To prove that $S$ is an equivalence relation, it must be shown that $S$ is
|
|
reflexive, symmetric, and transitive.
|
|
|
|
_Proof (that $S$ is reflexive):_
|
|
|
|
Let $x \in A$.
|
|
|
|
To prove that $S$ is reflexive, it must be shown that $x S x$. It is true that
|
|
$x$ is the same age as $x$. Thus $x S x$, and therefore $S$ is reflexive.
|
|
|
|
This is what was to be shown.
|
|
|
|
_Proof (that $S$ is symmetric):_
|
|
|
|
Let $x, y \in A$.
|
|
|
|
To prove that $S$ is symmetric, it must be shown that
|
|
$(x, y) \in S \to (y, x) \in S$.
|
|
|
|
Suppose $x S y$. By the definition of $S$, this means that $x$ is the same age
|
|
as $y$. By the symmetric property of equality, this means that $y$ is the same
|
|
age as $x$. It follows that $y R x$, and therefore $S$ is symmetric.
|
|
|
|
This is what was to be shown.
|
|
|
|
_Proof (that $S$ is transitive):_
|
|
|
|
Let $x, y, z \in A$.
|
|
|
|
To prove that $S$ is transitive, it must be shown that
|
|
$(x, y) \in S \wedge (y, z) \in S \to (x, z) \in S$.
|
|
|
|
Suppose $x S y$ and $y S z$. Then, by the definition of $S$, this means that $x$
|
|
is the same age as $y$ and $y$ is the same age as $z$. By the transitive
|
|
property of equality, this means that $x$ is the same age as $z$.
|
|
|
|
It follows that $(x, z) \in S$, and therefore $S$ is transitive.
|
|
|
|
This is what was to be shown.
|
|
|
|
_Conclusion:_
|
|
|
|
Since $S$ has been shown to be reflexive, symmetric, and transitive, it can be
|
|
concluded that $S$ is an equivalence relation.
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
There is one equivalence class for each student age (by year) at the college.
|
|
Each class consists of all students with that age.
|
|
|
|
20. $E$ is the relation defined on $\mathbb{Z}$ as follows:
|
|
|
|
$$ \text{For every } m, n \in \mathbb{Z}, m E n \Leftrightarrow 4 | (m - n) $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $m \in \mathbb{Z}$ and $n \in \mathbb{Z}$. Let $E$ be a relation defined
|
|
on $\mathbb{Z}$ as follows:
|
|
|
|
$$ \forall m, n \in \mathbb{Z}, m E n \Leftrightarrow 4 | (m - n) $$
|
|
|
|
To prove that $E$ is an equivalence relation, it must be shown that $E$ is
|
|
reflexive, symmetric, and transitive.
|
|
|
|
_Proof ($E$ is reflexive):_
|
|
|
|
Let $m \in \mathbb{Z}$.
|
|
|
|
To prove that $E$ is reflexive, it must be shown that $(m, m) \in E$.
|
|
|
|
By the definition for $E$, this means that:
|
|
|
|
$$ 4 | (m - m) $$
|
|
|
|
Since $m - m = 0$, this means that:
|
|
|
|
$$ 4 | 0 $$
|
|
|
|
This is true, since $0 = 4 \cdot 0$. It follows that $(m, m) \in E$, and
|
|
therefore $E$ is reflexive.
|
|
|
|
_Proof ($E$ is symmetric):_
|
|
|
|
Let $m \in \mathbb{Z}$ and $n \in \mathbb{Z}$.
|
|
|
|
To prove that $E$ is symmetric, it must be shown that
|
|
$(m, n) \in E \to (n, m) \in E$.
|
|
|
|
Since $(m, n) \in E$, this means that:
|
|
|
|
$$ 4 | (m - n) $$
|
|
|
|
By the definition of divisibility, this means that:
|
|
|
|
$$ m - n = 4k $$
|
|
|
|
for some integer $k$.
|
|
|
|
Now, consider:
|
|
|
|
$$ -1(m - n) = -1(4k) $$
|
|
|
|
$$ n - m = 4(-k) $$
|
|
|
|
Now, $-k$ is an integer (by the product of integers). It follows (by the
|
|
definition of divisibility), that:
|
|
|
|
$$ 4 | (n - m) $$
|
|
|
|
This means that $(n, m) \in E$, and therefore $E$ is symmetric.
|
|
|
|
_Proof ($E$ is transitive):_
|
|
|
|
Let $m \in \mathbb{Z}$, $n \in \mathbb{Z}$, and $p \in \mathbb{Z}$.
|
|
|
|
To prove that $E$ is transitive, it must be shown that
|
|
$(m, n) \in E \wedge (n, p) \in E \to (m, p) \in E$.
|
|
|
|
Suppose $(m, n) \in E$ and $(n, p) \in E$. By definition of $E$, this means
|
|
that:
|
|
|
|
$$ 4 | (m - n) $$
|
|
|
|
and
|
|
|
|
$$ 4 | (n - p) $$
|
|
|
|
By the definition of divisibility, this means that:
|
|
|
|
$$ m - n = 4k $$
|
|
|
|
$$ n - p = 4q $$
|
|
|
|
for some integers $k$ and $q$.
|
|
|
|
Subtracting the two yields:
|
|
|
|
$$ (m - n) - (n - p) = m - p $$
|
|
|
|
And then by substitution this is:
|
|
|
|
$$ m - p = 4k - 4q $$
|
|
|
|
By algebra:
|
|
|
|
$$ = 4(k - q) $$
|
|
|
|
Now, $k - q$ is an integer (by the difference of integers). It follows that
|
|
$4 | (m - p)$, and thus $(m, p) \in E$. Therefore, it can be concluded that $E$
|
|
is transitive.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $E$ is reflexive, symmetric, and transitive, it can
|
|
be concluded that $E$ is an equivalence relation. This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
Observe that for any $a \in \mathbb{Z}$, the equivalence class of $a$, ($[a]$),
|
|
is:
|
|
|
|
$$ [a] = \{x \in \mathbb{Z} | x E a\} = \{x \in \mathbb{Z} | 4 | x - a\} $$
|
|
|
|
By definition of divisiblity:
|
|
|
|
$$ = \{x \in \mathbb{Z} | x - a = 4k \text{ for some integer } k\} $$
|
|
|
|
By algebra:
|
|
|
|
$$ = \{x \in \mathbb{Z} | x = 4k + a \} $$
|
|
|
|
So, our equivalence classes are defined as follows:
|
|
|
|
$$ \{x \in \mathbb{Z} | x = 4k \}, \{x \in \mathbb{Z} | x = 4k + 1 \}, \{x \in \mathbb{Z} | x = 4k + 2 \}, \{x \in \mathbb{Z} | x = 4k + 3 \} $$
|
|
|
|
21. $R$ is the relation defined on $\mathbb{Z}$ as follows:
|
|
|
|
$$ \text{For every } m, n \in \mathbb{Z}, m R n \Leftrightarrow 7m - 5n \text{ is even} $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $m \in \mathbb{Z}$ and $n \in \mathbb{Z}$, such that $R$ is a relation
|
|
on $\mathbb{Z}$ defined as follows:
|
|
|
|
$$ \forall m, n \in \mathbb{Z}, m R n \Leftrightarrow 7m - 5n \text{ is even} $$
|
|
|
|
It must be shown that $R$ is an equivalence relation.
|
|
|
|
To show that $R$ is an equivalence relation, it must be shown that $R$ is
|
|
reflexive, symmetric, and transitive.
|
|
|
|
_Proof ($R$ is reflexive):_
|
|
|
|
Let $m \in \mathbb{Z}$.
|
|
|
|
To prove that $R$ is reflexive, it must be shown that $(m, m) \in R$. By the
|
|
definition of $R$, it then must be shown that:
|
|
|
|
$$ 7m - 5m \text{ is even} $$
|
|
|
|
Consider that:
|
|
|
|
$$ 7m - 5m = 2m $$
|
|
|
|
Since $m$ is an integer (by the supposition), it follows that $7m - 5m$ is even
|
|
(by the definition of even, since $7m - 5m = 2m$).
|
|
|
|
It follows that $(m, m) \in R$, and therefore $R$ is reflexive.
|
|
|
|
_Proof ($R$ is symmetric):_
|
|
|
|
Let $m, n \in \mathbb{Z}$.
|
|
|
|
To prove that $R$ is symmetric, it must be shown that
|
|
$(m, n) \in R \to (n, m) \in R$.
|
|
|
|
Suppose $(m, n) \in R$. By definition of $R$, this means that:
|
|
|
|
$$ 7m - 5n \text{ is even} $$
|
|
|
|
By definition of even, this means that:
|
|
|
|
$$ 7m - 5n = 2k $$
|
|
|
|
for some integer $k$.
