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@ -9,96 +9,380 @@ composite numbers.
a. Is $-17$ an odd integer?
$$ -17 = 2(-9) + 1 $$
Let $k = -9$, so our expression becomes by substitution:
$$ -17 = 2k + 1 $$
Since $-17$ can be represented by the form $2k + 1$ where $k = -9$ and $k$ is an
integer, by the definition of an odd number, $-17$ is an odd integer.
b. Is $0$ neither even nor odd?
No, $0$ can be represented as $0 = 2(0)$, and let $k = 0$, so $0 = 2k$, where
$k$ is an integer, and by definition of an even number, $0$ is even.
c. Is $2k - 1$ odd?
Yes, $2k - 1 = 2(k - 1) + 1$ where $k - 1$ is an integer by the difference of
integers. Let $m = k - 1$, so our expression becomes $2k - 1 = 2m + 1$, and
since $m$ is an integer, we can conclude that $2k - 1$ is an odd integer by
definition of odd integers.
2. Assume that $c$ is a particular integer.
a. Is $-6c$ an even integer?
Yes $-6c = 2(-3c)$, where $-3c$ is an integer by the product of integers, and
since $-6c$ can be expressed as $2 \cdot \text{ some integer}$, it is even by
the definition of even integers.
b. Is $8c + 5$ an odd integer?
c. Is $(c^1 + 1) - (c^2 - 1) - 2$ an even integer?
Yes, $8c + 5 = 8c + 4 + 1 = 2(4c + 2) + 1$ where $4c + 2$ is an integer by the
sum of products of integers. Since $8c + 5$ can be expressed as
$2(\text{some integer}) + 1$, we can conclude that $8c + 5$ is an odd integer by
the definition of odd integers.
c. Is $(c^2 + 1) - (c^2 - 1) - 2$ an even integer?
Yes, if evaluate the statement by laws of algebra, we get:
$$ (c^2 + 1) - (c^2 - 1) - 2 = c^2 + 1 - c^2 + 1 - 2 = 1 + 1 - 2 = 0 $$
And as established in 1b, $0$ is an even integer, so $(c^2 + 1) - (c^2 - 1) - 2$
can be expressed in the form of $2 \cdot \text{ some integer}$, so by the
definition of integers, $(c^2 + 1) - (c^2 - 1) - 2$ is an even integer.
3. Assume that $m$ and $n$ are particular integers?
a. Is $6m + 8n$ even?
Yes, $6m + 8n = 2(3m + 4n)$. $3m + 4n$ is an integer by the sum of products of
integers. Since $6m + 8n$ can be expressed as $2 \cdot \text{ some integer}$, by
the definition of even integers, $6m + 8n$ is even.
b. Is $10mn + 7$ odd?
$10mn + 7 = 10mn + 6 + 1 = 2(5mn + 3) + 1$. $5mn + 3$ is an integer by the
product and sum of integers. Since $10mn + 7$ can be expressed as
$2(\text{some integer}) + 1$, $10mn + 7$ is an odd integer.
c. If $m > n > 0$, is $m^2 - n^2$ composite?
Not necessarily. Consider $m = 3$ and $n = 2$, then $m^2 - n^2 = 9 - 4 = 5$,
which is a prime number.
4. Assume that $r$ and $s$ are particular integers.
a. Is $4rs$ even?
Yes, $4rs = 2(2rs)$, where $2rs$ is an integer by the product of integers. Since
$4rs = 2(\text{ some integer})$, by the definition of an even integer, $4rs$ is
an even integer.
b. Is $6r + 4s^2 + 3$ odd?
$6r + 4s^2 + 3 = 6r + 4s^2 + 2 + 1 = 2(3r + 2s^2 + 1) + 1$. $3r + 2s^2 + 1$ is
an integer by product and sum of integers. Let $k = 3r + 2s^2 + 1$, and so
$6r + 4s^2 + 3 = 2k + 1$. By definition of an odd integer, $6r + 4s^2 + 3$ is an
odd integer.
c. If $r$ and $s$ are both positive, is $r^2 + 2rs + s^2$ composite?
Since $r^2 + 2rs + s^2 = (r + s)^2 = (r + s)(r + s)$ and $r + s \geq 2$. And
since $r + s > 1$, the product of $(r + s)(r + s)$ is composite.
Prove the statements in 5-11.
