From 82a2fe137e2f61883d2195b15640520bd8f7ef82 Mon Sep 17 00:00:00 2001 From: tomit4 Date: Thu, 11 Jun 2026 09:13:09 -0700 Subject: [PATCH] :construction: Hands gave out... --- chapter_4/exercises.md | 54 ++++++++++++ chapter_4/notes.md | 164 +++++++++++++++++++++++++++++++++++++ chapter_4/test_yourself.md | 17 ++++ 3 files changed, 235 insertions(+) diff --git a/chapter_4/exercises.md b/chapter_4/exercises.md index 28d014e..1ec5f67 100644 --- a/chapter_4/exercises.md +++ b/chapter_4/exercises.md @@ -6595,3 +6595,57 @@ a. if $d \mid n$, then $n = \left\lfloor \dfrac{n}{d} \right\rfloor \cdot d$. b. if $n = \left\lfloor \dfrac{n}{d} \right\rfloor \cdot d$ then $d \mid n$. Omitted. + +--- + +**Exercise Set 4.7** + +Page 248 + +1. Fill in the blanks in the following proof by contradiction that there is no + least positive real number. + +**Proof:** Suppose not. That is, suppose that there is a least positive real +number $x$. _[We must deduce (a)]._ Consider the number $\dfrac{x}{2}$. Since +$x$ is a positive real number, $\dfrac{x}{2}$ is also (b). In addition, we can +deduce that $\dfrac{x}{2} < x$ by multiplying both sides of the inequality +$1 < 2$ by \(c\) and dividing (d). Hence $\dfrac{x}{2}$ is a positive real +number that is less than the least positive real number. This is a (e). _[Thus +the supposition is false, and so there is no least positive real number.]_ + +2. Is $\dfrac{1}{0}$ an irrational number? Explain. + +3. Use proof by contradiction to show that for every integer $n$, $3n + 2$ is + not divisible by $3$. + +4. Use proof by contradiction to show that for every integer $m$, $7m + 4$ is + not divisible by $7$. + +Carefully formulate the negations of each of the statements in 5-7. Then prove +each statement by contradiction. + +5. There is no greatest even integer. + +6. There is no greatest negative real number. + +7. There is no least positive rational number. + +8. Fill in the blanks for the following proof that the difference of any + rational number and any irrational number is irrational. + +**Proof (by contradiction):** + +Suppose not. That is, suppose that there exist (a) $x$ and (b) $y$ such that +$x - y$ is rational. By definition of rational, there exist integers $a$, $b$, +$c$, and $d$ with $b \neq 0$ and $d \neq 0$ so that $x = $ \(c\) and $x - y =$ +(d). By substitution, + +$$ \frac{a}{b} - y = \frac{c}{d} $$ + +Adding $y$ and subtracting $\dfrac{c}{d}$ on both sides gives + +$$ y = \text{(e)} \quad \text{ by substitution} $$ + +$$ = \frac{ad}{bd} - \frac{bc}{bd} $$ + +$$ = \frac{ad - bc}{bd} \quad \text{ by algebra} $$ diff --git a/chapter_4/notes.md b/chapter_4/notes.md index e8d45ba..c8f4450 100644 --- a/chapter_4/notes.md +++ b/chapter_4/notes.md @@ -814,3 +814,167 @@ Hence $$ 0 \leq r < d \quad \text{ by substitution} $$ _[This is what was to be shown.]_ + +--- + +Page 241 + +**Method of Proof by Contradiction** + +1. Suppose the statement to be proved is false. That is, suppose the negation of + the statement is true. + +2. Show that this supposition leads logically to a contradiction. + +3. Conclude that the statement to be proved is true. + +--- + +Page 242 + +**Theorem 4.7.1** + +There is no greatest integer. + +**Proof:** + +_[We take the negation of the theorem and suppose it to be true.]_ Suppose not. +That is, suppose there is a greatest integer $N$. _[We must deduce a +contradiction.]_ Then $N \geq n$ for every integer $n$. Let $M = N + 1$. Now $M$ +is an integer since it is a sum of integers. Also $M > N$ since $M = N + 1$. +Thus $M$ is an integer that is greater than $N$. So $N$ is the greatest integer +and $N$ is not the greatest integer, which is a contradiction. _[This +contradiction shows that the supposition is false and hence, that the theorem is +true.]_ + +--- + +**Theorem 4.7.2** + +There is no integer that is both even and odd. + +**Proof:** + +_[We take the negation of the theorem and suppose it to be true.]_ Suppose not. +That is, suppose that there is at least one integer $n$ that is both even and +odd. _[We must deduce a contradiction.]