|
|
|
|
Then, consider:
|
|
|
|
$$ 7n - 5m = (12 - 5)n - (12 - 7)m $$
|
|
|
|
$$ = 12n - 5n - 12m + 7m $$
|
|
|
|
$$ = 12n - 12m + (7m - 5n) $$
|
|
|
|
$$ = 12n - 12m + 2k $$
|
|
|
|
$$ = 2(6n - 6m + k) $$
|
|
|
|
Now, $6n - 6m + k$ is an integer (by the product, sum, and difference of
|
|
integers). It follows that $7n - 5m$ is even (by the definition of even).
|
|
Therefore $(n, m) \in R$, and therefore $R$ is symmetric.
|
|
|
|
_Proof ($R$ is transitive):_
|
|
|
|
Let $m, n, p \in \mathbb{Z}$.
|
|
|
|
To prove that $R$ is transitive, it must be shown that
|
|
$(m, n) \in R \wedge (n, p) \in R \to (m, p) \in R$.
|
|
|
|
Suppose $(m, n) \in R$ and $(n, p) \in R$. By definition of $R$, this means
|
|
that:
|
|
|
|
$$ 7m - 5n \text{ is even} $$
|
|
|
|
and
|
|
|
|
$$ 7n - 5p \text{ is even} $$
|
|
|
|
By the definition of even, this means that:
|
|
|
|
$$ 7m - 5n = 2r $$
|
|
|
|
and
|
|
|
|
$$ 7n - 5p = 2s $$
|
|
|
|
for some integers $r$ and $s$.
|
|
|
|
It must be shown that $7m - 5p \text{ is even}$. Consider:
|
|
|
|
$$ 7m - 5p = (7m - 5n + 5n) + (7n - 7n - 5p) $$
|
|
|
|
$$ = ((7m - 5n) + 5n) + (7n - (7n - 5p)) $$
|
|
|
|
$$ = (2r + 5n) + (7n - 2s) $$
|
|
|
|
$$ = 2r + 5n + 7n - 2s $$
|
|
|
|
$$ = 2r + 12n - 2s $$
|
|
|
|
$$ = 2(r + 6n - s) $$
|
|
|
|
Now, $r + 6n - s$ is an integer (by the product, sum, and difference of
|
|
integers). By the definition of even, this means that $7m - 5p$ is even. It
|
|
follows that $(m, p) \in R$, and therefore $R$ is transitive.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $R$ is reflexive, symmetric, and transitive, it can
|
|
be concluded that $R$ is an equivalence relation.
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
$$ \forall m, n \in \mathbb{Z}, m R n \Leftrightarrow 7m - 5n \text{ is even} $$
|
|
|
|
Consider $a \in \mathbb{Z}, then, by the definition of $r$, this means that:
|
|
|
|
$$ \{x \in \mathbb{Z} | x R a \} $$
|
|
|
|
By the definition of $R$:
|
|
|
|
$$ \{x \in \mathbb{Z} | 7x - 5a \text{ is even} \} $$
|
|
|
|
Since $7x - 5a$ is even, this means that both $7x$ and $5a$ are even, or both
|
|
$7x$ and $5a$ are odd. Since $7$ and $5$ are both odd (and odd times odd is odd,
|
|
and odd times even is even), this means that $7x$ and $5a$ have the same parity.
|
|
|
|
Thus there are two equivalency cases, one the set of all even integers, and the
|
|
other the set of all odd integers.
|
|
|
|
22. Let $A$ be the set of all statement forms in three variables $p$, $q$, and
|
|
$r$. $\mathbf{R}$ is the relation defined on $A$ as follows: For all $P$ and
|
|
$Q$ in $A$,
|
|
|
|
$$ P \mathbf{R} Q \Leftrightarrow P \text{ and } Q \text{ have the same truth table} $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $A$ is the set of all statement forms in three variables $p$, $q$, and
|
|
$r$. Let $\mathbf{R}$ be a relation on the set $A$ defined as follows:
|
|
|
|
$$ P \mathbf{R} Q \Leftrightarrow P \text{ and } Q \text{ have the same truth table} $$
|
|
|
|
To prove that $\mathbf{R}$ is an equivalence relation, it must be shown that
|
|
$\mathbf{R}$ is reflexive, symmetric, and transitive.
|
|
|
|
_Proof ($\mathbf{R}$ is reflexive):_
|
|
|
|
Let $P \in A$.
|
|
|
|
To prove that $\mathbf{R}$ is reflexive, it must be shown that
|
|
$(P, P) \in \mathbf{R}$. By the definition of $\mathbf{R}$, this means that $P$
|
|
and $P$ have the same truth table.
|
|
|
|
It is true that $P$ has the same truth table as itself.
|
|
|
|
It follows that $(P, P) \in \mathbf{R}$, and therefore $\mathbf{R}$ is
|
|
reflexive.
|
|
|
|
_Proof ($\mathbf{R}$ is symmetric):_
|
|
|
|
Let $P, Q \in A$.
|
|
|
|
To prove that $\mathbf{R}$ is symmetric, it must be shown that
|
|
$(P, Q) \in \mathbf{R} \to (Q, P) \in \mathbf{R}$.
|
|
|
|
Suppose $(P, Q) \in \mathbf{R}$, by the definition for $\mathbf{R}$, this means
|
|
that $P$ and $Q$ have the same truth tables.
|
|
|
|
It follows by the symmetric property of equality that $Q$ and $P$ have the same
|
|
truth tables.
|
|
|
|
Thus $(Q, P) \in \mathbf{R}$, and therefore $\mathbf{R}$ is symmetric.
|
|
|
|
_Proof ($\mathbf{R}$ is transitive):_
|
|
|
|
Let $P, Q, S \in A$.
|
|
|
|
To prove that $\mathbf{R}$ is transitive, it must be shown that
|
|
$(P, Q) \in \mathbf{R} \wedge (Q, S) \in \mathbf{R} \to (P, S) \in \mathbf{R}$.
|
|
|
|
Suppose $(P, Q) \in \mathbf{R}$ and $(Q, S) \in \mathbf{R}$. By the definition
|
|
of $\mathbf{R}$, this means that $P$ and $Q$ have the same truth tables, and
|
|
that $Q$ and $S$ have the same truth tables.
|
|
|
|
It follows, by the transitive property of equality, that $P$ and $S$ have the
|
|
same truth tables.
|
|
|
|
Thus $(P, S) \in \mathbf{R}$, and therefore $\mathbf{R}$ is transitive.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $\mathbf{R}$ is reflexive, symmetric, and
|
|
transitive, it can be concluded that $\mathbf{R}$ is an equivalence relation.
|
|
|
|
Q.E.D.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
There is an equivalence class corresponding to every possible truth table in 3
|
|
variables, $p, q, r$. There are 8 lines in every truth table, and each line has
|
|
2 options (true or false), so there are $2^8$ equivalence classes.
|
|
|
|
23. Let $P$ be a set of parts shipped to a company from various suppliers. $S$
|
|
is the relation defined on $P$ as follows: For every $x, y \in P$,
|
|
|
|
$$ x S y \Leftrightarrow x \text{ has the same part number and is shipped from the same supplier as } y $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $P$ is the set of all parts shipped to a company from various suppliers.
|
|
Let $S$ be a relation defined on $P$ as follows:
|
|
|
|
$$ \forall x, y \in P, x S y \Leftrightarrow x \text{ has the same part number and is shipped from the same supplier as } y $$
|
|
|
|
To prove that $S$ is an equivalence relation, it must be shown that $S$ is
|
|
reflexive, symmetric, and transitive.
|
|
|
|
_Proof ($S$ is reflexive):_
|
|
|
|
Let $x \in P$.
|
|
|
|
To prove that $S$ is reflexive, it must be shown that $(x, x) \in S$. By the
|
|
definition for $S$, this means it must be shown that $x$ has the same part
|
|
number and is shipped from the same supplier as $x$.
|
|
|
|
It is true that $x$ has the same part number as $x$ and that $x$ is shipped from
|
|
the same supplier as $x$.
|
|
|
|
Thus $(x, x) \in S$, and therefore $S$ is reflexive.
|
|
|
|
_Proof ($S$ is symmetric):_
|
|
|
|
Let $x, y \in P$.
|
|
|
|
To prove that $S$ is symmetric, it must be shown that
|
|
$(x, y) \in S \to (y, x) \in S$.
|
|
|
|
Suppose $(x, y) \in S$. By the definition for $S$, this means that $x$ has the
|
|
same part number as $y$ and $x$ is shipped from the same supplier as $y$.
|
|
|
|
It follows by the symmetry of equality that $y$ has the same part number as $x$
|
|
and $y$ is shipped from the same supplier as $x$.
|
|
|
|
Thus $(y, x) \in S$, and therefore $S$ is symmetric.
|
|
|
|
_Proof ($S$ is transitive):_
|
|
|
|
Let $x, y, z \in P$.
|
|
|
|
To prove that $S$ is transitive, it must be shown that
|
|
$(x, y) \in S \wedge (y, z) \in S \to (x, z) \in S$.