5. There are integers $m$ and $n$ such that $m > 1$ and $n > 1$ and
$\dfrac{1}{m} + \dfrac{1}{n}$ is an integer.
For example, let $m = 2$ and $n = 2$, then $\dfrac{1}{2} + \dfrac{1}{2} = 1$,
and $1$ is an integer.
6. There are distinct integers $m$ and $n$ such that
$\dfrac{1}{m} + \dfrac{1}{n}$ is an integer.
For example, let $m = -2$, and $n = 2$, then $\dfrac{1}{-2} + \dfrac{1}{2} = 0$,
and $0$ is an integer.
7. There are real numbers $a$ and $b$ such that
$$ \sqrt{a + b} = \sqrt{a} + \sqrt{b} $$
For example, let $a = 0$ and $b = 9$, then
$\sqrt{0 + 9} = \sqrt{9} = 3 = 0 + 3 = \sqrt{0} + \sqrt{9}$.
8. There is an integer $n > 5$ such that $2^n - 1$ is prime.
For example, let $n = 7$, then $2^7 - 1 = 127$, and $127$ is prime.
9. There is a real number $x$ such that $x > 1$ and $2^x > x^{10}$.
For example, let $x = \dfrac{1}{2}$, then $2^{\frac{1}{2}} \approx 1.414213562 >
0.0009765625 = \left(\frac{1}{2}\right)^{10}$
**Definition:** An integer $n$ is called a **perfect square** if, and only if,
$n = k^2$ for some integer $k$.
10. There is a perfect square that can be written as a sum of two other perfect
squares.
Let $n = 4$ and $m = 3$, and let $l = k^2$ be the sum of their squares:
$$ l = k^2 = n^2 + m^2 = (4)^2 + (3)^2 = 16 + 9 = 25 $$
$$ l = k^2 = 25 $$
So $l = 25$ can be written as $n^2 + m^2$ where $n = 4$ and $m = 3$, and since
both $n$ and $m$ are integers, we can say that $l$ is a perfect square by
definition of a perfect square and $l$ can be written as the sum of two other
perfect squares.
11. There is an integer $n$ such that $2n^2 - 5n + 2$ is prime.
For example let $n = 3$, then
$2n^2 - 5n + 2 = 2(3)^2 - 5(3) + 2 = 2(9) - 15 + 2 = 18 - 15 + 2 = 3 + 2 = 5$,
and $5$ is prime.
In 12-13, (a) write a negation for the given statement, and (b) use a
counterexample to disprove the given statement. Explain how the counterexample
actually shows that the given statement is false.
12. For all real numbers $a$ and $b$, if $a < b$ the $a^2 < b^2$.
(a)
Original:
$$ \forall a \in \mathbb{R} \forall b \in \mathbb{R}((a < b) \to (a^2 < b^2)) $$
Negation:
$$ \exists a \in \mathbb{R} \exists b \in \mathbb{R}((a < b) \wedge (a^2 \geq b^2)) $$
There exist real numbers $a$ and $b$ such that $a < b$ and $a^2 \geq b^2$.
(b)
Counterexample:
Let $a = -2$ and let $b = -1$. The hypothesis $a < b$ holds as $-2 < -1$ is
true, but the conclusion of the original statement $a^2 < b^2$ is false as
$(-2)^2 = 4 \cancel{<} 1 = (-1)^2$.
Since the original statement claims that the implication holds true for all real
numbers $a$ and $b$, a single counterexample is sufficient to show that the
statement is false.
13. For every integer $n$, if $n$ is odd then $\dfrac{n - 1}{2}$ is odd.
(a)
Original:
Let $P(n) = n \text{ is odd}$
Let $Q(m) = m \text{ is odd}$
$$ \forall n \in \mathbb{Z}(P(n) \to Q\left(\frac{n - 1}{2}\right)) $$
Negation:
$$ \exists n \in \mathbb{Z}(P(n) \wedge \neg Q\left(\frac{n - 1}{2}\right)) $$
There exists some integer $n$ such that $n$ is odd and $\dfrac{n - 1}{2}$ is not
odd.