_ By definition of even, $n = 2a$ for +some integer $a$, and by definition of odd, $n = 2b + 1$ for some integer $b$. +Consequently, + +$$ 2a = 2b + 1 \quad \text{ by equating the two expressions for } n $$ + +and so + +$$ +2a - 2b = 1 \\ +2(a - b) = 1 \\ +a j b = \frac{1}{2} \quad \text{ by algebra} +$$ + +Now since $a$ and $b$ are integers, the difference $a - b$ must also be an +integer. But $a - b = \dfrac{1}{2}$, and $\dfrac{1}{2}$ is not an integer. Thus +$a - b$ is an integer and $a - b$ is not an integer, which is a contradiction. +_[This contradiction shows that the supposition is false and, hence, that the +theorem is true.]_ + +--- + +Page 244 + +**Theorem 4.7.3** + +The sum of any rational number and any irrational number is irrational. + +**Proof:** + +_[We take the negation of the theorem and suppose it to be true.]_ Suppose not. +That is, suppose there is a rational number $r$ and an irrational number $s$ +such that $r + s$ is rational. _[We must deduce a contradiction.]_ By definition +of rational, $r = \dfrac{a}{b}$ and $r + s = \dfrac{c}{d}$ for some integers +$a$, $b$, $c$, and $d$ with $b \neq 0$ and $d \neq 0$. By substitution, + +$$ \frac{a}{b} + s = \frac{c}{d} $$ + +and so, + +$$ s = \frac{c}{d} - \frac{a}{b} \quad \text{ by subtracting } \frac{a}{b} \text{ from both sides} $$ + +$$ = \frac{bc - ad}{bd} $$ + +Now $bc - ad$ and $bd$ are both integers _[since $a$, $b$, $c$, and $d$ are +integers and since products and differences of integers are integers]_, and +$bd \neq 0$ _[by the zero product property.]_ Hence $s$ is a quotient of the two +integers $bc - ad$ and $bd$ with $bd \neq 0$. Thus, by definition of rational, +$s$ is rational, which contradicts the supposition that $s$ is irrational. +_[Hence the supposition is false and the theorem is true.]_ + +--- + +Page 245 + +**Method of Proof by Contraposition** + +1. Express the statement to be proved in the form + +$$ \forall x \text{ in } D, \text{ if } P(x) \text{ then } Q(x) $$ + +(This step may be done mentally.) + +2. Rewrite this statement in the contrapositive form + +$$ \forall x \text{ in } D, \text{ if } Q(x) \text{ is false then } P(x) \text{ is false} $$ + +(This step may be done mentally.) + +3. Prove the contrapositive by a direct proof. + +a. Suppose $x$ is a (particular but arbitrarily chosen) element of $D$ such that +$Q(x)$ is false. + +b. Show that $P(x)$ is false. + +--- + +Page 245 + +**Proposition 4.7.4** + +For every integer $n$, if $n^2$ is even then $n$ is even. + +**Proof (by contraposition):** + +Suppose $n$ is any odd integer. _[We must show that $n^2$ is odd.]_ By +definition of odd, $n = 2k + 1$ for some integer $k$. By substitution and +algebra, + +$$ n^2 =(2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 $$ + +Now $2k^2 + 2k$ is an integer because products and sums of integers are +integers. So $n^2 = 2 \cdot (\text{an integer}) + 1$, and thus, by definition of +odd, $n^2$ is odd _[as was to be shown]._ + +--- + +Page 246 + +**Proposition 4.7.4** + +For every integer $n$, if $n^2$ is even then $n$ is even. + +**Proof (by contradiction):** + +_[We take the negation of the theorem and suppose it to be true.]_ Suppose not. +That is, suppose there is an integer $n$ such that $n^2$ is even and $n$ is not +even. _[We must deduce a contradiction.]_ By the quotient-remainder theorem with +divisor equal to $2$, any integer is even or odd. Hence, since $n$ is not even +it is odd, and thus, by definition of odd, $n = 2k + 1$ for some integer $k$. By +substitution and algebra, + +$$ n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 $$ + +Now $2k^2 + 2k$ is an integer because products and sums of integers are +integers. So $n^2 = 2 \cdot (\text{an integer}) + 1$, and thus, by definition of +odd, $n^2$ is odd. Therefore, $n^2$ is both even and odd. This contradicts +Theorem 4.7.2, which states that no integer can be both even and odd. _[This +contradiction shows that the supposition is false and, hence, that the +proposition is true.]_ diff --git a/chapter_4/test_yourself.md b/chapter_4/test_yourself.md index 81417a0..42d8d80 100644 --- a/chapter_4/test_yourself.md +++ b/chapter_4/test_yourself.md @@ -200,3 +200,20 @@ $n \leq x < n + 1$ that ______. $n - 1 < x \leq n$ + +--- + +**Test Yourself** + +Page 248 + +1. To prove a statement by contradiction, you suppose that ______ and you show + that ______. + +2. A proof by contraposition of a statement of the form + "$\forall x \in D, \text{ if } P(x) \text{ then } Q(x)$" is a direct proof of + ______. + +3. To prove a statement of the form + "$\forall x \in D, \text{ if } P(x) \text{ then } Q(x)$" by contraposition, + you suppose that ______ and you show that ______.