|
|
|
|
Suppose $(x, y) \in S$ and $(y, z) \in S$. By the definition for $S$, this means
|
|
that:
|
|
|
|
$x$ has the same part number and is shipped from the same supplier as $y$.
|
|
|
|
and that:
|
|
|
|
$y$ has the same part number and is shipped from the same supplier as $z$.
|
|
|
|
By the definition of the transitivity of equality, this means that $x$ has the
|
|
same part number and is shipped from the same supplier as $z$.
|
|
|
|
Thus $(x, z) \in S$, and therefore $S$ is transitive.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $S$ is reflexive, symmetric, and transitive, it can
|
|
be concluded that $S$ is an equivalence relation. This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
The number of distinct equivalence classes is grouped based off of parts that
|
|
all have the same part number and are shipped from the same supplier (_i.e._ the
|
|
equivalence classes are sets of all parts with the same part number and
|
|
supplier.)
|
|
|
|
24. Let $A$ be the set of identifiers in a computer program. It is common for
|
|
identifiers to be used for only a short part of the execution time of a
|
|
program and not to be used again to execute other parts of the program. In
|
|
such cases, arranging for identifiers to share memory locations makes
|
|
efficient use of a computer's memory capacity. Define a relation $R$ on $A$
|
|
as follows: For all identifiers $x$ and $y$,
|
|
|
|
$$ x R y \Leftrightarrow \text{ the values of } x \text{ and } y \text{ are stored in the same memory location during execution of the program} $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $A$ is the set of identifiers in a computer program. Let $R$ be a
|
|
relation on the set $A$ such that it is defined as follows:
|
|
|
|
$$ \forall x, y \in A, x R y \Leftrightarrow \text{ the values of } x \text{ and } y \text{ are stored in the same memory location during execution of the program} $$
|
|
|
|
To prove that $R$ is an equivalence relation, it must be shown that $R$ is
|
|
reflexive, symmetric, and transitive.
|
|
|
|
_Proof ($R$ is reflexive):_
|
|
|
|
Let $x \in A$.
|
|
|
|
To prove that $R$ is reflexive, it must be shown that $(x, x) \in R$.
|
|
|
|
By definition of $R$, this means that it must be shown that the values of $x$
|
|
and $x$ are stored in the same memory location during execution of the program.
|
|
|
|
It is true that $x$ and $x$ are stored in the same memory location during
|
|
execution of the program (since $x$ is the same identifier as $x$.)
|
|
|
|
Thus $(x, x) \in R$ and therefore $R$ is reflexive.
|
|
|
|
_Proof ($R$ is symmetric):_
|
|
|
|
Let $x, y \in A$.
|
|
|
|
To prove that $R$ is symmetric, it must be shown that
|
|
$(x, y) \in R \to (y, x) \in R$.
|
|
|
|
Suppose $(x, y) \in R$. Then, by definition of $R$, this means that the values
|
|
of $x$ and $y$ are stored in the same memory location during the execution of
|
|
the program.
|
|
|
|
By the symmetric property of equality, this means that the values of $y$ and $x$
|
|
are stored in the same memory location during the execution of the program.
|
|
|
|
Thus, $(y, x) \in R$, and therefore $R$ is symmetric.
|
|
|
|
_Proof ($R$ is transitive):_
|
|
|
|
Let $x, y, z \in A$.
|
|
|
|
To prove that $R$ is transitive, it must be shown that
|
|
$(x, y) \in R \wedge (y, z) \in R \to (x, z) \in R$.
|
|
|
|
Suppose $(x, y) \in R$ and $(y, z) \in R$. By the definition for $R$, this means
|
|
that:
|
|
|
|
The values of $x$ and $y$ are stored in the same memory location during
|
|
execution of the program.
|
|
|
|
and that:
|
|
|
|
The values of $y$ and $z$ are stored in the same memory location during
|
|
execution of the program.
|
|
|
|
By the transitive property of equality, this means that the values of $x$ and
|
|
$z$ are stored in the same memory location during execution of the program.
|
|
|
|
Thus $(x, z) \in R$, and therefore $R$ is transitive.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $R$ is reflexive, symmetric, and transitive, it can
|
|
be concluded that $R$ is an equivalence relation.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
The number of equivalence classes is based off the number of identifiers in a
|
|
computer program that are stored in the same memory location during execution of
|
|
the program.
|
|
|
|
25. $A$ is the "absolute value" relation defined on $\mathbb{R}$ as follows:
|
|
|
|
$$ \text{For every } x, y \in \mathbb{R}, x A y \Leftrightarrow |x| = |y| $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $A$ is the "absolute value" relation on $\mathbb{R}$, defined as
|
|
follows:
|
|
|
|
$$ \forall x, y \in \mathbb{R}, x A y \Leftrightarrow |x| = |y| $$
|
|
|
|
To prove that $A$ is an equivalence relation, it must be shown that $A$ is
|
|
reflexive, symmetric, and transitive.
|
|
|
|
_Proof ($A$ is reflexive):_
|
|
|
|
Let $x \in \mathbb{R}$.
|
|
|
|
To prove that $A$ is reflexive, it must be shown that $(x, x) \in A$.
|
|
|
|
By definition for $A$, this means that it must be proved that:
|
|
|
|
$$ |x| = |x| $$
|
|
|
|
It is trivially true that $|x| = |x|$.
|
|
|
|
Thus $(x, x) \in A$, and therefore $A$ is reflexive.
|
|
|
|
_Proof ($A$ is symmetric):_
|
|
|
|
Let $x, y \in \mathbb{R}$.
|
|
|
|
To prove that $A$ is symmetric, it must be shown that
|
|
$(x, y) \in A \to (y, x) \in A$.
|
|
|
|
Suppose $(x, y) \in A$. By the definition for $A$, this means that:
|
|
|
|
$$ |x| = |y| $$
|
|
|
|
By the symmetric property of equality, it follows that:
|
|
|
|
$$ |y| = |x| $$
|
|
|
|
Thus $(y, x) \in A$, and therefore $A$ is symmetric.
|
|
|
|
_Proof ($A$ is transitive):_
|
|
|
|
Let $x, y, z \in \mathbb{R}$.
|
|
|
|
To prove that $A$ is transitive, it must be shown that
|
|
$(x, y) \in A \wedge (y, z) \in A \to (x, z) \in A$.
|
|
|
|
Suppose $(x, y) \in A$ and $(y, z) \in A$. By the definition for $A$, this means
|
|
that:
|
|
|
|
$$ |x| = |y| $$
|
|
|
|
and that:
|
|
|
|
$$ |y| = |z| $$
|
|
|
|
It follows, by the transitive property of equality that $|x| = |z|$.
|
|
|
|
Thus $(x, z) \in A$, and therefore $A$ is transitive.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $A$ is reflexive, symmetric, and transitive, it can
|
|
be concluded that $A$ is an equivalence relation. This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
Let $a \in \mathbf{R}$, then by the definition of absolute value:
|
|
|
|
$$ |-a| = |a| $$
|
|
|
|
with the exception of $0$, since $0 \in \mathbb{R}$, but there is no $-0$.
|
|
|
|
Thus the equivalence classes are all sets of all real numbers and their
|
|
corresponding negative counterpart, and also the set $\{0\}$.
|
|
|
|
26. $D$ is the relation defined on $\mathbb{Z}$ as follows: For every
|
|
$m, n \in \mathbb{Z}$,
|
|
|
|
$$ m D n \Leftrightarrow 3 | (m^2 - n^2) $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $D$ is a relation on $\mathbb{Z}$ defined as follows:
|
|
|
|
$$ \forall m, n \in \mathbb{Z}, m D n \Leftrightarrow 3 | (m^2 - n^2) $$
|
|
|
|
To prove that $D$ is an equivalence relation, it must be shown that $D$ is
|
|
reflexive, symmetric, and transitive.
|
|
|
|
_Proof ($D$ is reflexive):_
|
|
|
|
Let $x \in \mathbb{Z}$.
|
|
|
|
To prove that $D$ is reflexive, it must be shown that $(x, x) \in D$. By the
|
|
definition for $D$, this means it must be shown that:
|
|
|
|
$$ 3 | (x^2 - x^2) $$
|
|
|
|
Now, $x^2 - x^2 = 0$, and it is true that $3 | 0$, since $0 = 3 \cdot 0$. Thus
|
|
$(x, x) \in D$, and it can be concluded that $D$ is reflexive.
|
|
|
|
_Proof ($D$ is symmetric):_
|
|
|
|
Let $x, y \in \mathbb{Z}$.
|
|
|
|
To prove that $D$ is symmetric, it must be shown that
|
|
$(x, y) \in D \to (y, x) \in D$. By definition of $D$, this means it must be
|
|
shown that:
|
|
|
|
$$ [3 | (x^2 - y^2)] \to [3 | (y^2 - x^2)] $$
|
|
|
|
Suppose $3 | (x^2 - y^2)$. By the definition of divisibility, this means that:
|
|
|
|
$$ x^2 - y^2 = 3k $$
|
|
|
|
for some integer $k$.