(b)
Counterexample:
Let $n = 1$. $n$ is odd as $1$ can be expressed as $n = 1 = 2(k) + 1$, where
$k = 0$. This means that $1$ is odd by the definition of an odd integer, and the
hypothesis of the original statement is true. The conclusion of the original
statement, however, is false, as
$\dfrac{n - 1}{2} = \dfrac{1 - 1}{2} = \dfrac{0}{2} = 0$, and $0$ is not odd.
Since the original statement claims that the implication holds true for all
integers $n$, a single counterexample is sufficient to show that the statement
is false.
Disprove each of the statements in 14-16 by giving a counterexample. In each
case explain how the counterexample actually disproves the statement.
14. For all integers $m$ and $n$, if $2m + n$ is odd then $m$ and $n$ are both
odd.
Let $m = 2$ and let $n = 1$, the hypothesis $2m + n$ is odd is true as
$2(2) + 1 = 5$, and $5$ is odd, but the conclusion that both $m$ and $n$ are odd
is false, as $m$ is even.
15. For every integer $p$, if $p$ is prime then $p^2 - 1$ is even.
Let $p = 2$. The hypothesis holds true as $2$ is prime, but the conclusion
"$p^2 - 1$ is even" is false for this $p$ as $(2)^2 - 1 = 4 - 1 = 3$, and $3$ is
not even.
16. For every integer $n$, if $n$ is even then $n^2 + 1$ is prime.
Let $n = 0$, the hypothesis "$n$ is even" holds true for this $n$ as $0$ is
even. The conclusion "$n^2 + 1$ is prime" fails for this $n$ as
$0^2 + 1 = 0 + 1 = 1$, and $1$ is not prime.
In 17-20, determine whether the property is true for all integers, true for no
integers, or true for some integers and false for other integers. Justify your
answers.
17. $(a + b)^2 = a^2 + b^2$
This property is true for some integers and not others.
For example where it is true, consider $a = 0$ and $b = 1$, then
$(0 + 1)^2 = 1^2 = 1 = 0 + 1 = (0)^2 + (1)^2$ holds true for at least two
integers.
For example where it is false, consider $a = 1$ and $b = 1$, then
$(1 + 1)^2 = 2^2 = 4 \neq 2 = 1 + 2 = (1)^2 + (1)^2$. Since this provides a
counterexample, this property cannot hold true for all integers.
Therefore, this property holds true for some integers and not others.
18. $\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{a + c}{b + d}$
This is true for $a = c = 0$ and $b = d = 1$as:
$$ \frac{0}{1} + \frac{0}{1} = 0 + 0 = 0 = \frac{0}{2} = \frac{0 + 0}{1 + 1} $$
This is false for $a = b = c = d = 1$, as:
$$ \frac{1}{1} + \frac{1}{1} = 1 + 1 = 2 \neq 1 = \frac{2}{2} = \frac{1 + 1}{1 + 1} $$
Therefore, this property holds true for some integers and not others.
19. $-a^n = (-a)^n$
This is true for $a = -1$ and $n = 1$.
$$ -(-1)^1 = (-(-1))^1 $$
$$ -(-1) = (-(-1)) $$
$$ 1 = 1 $$
This is false for $a = -1$ and $n = 2$.
$$ -(-1)^2 = (-(-1))^2 $$
$$ -(1) = (1)^2 $$
$$ -1 \neq 1 $$
Therefore, this property holds true for some integers and not others.
20. The average of any two odd integers is odd.
Let $m$ and $n$ be odd integers. Let $m = 2k + 1$ and $n = 2p + 1$ where $k$ and
$p$ are any integers.
We are asserting that $\dfrac{m + n}{2}$ is odd. By substitution, we can express
this as:
$$ \frac{m + n}{2} = \frac{(2k + 1) + (2p + 1)}{2} = \frac{2k + 2p + 2}{2} = \frac{2(k + p + 1)}{2} = k + p + 1 $$
In order to prove that $k + p + 1$ is odd, we need to be able to express it in
the form of $2(\text{some integer}) + 1$ by the definition of an odd integer.
An example where this is true is if $k = 2$ and $p = 4$, then
$k + p + 1 = 2 + 4 + 1 = 7$, and $7$ is an odd integer.
A counterexample where this is false is if $k = 3$ and $p = 4$, then
$k + p + 1 = 3 + 4 + 1 = 8$, and $8$ is not an odd integer.
Therefore, this property holds true for some integers and not others.
Prove the statement in 21 and 22 by the method of exhaustion.