|
|
|
|
Now, consider that:
|
|
|
|
$$ y^2 - x^2 = -1(x^2 - y^2) $$
|
|
|
|
Then, by substitution:
|
|
|
|
$$ = -1(3k) $$
|
|
|
|
$$ = 3(-k) $$
|
|
|
|
Now, $-k$ is an integer (by the product of integers), thus $3 | (y^2 - x^2)$,
|
|
and hence $(y, x) \in D$, and therefore $D$ is symmetric.
|
|
|
|
_Proof ($D$ is transitive):_
|
|
|
|
Let $x, y, z \in \mathbb{Z}$.
|
|
|
|
To prove that $D$ is transitive, it must be shown that
|
|
$[(x, y) \in D \wedge (y, z) \in D] \to [(x, z) \in D]$.
|
|
|
|
Suppose $(x, y) \in D$ and $(y, z) \in D$. Then, by the definition for $D$, this
|
|
means:
|
|
|
|
$$ 3 | (x^2 - y^2) $$
|
|
|
|
and also:
|
|
|
|
$$ 3 | (y^2 - z^2) $$
|
|
|
|
(It must be shown that $3 | (x^2 - z^2)$.)
|
|
|
|
By the definition of divisibility, this means that:
|
|
|
|
$$ x^2 - y^2 = 3k $$
|
|
|
|
and also that:
|
|
|
|
$$ y^2 - z^2 = 3p $$
|
|
|
|
for some integers $k$ and $p$.
|
|
|
|
Now, if one adds $x^2 - y^2$ and $y^2 - z^2$, this yields:
|
|
|
|
$$ x^2 - y^2 + y^2 - z^2 = x^2 - z^2 $$
|
|
|
|
Then, by substitution:
|
|
|
|
$$ x^2 - z^2 = (3k) + (3p) $$
|
|
|
|
$$ = 3(k + p) $$
|
|
|
|
Now, $k + p$ is an integer (by the sum of integers). Thus $3 | (x^2 - z^2)$ (by
|
|
the definition of divisibility). It follows that $(x, z) \in D$, and therefore
|
|
$D$ is transitive.
|
|
|
|
_Conclusion:_
|
|
|
|
Since $D$ has been shown to be reflexive, symmetric, and transitive, it follows
|
|
that $D$ is an equivalence relation. This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
There are two distinct equivalence classes:
|
|
|
|
$$ [0] = \{\dots, -6, -3, 0, 3, 6, \dots\}, [1] = \{\dots, -5, -4, -2, -1, 1, 2, 4, 5, \dots\} $$
|
|
|
|
27. $R$ is the relation defined on $\mathbb{Z}$ as follows: For every
|
|
$(m, n) \in \mathbb{Z}$,
|
|
|
|
$$ m R n \Leftrightarrow 4 | (m^2 - n^2) $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ is a relation defined on $\mathbb{Z}$ as follows:
|
|
|
|
$$ \forall m, n \in \mathbb{Z}, m R n \Leftrightarrow 4 | (m^2 - n^2) $$
|
|
|
|
To prove that $R$ is an equivalence relation, it must be shown that $R$ is
|
|
reflexive, symmetric, and transitive.
|
|
|
|
_Proof ($R$ is reflexive):_
|
|
|
|
Let $x \in \mathbb{Z}$.
|
|
|
|
To prove that $R$ is reflexive, it must be shown that $(x, x) \in R$. By the
|
|
definition for $R$, this means it must be shown that:
|
|
|
|
$$ 4 | (x^2 - x^2) $$
|
|
|
|
Since $x^2 - x^2 = 0$, this means it must be shown that $4 | 0$. Now, $4 | 0$
|
|
because $0 = 4 \cdot 0$. Therefore $(x, x) \in R$, and it can be concluded that
|
|
$R$ is reflexive.
|
|
|
|
_Proof ($R$ is symmetric):_
|
|
|
|
Let $x, y \in \mathbb{Z}$.
|
|
|
|
To prove that $R$ is symmetric, it must be shown that
|
|
$(x, y) \in R \to (y, x) \in R$.
|
|
|
|
Suppose $(x, y) \in R$, then, by definition for $R$, this means:
|
|
|
|
$$ 4 | (x^2 - y^2) $$
|
|
|
|
By the definition of divisibility, this means that:
|
|
|
|
$$ x^2 - y^2 = 4k $$
|
|
|
|
for some integer $k$.
|
|
|
|
Now, consider that:
|
|
|
|
$$ y^2 - x^2 = -1(x^2 - y^2) $$
|
|
|
|
Then, by substitution:
|
|
|
|
$$ y^2 - x^2 = -1(4k) $$
|
|
|
|
$$ = 4(-k) $$
|
|
|
|
Now, $-k$ is an integer (by the product of integers). Hence $4 | (y^2 - x^2)$,
|
|
and it follows that $(y, x) \in R$, and therefore $R$ is symmetric.
|
|
|
|
_Proof ($R$ is transitive):_
|
|
|
|
Let $x, y, z \in \mathbb{Z}$.
|
|
|
|
To prove that $R$ is transitive, it must be shown that
|
|
$[(x, y) \in R \wedge (y, z) \in R] \to (x, z) \in R$.
|
|
|
|
Suppose $(x, y) \in R$ and $(y, z) \in R$. By the definition for $R$, this means
|
|
that:
|
|
|
|
$$ 4 | (x^2 - y^2) $$
|
|
|
|
and also that:
|
|
|
|
$$ 4 | (y^2 - z^2) $$
|
|
|
|
Now, by the definition for divisibility, this means that:
|
|
|
|
$$ x^2 - y^2 = 4k $$
|
|
|
|
and also that:
|
|
|
|
$$ y^2 - z^2 = 4p $$
|
|
|
|
for some integers $k$ and $p$.
|
|
|
|
Now, consider that:
|
|
|
|
$$ x^2 - z^2 = x^2 - y^2 + y^2 - z^2 $$
|
|
|
|
Then, by substitution:
|
|
|
|
$$ x^2 - z^2 = 4k + 4p $$
|
|
|
|
$$ x^2 - z^2 = 4(k + p) $$
|
|
|
|
Now, $k + p$ is an integer (by the sum of integers), and so it follows that
|
|
$4 | (x^2 - z^2)$. This means that $(x, z) \in R$, and therefore $R$ is
|
|
transitive.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $R$ is reflexive, symmetric, and transitive, it can
|
|
be concluded that $R$ is an equivalence relation. This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
There are two distinct equivalence classes:
|
|
|
|
$$ [0] = \{\dots, -8, -4, -2, 0, 2, 4, 8, \dots\} = \text{ the set of all even integers } $$
|
|
|
|
$$ [1] = \{\dots, -9, -5, -1, 1, 5, 9\dots\} = \text{ the set of all odd integers } $$
|
|
|
|
28. $I$ is the relation defined on $\mathbb{R}$ as follows:
|
|
|
|
$$ \text{For every } x, y \in \mathbb{R}, m I n \Leftrightarrow x - y \text{ is an integer} $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $I$ is a relation defined on $\mathbb{R}$ as follows:
|
|
|
|
$$ \forall x, y \in \mathbb{R}, m I n \Leftrightarrow (x - y) \in \mathbb{Z} $$
|
|
|
|
To prove that $I$ is an equivalence relation, it must be shown that $I$ is
|
|
reflexive, symmetric, and transitive.
|
|
|
|
_Proof ($I$ is reflexive):_
|
|
|
|
Let $x \in \mathbb{R}$.
|
|
|
|
To prove that $I$ is reflexive, it must be shown that $(x, x) \in I$. By the
|
|
definition for $I$, this means it must be shown that:
|
|
|
|
$$ (x - x) \in \mathbb{Z} $$
|
|
|
|
Now, $x - x = 0$, and $0 \in \mathbb{Z}$. Thus $(x, x) \in I$, and therefore $I$
|
|
is reflexive.
|
|
|
|
_Proof ($I$ is symmetric):_
|
|
|
|
Let $x, y \in \mathbb{R}$.
|
|
|
|
To prove that $I$ is symmetric, it must be shown that
|
|
$(x, y) \in I \to (y, x) \in I$.
|
|
|
|
Suppose $(x, y) \in I$, by the definition for $I$, this means that:
|
|
|
|
$$ (x - y) \in \mathbb{Z} $$
|
|
|
|
Now, consider:
|
|
|
|
$$ y - x = -1(x - y) $$
|
|
|
|
Now, $-1(x - y)$ is an integer (by the product of integers), and thus
|
|
$(y - x) \in \mathbb{Z}$. Thus $(y, x) \in I$, and therefore $I$ is symmetric.
|
|
|
|
_Proof ($I$ is transitive):_
|
|
|
|
Let $x, y, z \in \mathbb{R}$.
|
|
|
|
To prove that $I$ is transitive, it must be shown that
|
|
$[(x, y) \in I \wedge (y, z) \in I] \to (x, z) \in I$.
|
|
|
|
Suppose $(x, y) \in I$ and $(y, z) \in I$. By the definition for $I$, this means
|
|
that:
|
|
|
|
$$ (x - y) \in \mathbb{Z} $$
|
|
|
|
and also that:
|
|
|
|
$$ (y - z) \in \mathbb{Z} $$
|
|
|
|
Now, consider that:
|
|
|
|
$$ x - z = (x - y) + (y - z) $$
|
|
|
|
Thus, $(x - z) \in \mathbb{Z}$ (by the sum of integers). It follows that
|
|
$(x, z) \in I$, and therefore $I$ is transitive.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $I$ is reflexive, symmetric, and transitive, it can
|
|
be concluded that $I$ is an equivalence relation. This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
There is one class for each real number $x$ with $0 \leq x < 1$. The distinct
|
|
classes are all sets of the form
|
|
$[x] = y \in \mathbb{R}, | y = n + x \text{ for some integer } n$, where $x$ is
|
|
a real number such that $0 \leq x < 1$.
|
|
|
|
29. Define $P$ on the set $\mathbb{R} \times \mathbb{R}$ of ordered pairs of
|
|
real numbers as follows: For every
|
|
$(w, x), (y, z) \in \mathbb{R} \times \mathbb{R}$,
|
|
|
|
$$ (w, x) P (y, z) \Leftrightarrow w = y $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $P$ is a relation on $\mathbb{R} \times \mathbb{R}$, defined as:
|
|
|
|
$$ \forall (w, x), (y, z) \in \mathbb{R} \times \mathbb{R}, (w, x) P (y, z) \Leftrightarrow w = y $$
|
|
|
|
To prove that $P$ is an equivalence relation, it must be shown that $P$ is
|
|
reflexive, symmetric, and transitive.
|
|
|
|
_Proof ($P$ is reflexive):_
|
|
|
|
Let $(w, x) \in \mathbb{R} \times \mathbb{R}$.
|
|
|
|
To prove that $P$ is reflexive, it must be shown that $[(w, x), (w, x)] \in P$.
|
|
By the definition for $P$, this means it must be shown that:
|
|
|
|
$$ w = w $$
|
|
|
|
This is trivially true. Thus $[(w, x), (w, x)] \in P$, and therefore $P$ is
|
|
reflexive.
|
|
|
|
_Proof ($P$ is symmetric):_
|
|
|
|
Let $(w, x), (y, z) \in \mathbb{R} \times \mathbb{R}$.
|
|
|
|
TO prove that $P$ is symmetric, it must be shown that
|
|
$[(w, x), (y, z)] \in P \to [(y, z), (w, x)] \in P$.
|
|
|
|
Suppose $[(w, x), (y, z)] \in P$. By definition for $P$, this means that:
|
|
|
|
$$ w = y $$
|
|
|
|
This means that $y = w$, by the symmetric property of equality. This means that
|
|
$[(y, z), (w, x)] \in P]$, and therefore $P$ is symmetric.
|
|
|
|
_Proof ($P$ is transitive):_
|
|
|
|
Let $(w, x), (y, z), (a, b) \in \mathbb{R} \times \mathbb{R}$.
|
|
|
|
To prove that $P$ is transitive, it must be shown that
|
|
$[[(w, x), (y, z)] \in P \wedge [(y, z), (a, b)] \in P \to [(w, x), (a, b)] \in P$.
|
|
|
|
Suppose $[(w, x), (y, z)] \in P$ and $[(y, z), (a, b)] \in P$. By the definition
|
|
for $P$, this means that:
|
|
|
|
$$ w = y $$
|
|
|
|
And also that:
|
|
|
|
$$ y = a $$
|
|
|
|
By the transitive property of equality, this means that $w = a$. It follows that
|
|
$[(w, x), (a, b)] \in P$, and therefore $P$ is transitive.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $P$ is reflexive, symmetric, and transitive, it
|
|
follows that $P$ is an equivalence relation. This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
There is one equivalence class for each real number. The distinct equivalence
|
|
classes are all sets of ordered pairs
|
|
$(x, y) \in \mathbb{R} \times \mathbb{R}, | x = a$ for each real number $a$.
|
|
|
|
30. Define $Q$ on the set $\mathbb{R} \times \mathbb{R}$ as follows: For every
|
|
$(w, x), (y, z) \in \mathbb{R} \times \mathbb{R}$,
|
|
|
|
$$ (w, x) Q (y, z) \Leftrightarrow x = z $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
Omitted.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
Omitted.
|
|
|
|
31. Let $P$ be the set of all points in the Cartesian plane except the origin.
|
|
$R$ is the relation defined on $P$ as follows: For every $p_1$ and $p_2$ in
|
|
$P$,
|
|
|
|
$$ p_1 R p_2 \Leftrightarrow p_1 \text{ and } p_2 \text{ lie on the same half-line emanating from the origin} $$
|
|
|
|
(1) Prove that the relation is an equivalence relation.
|
|
|
|
Omitted.
|
|
|
|
(2) Describe the distinct equivalence classes of each relation.
|
|
|
|
Omitted.
|
|
|
|
32. Let $A$ be the set of all straight lines in the Cartesian plane. Define a
|
|
relation $\mid \mid$ on $A$ as follows: For every $l_1$ and $l_2$ in $A$,
|
|
|
|
$$ l_1 \parallel l_2 \Leftrightarrow l_1 \text{ is parallel to } l_2 $$
|
|
|
|
Then $\parallel$ is an equivalence relation on $A$. Describe the equivalence
|
|
classes of this relation.
|
|
|
|
Every possible slope is an equivalence class, including vertical lines
|
|
(undefined).
|
|
|
|
33. Let $A$ be the set of points in the rectangle with $x$ and $y$ coordinates
|
|
between $0$ and $1$. That is,
|
|
|
|
$$ A = \{(x, y) \in \mathbb{R} \times \mathbb{R} | 0 \leq x \leq 1 \text{ and } 0 \leq y \leq 1\} $$
|
|
|
|
Define a relation $R$ on $A$ as follows: For all $(x_1, y_1) and $(x_2, y_2)$ in
|
|
$A$,
|
|
|
|
$$ (x_1, y_1) R (x_2, y_2) \Leftrightarrow (x_1, y_1) = (x_2, y_2) $$
|
|
|
|
or
|
|
|
|
$$ x_1 = 0 \text{ and } x_2 = 1 \text{ and } y_1 = y_2 $$
|
|
|
|
or
|
|
|
|
$$ x_1 = 1 \text{ and } x_2 = 0 \text{ and } y_1 = y_2 $$
|
|
|
|
or
|
|
|
|
$$ y_1 = 0 \text{ and } y_2 = 1 \text{ and } x_1 = x_2 $$
|
|
|
|
or
|
|
|
|
$$ y_1 = 1 \text{ and } y_2 = 0 \text{ and } x_1 = x_2 $$
|
|
|
|
In other words, all points along the top edge of the rectangle are related to
|
|
the points along the bottom edge directly beneath them, and all points directly
|
|
opposite each other along the left and right edges are related to each other.
|
|
The points in the interior of the rectangle are not related to anything other
|
|
than themselves. Then $R$ is an equivalence relation on $A$. Imagine gluing
|
|
together all the points that are in the same equivalence class. Describe the
|
|
resulting figure.
|
|
|
|
Gluing the top and the bottom edges of the rectangle together forms a cylinder,
|
|
and then gluing the left and right edges of the rectangle together forms a
|
|
doughnut shape (a torus).
|
|
|
|
34. The documentation for the computer language Java recommends that when an
|
|
"equals method" is defined for an object, it be an equivalence relation.
|
|
That is, if $R$ is defined as follows:
|
|
|
|
$$ x R y \Leftrightarrow \text{x.equals}(y) \text{ for all objects in the class} $$
|
|
|
|
then $R$ should be an equivalence relation. Suppose that in trying to optimize
|
|
some of the mathematics of a graphics application, a programmer creates an
|
|
object called a point, consisting of two coordinates in the plane. The
|
|
programmer defines an equals method as follows: If $p$ and $q$ are any points,
|
|
then
|
|
|
|
$$ \text{p.equals}(q) \Leftrightarrow \text{ the distance from } p \text{ to } q \text{ is less than or equal to } c $$
|
|
|
|
where $c$ is a small positive number that depends on the resolution of the
|
|
computer display. Is the programmer's equals method an equivalence relation?
|
|
Justify your answer.
|
|
|
|
No. If points $p$, $q$, and $r$ all lie on a straight line with $q$ in the
|
|
middle, and if $p$ is $c$ units from $q$ and $q$ is $c$ units from $r$, then $p$
|
|
is more than $c$ units from $r$. In other words, the programmer's equals method
|
|
is not an equivalence relation because it is not transitive.