21. Every positive even integer less than 26 can be expressed as a sum of three
of fewer perfect squares. (For instance, $10 = 1^2 + 3^2$ and $16 = 4^2$.)
or fewer perfect squares. (For instance, $10 = 1^2 + 3^2$ and $16 = 4^2$.)
22. For each integer $n$ with $1 \leq n \leq 10$, $n^2 -n + 11$ is a prime
Let's first establish all positive even integers less than 26:
$$ \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\} $$
$$ 2 = 1^2 + 1^2 $$
$$ 4 = 2^2 $$
$$ 6 = 2^2 + 1^2 + 1^2 $$
$$ 8 = 2^2 + 2^2 $$
$$ 10 = 3^2 + 1^2 $$
$$ 12 = 2^2 + 2^2 + 2^2 $$
$$ 14 = 3^2 + 2^2 + 1^2 $$
$$ 16 = 4^2 $$
$$ 18 = 4^2 + 1^2 + 1^2 $$
$$ 20 = 4^2 + 2^2 $$
$$ 22 = 3^2 + 3^2 + 2^2 $$
$$ 24 = 4^2 + 2^2 + 2^2 $$
22. For each integer $n$ with $1 \leq n \leq 10$, $n^2 - n + 11$ is a prime
number.
Let's establish all possible values for $n$:
$$ \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} $$
$n = 1$:
$$ (1)^2 - (1) + 11 = 1 - 1 + 11 = 11 \text{ is prime} $$
$n = 2$:
$$ (2)^2 - (2) + 11 = 4 - 2 + 11 = 13 \text{ is prime} $$
$n = 3$:
$$ (3)^2 - (3) + 11 = 9 - 3 + 11 = 17 \text{ is prime} $$
$n = 4$:
$$ (4)^2 - (4) + 11 = 16 - 4 + 11 = 23 \text{ is prime} $$
$n = 5$:
$$ (5)^2 - (5) + 11 = 25 - 5 + 11 = 31 \text{ is prime} $$
$n = 6$:
$$ (6)^2 - (6) + 11 = 36 - 6 + 11 = 41 \text{ is prime} $$
$n = 7$:
$$ (7)^2 - (7) + 11 = 49 - 7 + 11 = 53 \text{ is prime} $$
$n = 8$:
$$ (8)^2 - (8) + 11 = 64 - 8 + 11 = 67 \text{ is prime} $$
$n = 9$:
$$ (9)^2 - (9) + 11 = 81 - 9 + 11 = 83 \text{ is prime} $$
$n = 10$:
$$ (10)^2 - (10) + 11 = 100 - 10 + 11 = 101 \text{ is prime} $$
Each of the statements in 23-26 is true. For each, (a) rewrite the statement
with the quantification implicit as If _____, then _____, and (b) write the
first sentence of a proof (the "starting point") and the last sentence of a
@ -107,12 +391,45 @@ the statements in order to be able to do these exercises.)
23. For every integer $m$, if $m > 1$ then $0 < \dfrac{1}{m} < 1$.
(a) If an integer is greater than 1, then its reciprocal is between 0 and 1.
(b)
Starting Point: Suppose $m$ is any integer such that $m > 1$.
To Show: $0 < \dfrac{1}{m} < 1$
24. For every real number $x$, if $x > 1$ then $x^2 > x$.
(a) If a real number is greater than 1, then it's square is greater than itself.
(b)
Starting Point: Suppose $x$ is any real number such that $x > 1$.
To Show: $x^2 > x$.
25. For all integers $m$ and $n$, if $mn = 1$ then $m = n = 1$ or $m = n = -1$.
(a) If the product of any two integers is equal to 1, then both integers either
equal 1 or -1.
(b)
Starting Point: Suppose $m$ and $n$ are any integers such that $mn = 1$.
To Show: $m = n = 1$ or $m = n = -1$.
26. For every real number $x$, if $0 < x < 1$ then $x^2 < x$.
(a) If a real number is between 0 and 1, then its square is less than itself.
(b)
Starting Point: Suppose $x$ is any real number such that $0 < x < 1$.
To Show: $x^2 < x$.
27. Fill in the blanks in the following proof.
**Theorem:** For every odd integer $n$, $n^2$ is odd.
@ -134,6 +451,14 @@ Because we have not assumed anything about $n$ except that it is an odd integer,
it follows from the principle of ___ (d) ___ that for _every_ odd integer $n$,
$n^2$ is odd.
a. odd integer.
b. $2k + 1$
c. $n^2$
d. universal generalization
In each of 28-31:
a. Rewrite the theorem in three different ways:
@ -170,6 +495,26 @@ integers and because __ \(c\) __.
Hence $m + n = 2u$, where $u$ is an integer, and so, by __ (d) __, $m + n$ is
even _[as was to be shown]._
a.
**Theorem:** the sum of any two odd integers is even.
$\forall$ integers $m$ and $n$, if $m$ and $n$ are odd, then $m + n$ is even.
$\forall$ odd integers $m$ and $n$, $m + n$ is even.
If any two integers are odd, then their sum is even.
b.
(a) the definition of an odd integer
(b) substitution
(\c\) any sum of integers is an integer
(d)$ the definition of an even integer
29.
**Theorem:** The negative of any integer is even.
@ -193,6 +538,26 @@ Let $r = -k$. Then $r$ is an integer because $(-1)$ and $k$ are integers and __
Hence $-n = 2r$, where $r$ is an integer, and so $-n$ is even by __ (d) __ _[as
was to be shown]._
a.
**Theorem:** The negative of any integer is even.
$\forall$ integers $n$, if $n$ is negative, then $n$ is even.
$\forall$ negative integers $n$, $n$ is even.
If an integer is negative, then it is even.
b.
(a) the definition of an even integer
(b) substitution
\(c\) the product of any two integers is an integer
(d) the definition of an even integer
30.
**Theorem 4.1.2:** The sum of any even integer and any odd integer is odd.
@ -206,6 +571,27 @@ $$ m + n = \text{\_\_ (c) \_\_} = 2(r + s) + 1 $$
Since $r$ and $s$ are both integers, so is their sum $r + s$. Hence $m + n$ has
the form twice some integer plus one, and so __ (d) __ by definition of odd.
a.
**Theorem 4.1.2:** The sum of any even integer and any odd integer is odd.
$\forall$ integers $m$ and $n$, if $m$ is an even integer and $n$ is an odd
integer, then $m + n$ is odd.
$\forall$ even integers $m$ and odd integers $n$, $m + n$ is odd.
If $m$ is an even integer and $n$ is any odd integer, then $m + n$ is odd.
b.
(a) any odd integer
(b) integer $r$
\(c\) $2r + (2s + 1)$
(d) $m + n$ is odd
31.
**Theorem:** Whenever $n$ is an odd integer, $5n^2 + 7$ is even.
@ -231,3 +617,23 @@ integers are integers.
Hence $5n^2 + 7 = 2t$, where $t$ is an integer, and thus __ (d) __ by definition
of even _[as was to be shown]._
a.
**Theorem:** Whenever $n$ is an odd integer, $5n^2 + 7$ is even.
$\forall$ integers $n$, if $n$ is an odd integer, then $5n^2 + 7$ is even.
$\forall$ odd integers $n$, $5n^2 + 7$ is even.
If $n$ is an odd integer, then $5n^2 + 7$ is even.
b.
(a) $2k + 1$
(b) $5(2k + 1)^2 + 7$
\(c\) $10k^2 + 10k + 6$
(d) $5n^2 + 7$ is even

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@ -4,16 +4,31 @@ Page 194
1. An integer is even if, and only if, ______.
it equals twice some integer.
2. An integer is odd if, and only if, ______.
it equals twice some integer plus 1.
3. An integer $n$ is prime if, and only if, ______.
$n$ is greater than $1$ and if $n$ equals the product of any two positive
integers, then one of the integers equals $1$ and the other equals $n$.
4. The most common way to disprove a universal statement is to find ______.
a counterexample.
5. According to the method of generalizing from the generic particular, to show
that every element of a set satisfies a certain property, suppose $x$ is a
______, and show that ______.
particular but arbitrarily chosen element of the set; $x$ satisfies the given
property.
6. To use the method of direct proof to prove a statement of the form, "For
every $x$ in a set $D$, if $P(x)$ then $Q(x)$," one supposes that ______ and
one shows that ______.
$x$ is a particular but arbitrarily chosen element of the set $D$ that makes the
hypothesis $P(x)$ true; $x$ makes the conclusion $Q(x)$ true.