|
|
|
|
35. Find an additional representative circuit for the input/output table of
|
|
Example 8.3.9.
|
|
|
|
Omitted.
|
|
|
|
Let $R$ be an equivalence relation on a set $A$. Prove each of the statements in
|
|
36-41 directly from the definitions of equivalence relation and equivalence
|
|
class without using the results of Lemma 8.3.2, Lemma 8.3.3, or Theorem 8.3.4.
|
|
|
|
36. For every $a$ in $a$, $a \in [a]$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ is an equivalence relation on a set $A$, and let $a \in A$.
|
|
|
|
Since $R$ is an equivalence relation, this means that $R$ is reflexive, or, in
|
|
other words, every element in $A$ is related to itself by $R$. In particular,
|
|
$a R a$, and hence, by definition of an equivalence class, $a \in [a]$. This is
|
|
what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
37. For every $a$ and $b$ in $A$, if $b \in [a]$ then $a R b$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ is an equivalence relation on a set $A$, and let $a, b \in A$.
|
|
|
|
Let $b \in [a]$.
|
|
|
|
By definition of class, this means that:
|
|
|
|
$$ b \in [a] \Leftrightarrow b R a $$
|
|
|
|
Since $R$ is an equivalence relation, $R$ is symmetric. By the definition of
|
|
symmetry, this means that:
|
|
|
|
$$ a R b $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
38. For every $a$, $b$, and $c$ in $A$, if $b R c$ and $c \in [a]$ then
|
|
$b \in [a]$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ is an equivalence relation on a set $A$, and let $a, b, c \in A$.
|
|
|
|
Let $b R c$ and let $c \in [a]$.
|
|
|
|
We must show that $b \in [a]$.
|
|
|
|
By the definition of class, since $c \in [a]$, this means that $c R a$. Since
|
|
$R$ is an equivalence relation, and therefore transitive, and also since
|
|
$b R c$, it follows, by the definition of transitive, that $b R c$ and $c R a$.
|
|
In other words $b R a$, and by definition of class, this means that $b \in [a]$.
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
39. For every $a$ and $b$ in $A$, if $[a] = [b]$ then $a R b$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ is an equivalence relation on a set $A$, and let $a, b \in A$.
|
|
|
|
Let $[a] = [b]$.
|
|
|
|
Since $R$ is reflexive (by the definition of equivalence relation), it follows
|
|
that $a \in [a]$ and $b \in [b]$.
|
|
|
|
By the supposition, $[a] = [b]$, and so it follows that $a \in [b]$. By the
|
|
definition of class, this means that $a R b$. This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
40. For every $a$, $b$, and $x$ in $A$, if $a R b$ and $x \in [a]$ then
|
|
$x \in [b]$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ is an equivalence relation on a set $A$, and let $a, b, x \in A$.
|
|
|
|
Let $a R b$ and $x \in [a]$.
|
|
|
|
It must be shown that $x \in [b]$.
|
|
|
|
Since $x \in [a]$, by the definition of equivalence class, this means that
|
|
$x R a$. Since $a R b$, by the definition of transitivity (since $R$ is an
|
|
equivalence relation and therefore transitive), it follows that $x R b$. By the
|
|
definition of equivalence class, this means that $x \in [b]$. This is what was
|
|
to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
41. For every $a$ and $b$ in $A$, if $a \in [b]$ then $[a] = [b]$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $R$ is an equivalence relation on a set $A$, and let $a, b \in A$.
|
|
|
|
Let $a \in [b]$.
|
|
|
|
To prove that $[a] = [b]$, it must be shown that $[a] \subseteq [b]$, and that
|
|
$[b] \subseteq [a]$.
|
|
|
|
_Proof ($[a] \subseteq [b]$):_
|
|
|
|
Let $x \in [a]$.
|
|
|
|
By the definition of equivalence class, this means that $x R a$. Since
|
|
$a \in [b]$, this means that $a R b$. Since $R$ is transitive (because $R$ is an
|
|
equivalence relation), this means that $x R b$. By the definition of class, this
|
|
means that $x \in [b]$. It follows that $[a] \subseteq [b]$. This is what was to
|
|
be shown.
|
|
|
|
_Proof ($[b] \subseteq [a]$):_
|
|
|
|
Let $x \in [b]$.
|
|
|
|
By the definition of equivalence, class this means that $x R b$. Since
|
|
$a \in [b]$, this means that $a R b$. Since $R$ is symmetric (because $R$ is an
|
|
equivalence relation), this means that $b R a$. Then, since $R$ is transitive
|
|
(again, because $R$ is an equivalence relation), it follows that $x R a$. By the
|
|
definition of equivalence class, this means that $x \in [a]$. It follows that
|
|
$[b] \subseteq [a]$. This is what was to be shown.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $[a] \subseteq [b]$ and also that
|
|
$[b] \subseteq [a]$, it follows (by the definition for subset), that
|
|
$[a] = [b]$. This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
42. Let $R$ be the relation defined in Example 8.3.12.
|
|
|
|
a. Prove that $R$ is reflexive.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $A$ is the set of all ordered pairs of integers for which the second
|
|
element of the pair is nonzero:
|
|
|
|
$$ A = \mathbb{Z} \times (\mathbb{Z} - \{0\}) $$
|
|
|
|
Then, define a relation $R$ on $A$ as follows:
|
|
|
|
$$ \forall (a, b), (c, d) \in A, (a, b) R (c, d) \Leftrightarrow ad = bc $$
|
|
|
|
Let $(x, y) \in A$.
|
|
|
|
To prove that $R$ is reflexive, it must be shown that $[(x, y), (x, y)] \in R$.
|
|
By the definition of $R$, this means it must be shown that:
|
|
|
|
$$ xy = yx $$
|
|
|
|
By the commutative law of product, this is true. Therefore
|
|
$[(x, y), (x, y)] \in R$, and $R$ is reflexive.
|
|
|
|
Q.E.D.
|
|
|
|
b. Prove that $R$ is symmetric.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $A$ is the set of all ordered pairs of integers for which the second
|
|
element of the pair is nonzero:
|
|
|
|
$$ A = \mathbb{Z} \times (\mathbb{Z} - \{0\}) $$
|
|
|
|
Then, define a relation $R$ on $A$ as follows:
|
|
|
|
$$ \forall (a, b), (c, d) \in A, (a, b) R (c, d) \Leftrightarrow ad = bc $$
|
|
|
|
Let $(x, y), (z, a) \in A$.
|
|
|
|
To prove that $R$ is symmetric, it must be shown that
|
|
$[(x, y), (z, a)] \in R \to [(z, a), (x, y)] \in R$.
|
|
|
|
Suppose $[(x, y), (z, a)] \in R$. By the definition for $R$, this means that:
|
|
|
|
$$ xa = yz $$
|
|
|
|
(It must be shown that $zy = ax$.)
|
|
|
|
By the commutative property for product, and the symmetric property of equality,
|
|
$xa = yz$ can be rewritten as:
|
|
|
|
$$ zy = ax $$
|
|
|
|
It follows that $[(z, a), (x, y)] \in R$, and therefore $R$ is symmetric. This
|
|
is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
c. List four distinct elements in $[(1, 3)]$.
|
|
|
|
$$ (2, 6), (-2, -6), (3, 9), (-3, -9) $$
|
|
|
|
d. List four distinct elements in $[(2, 5)]$.
|
|
|
|
$$ (4, 10), (6, 15), (8, 20), (10, 25) $$
|
|
|
|
43. In Example 8.3.12, define operations of addition $(+)$ and multiplication
|
|
$(\cdot)$ as follows: For every $(a, b), (c, d) \in A$,
|
|
|
|
$$ [(a, b)] + [(c, d)] = [(ad + bc, bd)] $$
|
|
|
|
$$ [(a, b)] \cdot [(c, d)] = [(ac, bd)] $$
|
|
|
|
a. Prove that this addition is well defined. That is, show that if
|
|
$[(a, b)] = [(a', b')]$ and $[(c, d)] = [(c', d')]$, then
|
|
$[(ad + bc), bd] = [(a'd' + b'c', b'd')]$.
|
|
|
|
Omitted.
|
|
|
|
b. Prove that this multiplication is well defined. That is, show that if
|
|
$[(a, b)] = [(a', b')]$ and $[(c, d)] = [(c', d')]$, then
|
|
$[(ac, bd)] = [(a'c', b'd')]$.
|
|
|
|
Omitted.
|
|
|
|
c. Show that $[(0, 1)]$ is an identity element for addition. That is, show that
|
|
for any $(a, b) \in A$,
|
|
|
|
$$ [(a, b)] + [(0, 1)] = [(0, 1)] + [(a, b)] = [(a, b)] $$
|
|
|
|
Omitted.
|
|
|
|
d. Find an identity element for multiplication. That is, find $(i, j)$ in $A$ so
|
|
that for every $(a, b)$ in $A$,
|
|
$[(a, b)] \cdot [(i, j)] = [(i, j)] \cdot [(a, b)] = [(a, b)]$.
|
|
|
|
Omitted.
|
|
|
|
e. For any $(a, b) \in A$, show that $[(-a, b)]$ is an inverse for $[(a, b)]$
|
|
for addition. That is, show that
|
|
$[(-a, b)] + [(a, b)] = [(a, b)] + [(-a, b)] = [(0, 1)]$.
|
|
|
|
Omitted.
|
|
|
|
f. Given any $(a, b) \in A$ with $a \neq 0$, find an inverse for $[(a, b)]$ for
|
|
multiplication. That is, find $(c, d)$ in $A$ so that
|
|
$[(a, b)] \cdot [(c, d)] = [(c, d)] \cdot [(a, b)] = [(i, j)]$, where $[(i, j)]$
|
|
is the identity element you found in part (d).
|
|
|
|
Omitted.
|
|
|
|
44. Let $A = \mathbb{Z}^+ \times \mathbb{Z}^+$. Define a relation $R$ on $A$ as
|
|
follows: For every $(a, b)$ and $(c, d)$ in $A$,
|
|
|
|
$$ (a, b) R (c, d) \Leftrightarrow a + d = c + b $$
|
|
|
|
a. Prove that $R$ is reflexive.
|
|
|
|
Omitted.
|
|
|
|
b. Prove that $R$ is symmetric.
|
|
|
|
Omitted.
|
|
|
|
c. Prove that $R$ is transitive.
|
|
|
|
Omitted.
|
|
|
|
d. List five elements in $[(1, 1)]$.
|
|
|
|
Omitted.
|
|
|
|
e. List five elements in $[(3, 1)]$.
|
|
|
|
Omitted.
|
|
|
|
f. List five elements in $[(1, 2)]$.
|
|
|
|
Omitted.
|
|
|
|
g. Describe the distinct equivalence classes of $R$.
|
|
|
|
45. The following argument claims to prove that the requirement that an
|
|
equivalence relation be reflexive is redundant. In other words, it claims to
|
|
show that if a relation is symmetric and transitive, then it is reflexive.
|
|
Find the mistake in the argument.
|
|
|
|
"**Proof:** Let $R$ be a relation on a set $A$ and suppose $R$ is symmetric and
|
|
transitive. For any two elements $x$ and $y$ in $A$, if $x R y$ then $y R x$
|
|
since $R$ is symmetric. Thus it follows by transitivity that $x R x$, and hence
|
|
$R$ is reflexive."
|
|
|
|
The mistake in the argument is that just because $R$ is symmetric and transitive
|
|
does not necessarily mean it is reflexive. Recall that for $R$ to be reflexive,
|
|
$\forall x \in A, x R x$. Consider, however, a set where the relation is both
|
|
symmetric and transitive, but not reflexive:
|
|
|
|
$$ A = \{1, 2\} $$
|
|
|
|
$$ R = \{(1, 1)\} $$
|
|
|
|
Now, $R$ is symmetric, since $(1, 1) \to (1, 1)$, and $R$ is transitive, since
|
|
$(1, 1) \wedge (1, 1) \to (1, 1)$, and while $(1, 1)$ is reflexive, there is no
|
|
ordered pair in the set where $2 R 2$, so therefore $R$ is not reflexive, even
|
|
though $R$ is symmetric and transitive.
|
|
|
|
46. Let $R$ be a relation on a set $A$ and suppose $R$ is symmetric and
|
|
transitive. Prove the following: If for every $x$ in $A$ there is a $y$ in
|
|
$A$ such that $x R y$, then $R$ is an equivalence relation.
|
|
|
|
Omitted.
|
|
|
|
47. Refer to the quote at the beginning of this section to answer the following
|
|
questions.
|
|
|
|
a. What is the name of the Knight's song called?
|
|
|
|
Omitted.
|
|
|
|
b. What is the name of the Knight's song?
|
|
|
|
Omitted.
|
|
|
|
c. What is the Knight's song called?
|
|
|
|
Omitted.
|
|
|
|
d. What _is_ the Knight's song?
|
|
|
|
Omitted.
|
|
|
|
e. What is your (full, legal) name?
|
|
|
|
Omitted.
|
|
|
|
f. What are you called?
|
|
|
|
Omitted.
|
|
|
|
g. What _are_ you? (Do not answer this on paper; just think about it.)
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
Page 567
|
|
|
|
**Exercise Set 8.4**
|
|
|
|
1.
|
|
|
|
a. Use the Caesar cipher to encrypt the message WHERE SHALL WE MEET.
|
|
|
|
b. Use the Caesar cipher to decrypt the message LQ WKH FDIHWHULD.
|
|
|
|
2.
|
|
|
|
a. Use the Caesar cipher to encrypt the message AN APPLE A DAY.
|
|
|
|
b. Use the Caesar cipher to decrypt the message NHHSV WKH GRFWRU DZDB.
|
|
|
|
3. Let $a = 25, $b = 19$, and $n = 3$.
|
|
|
|
a. Verify that $3 | (25 - 19)$.
|
|
|
|
b. Explain why $25 \equiv 19 (\mod 3)$.
|
|
|
|
c. What value of $k$ has the property that $25 = 19 + 3k$?
|
|
|
|
d. What is the (nonnegative) remainder obtained when $25$ is divided by $3$?
|
|
When $19$ is divided by $3$?
|
|
|
|
e. Explain why $25 \mod 3 = 19 \mod 3$.
|
|
|
|
4. Let $a = 68$, $b = 33$, and $n = 7$.
|
|
|
|
a. Verify that $7 | (68 - 33)$.
|
|
|
|
b. Explain why $68 \equiv 33(\mod 7)$.
|
|
|
|
c. What value of $k$ has the property that $68 = 33 + 7k$?
|
|
|
|
d. What is the (nonnegative) remainder obtained when $68$ is divided by $7$?
|
|
When $33$ is divided by $7$?
|
|
|
|
e. Explain why $68 \mod 7 = 33 \mod 7$.
|
|
|
|
5. Prove the transitivity of modular congruence. That is, prove that for all
|
|
integers $a$, $b$, $c$, and $n$ with $n > 1$, if $a \equiv b(\mod n)$ and
|
|
$b \equiv c(\mod n)$ then $a \equiv c(\mod n)$.
|
|
|
|
6. Prove that the distinct equivalence classes of the relation of congruence
|
|
modulo $n$ are the sets $[0], [1], [2], \dots, [n - 1]$, where for each
|
|
$a = 0, 1, 2, \dots, n - 1$,
|
|
|
|
$$ [a] = \{m \in \mathbb{Z} | m \equiv a (\mod n)\} $$
|
|
|
|
7. Verify the following statements.
|
|
|
|
a. $128 \equiv 2(\mod 7)$ and $61 \equiv 5(\mod 7)$
|
|
|
|
b. $(128 + 61) \equiv (2 + 5)(\mod 7)$
|
|
|
|
c. $(128 - 61) \equiv (2 - 5)(\mod 7)$
|
|
|
|
d. $(128 \cdot 61) \equiv (2 \cdot 5)(\mod 7)$
|
|
|
|
e. $128^2 = 2^2(\mod 7)$
|
|
|
|
8. Verify the following statements.
|
|
|
|
a. $45 \equiv 3(\mod 6)$ and $104 \equiv 2(\mod 6)$
|
|
|
|
b. $(45 + 104) \equiv (3 + 2)(\mod 6)$
|
|
|
|
c. $(45 - 104) \equiv (3 - 2)(\mod 6)$
|
|
|
|
d. $(45 \cdot 104) \equiv (3 \cdot 2)(\mod 6)$
|
|
|
|
e. $45^2 \equiv 3^2(\mod 6)$
|
|
|
|
In 9-11, prove each of the following statements, assuming that $a$, $b$, $c$,
|
|
$d$, and $n$ are integers with $n > 1$ and that $a \equiv c(\mod n)$ and
|
|
$b \equiv d(\mod n)$.
|
|
|
|
9.
|
|
|
|
a. $(a + b) \equiv (c + d)(\mod n)$
|
|
|
|
b. $(a - b) \equiv (c - d)(\mod n)$
|
|
|
|
10. $a^2 \equiv c^2(\mod n$
|
|
|
|
11. $a^m \equiv c^m(\mod n)$ for every integer $m \geq 1$ (Use mathematical
|
|
induction on $m$.)
|
|
|
|
12.
|
|
|
|
a. Prove that for every integer $n \geq 0$, $10^n \equiv 1(\mod 9)$.
|
|
|
|
b. Use part (a) to prove that a positive integer is divisible by $9$ if, and
|
|
only if, the sum of its digits is divisible by $9$.
|
|
|
|
13.
|
|
|
|
a. Prove that for every integer $n \geq 1$, $10^n \equiv (-1)^n(\mod 11)$ .
|
|
|
|
b. Use part (a) to prove that a positive integer is divisible by $11$ if, and
|
|
only if, the alternating sum of its digits is divisible by $114. (For instance,
|
|
the alternating sum of the digits of $82,379$ is $8 - 2 + 3 - 7 + 9 = 11$ and
|
|
$82,379 = 11 \cdot 7489$.)
|
|
|
|
14. Use the technique of Example 8.4.4 to find $14^2 \mod 55$, $14^4 \mod 55$,
|
|
$14^8 \mod 55$, and $14^{16} \mod 55$.
|
|
|
|
15. Use the result of exercise 14 and the technique of Example 8.4.5 to find
|
|
$14^{27} \mod 55$.
|
|
|
|
In 16-18, use the techniques of Example 8.4.4 and Example 8.4.5 to find the
|
|
given numbers.
|
|
|
|
16. $675^{307} \mod 713$
|
|
|
|
17. $89^{307} \mod 713$
|
|
|
|
18. $48^{307} \mod 713$
|
|
|
|
In 19-24, use the RSA cipher from Examples 8.4.9 and 8.4.10. In 19-21, translate
|
|
the message into its numeric equivalent and encrypt it. In 22-24, decrypt the
|
|
cipher-text and translate the result into letters of the alphabet to discover
|
|
the message.
|
|
|
|
19. HELLO
|
|
|
|
20. WELCOME
|
|
|
|
21. EXCELLENT
|
|
|
|
22. 13 20 20 09
|
|
|
|
23. 08 05 15
|
|
|
|
24. 51 14 49 15
|
|
|
|
25. Use Theorem 5.2.2 to prove that if $a$ and $n$ are positive integers and
|
|
$a^{n - 1}$ is prime, then $a = 2$ and $n$ is prime.
|
|
|
|
In 26 and 27, use the extended Euclidean algorithm to find the greatest common
|
|
divisor of the given numbers and express it as a linear combination of the two
|
|
numbers.
|
|
|
|
26. 6664 and 765
|
|
|
|
27. 4158 and 1568
|
|
|
|
Exercises 28 and 29 refer to the following formal version of the extended
|
|
Euclidean algorithm.
|
|
|
|
**Algorithm 8.4.1 Extended Euclidean Algorithm**
|
|
|
|
_[Given integers $A$ and $B$ with $A > B > 0$, this algorithm computes
|
|
$\text{gcd}(A, B) and finds integers $s$ and $t$ such that
|
|
$sA + tB = \text{gcd}(A, B)$.]_
|
|
|
|
**Input:** $A$, $B$ _[integers with $A > B > 0$]_
|
|
|
|
**Algorithm Body:**
|
|
|
|
$a := A, b := B, s := 1, t := 0, u := 0, v := 1\\ \textit{[pre-codndition: } a =
|
|
sA + tB \textit{ and } b = uA + vB,\\ \text{gcd}(a, b) = \text{gcd}(A, B)
|
|
\textit{]}\\ \textbf{while} (b \neq 0) \\ \ \ \textit{[loop invariant: } a =
|
|
sA + tB \textit{ and } b = uA + vB,\\ \ \ \text{gcd}(a, b) = \text{gcd}(A, B)\\ \ \ r:= a \mod b, q := a \text{ div } b\\ \ \ a := b, b := r\\ \ \ \textit{newu } := s - uq, \textit{newv } := t - vq\\ \ \ s := u, t := v\\ \ \ u:= \textit{newu}, v := \textit{newv}\\ \textbf{end while}\\ gcd := a\\ \textit{[post condition: } \text{gcd}(A, B) = a = sA + tB \textit{]}$
|
|
|
|
**Output:** $\text{gcd}\textit{[a positive integer]}, s, t \textit{[integers]}$
|
|
|
|
In 28 and 29, for the given values of $A$ and $B$, make a table showing the
|
|
values of $s$, $t$, and $sA + tB$ before the start of the while loop and after
|
|
each iteration of the loop
|
|
|
|
28. $A = 330$, $B = 156$
|
|
|
|
29. $A = 284$, $B = 168$
|
|
|
|
30. Finis the proof of Theorem 8.4.5 by proving that if $a$, $b$, and $c$ are as
|
|
in the proof, then $c | b$.
|
|
|
|
31.
|
|
|
|
a. Find an inverse for $210$ modulo $13$.
|
|
|
|
b. Find a positive inverse for $210$ modulo $13$.
|
|
|
|
c. Find a positive solution for the congruence $210x \equiv 8 (\mod 13)$.
|
|
|
|
32.
|
|
|
|
a. Find an inverse for $41$ modulo $660$.
|
|
|
|
b. Find the least positive solution for the following congruence:
|
|
$41x \equiv 125(\mod 660)$.
|
|
|
|
33. Use Theorem 8.4.5 to prove that for all integers $a$, $b$, and $c$, if
|
|
$\text{gcd}(a, b) = 1$ and $a | c$ and $b | c$, then $ab | c$.
|
|
|
|
34. Give a counterexample to show that the statement of exercise 33 is false if
|
|
the hypothesis that $\text{gcd}(a, b) = 1$ is removed.
|
|
|
|
35. Corollary 8.4.7 guarantees the existence of an inverse modulo $n$ for an
|
|
integer $a$ when $a$ and $n$ are relatively prime. Use Euclid's lemma to
|
|
prove that the inverse is unique modulo $n$. In other words, show that if
|
|
$s$ and $t$ are any two integers whose product with $a$ is congruent to $1$
|
|
modulo $n$, then $s$ and $t$ are congruent to each other modulo $n$.
|
|
|
|
In 36, 37, 39, and 40, use the RSA cipher with public key
|
|
$n = 713 = 23 \cdot 31$ and $e = 43$. In 36 and 37, encode the messages into
|
|
their numeric equivalents and encrypt them. In 39 and 40, decrypt the given
|
|
ciphertext and find the original messages.
|
|
|
|
36. HELP
|
|
|
|
37. COME
|
|
|
|
38. Find the least positive inverse for $43$ modulo $660$.
|
|
|
|
39. 675 089 089 048
|
|
|
|
40. 028 018 675 129
|
|
|
|
41.
|
|
|
|
a. Use mathematical induction and Euclid's lemma to prove that for every
|
|
positive integer $s$, if $p$ and $q_1, q_2, \dots, q_s$ are 0rime numbers and
|
|
$p | q_1q_2 \cdots q_s$, then $p = q_i$ for some $i$ with $1 \leq i \leq s$.
|
|
|
|
b. The uniqueness part of the unique factorization theorem for the integers says
|
|
that given any integer $n$, if
|
|
|
|
$$ n = p_1p_2 \cdots p_r = q_1q_2 \cdots q_s $$
|
|
|
|
for some positive integers $r$ and $s$ and prime numbers
|
|
$p_1 \leq p_2 \leq \cdots \leq p_r$ and $q_1 \leq q_2 \leq \cdots \leq q_s$,
|
|
then $r = s$ and $p_i = q_i$ for every integer $i$ with $1 \leq i \leq r$.
|
|
|
|
Use the result of part (a) to fill in the details of the following sketch of a
|
|
proof:
|
|
|
|
Suppose that $n$ is an integer with two different prime factorizations:
|
|
$n = p_1p_2 \cdots p_t = q_1q_2 \cdots q_u$. All the prime factors that appear
|
|
on both sides can be cancelled (as many times as they appear on both sides) to
|
|
arrive at the situation where $p_1p_2 \cdots p_r = q_1q_2 \cdots q_s$,
|
|
$p_1 \leq p_2 \leq \cdots \leq p_r$, $q_1 \leq q_2 \leq \cdots \leq q_s$, and
|
|
$p_i \neq q_j$ for any integers $i$ and $j$. Then use part (a) to deduce a
|
|
contradiction, and conclude that the prime factorization of $n$ is unique
|
|
except, possibly, for the order in which the prime factors are written.
|
|
|
|
42. According to Fermat's little theorem, if $p$ is a prime number and $a$ and
|
|
$p$ are relatively prime, then $a^{p - 1} \equiv 1 (\mod p)$. Verify that
|
|
this theorem gives correct results for the following:
|
|
|
|
a. $a = 15$ and $p = 7$
|
|
|
|
b. $a = 8$ and $p = 11$
|
|
|
|
43. Fermat's little theorem can be used to show that a number is not prime by
|
|
finding a number $a$ relatively prime to $p$ with the property that
|
|
$a^{p - 1} \cancel{\equiv} 1(\mod p)$. However, it cannot be used to show
|
|
that a number _is_ prime. Find an example to illustrate this fact. That is,
|
|
find integers $a$ and $p$ such that $a$ and $p$ are relatively prime and
|
|
$a^{p - 1} \equiv 1(\mod p)$ but $p$ is not prime.
|