From 58293523171ae7a65e23ea85f4d95c40d1f94375 Mon Sep 17 00:00:00 2001 From: tomit4 Date: Sun, 7 Jun 2026 03:08:22 -0700 Subject: [PATCH] :construction: Fin 4.2 --- chapter_4/exercises.md | 1049 +++++++++++++++++++++++++++++++++++- chapter_4/test_yourself.md | 20 + 2 files changed, 1065 insertions(+), 4 deletions(-) diff --git a/chapter_4/exercises.md b/chapter_4/exercises.md index d74fafa..8b5fdff 100644 --- a/chapter_4/exercises.md +++ b/chapter_4/exercises.md @@ -651,38 +651,406 @@ for writing proofs of universal statements. 1. For every integer $n$, if $n$ is odd then $3n + 5$ is even. +**Theorem:** Suppose $n$ is any odd integer. + +**Proof:** + +Since $n$ is odd, $n = 2k + 1$ for some integer $k$. + +Then + +$$ 3n + 5 = 3(2k + 1) + 5 \quad \text{ by substitution} $$ + +$$ \quad = 6k + 3 + 5 $$ + +$$ \quad = 6k + 8 $$ + +$$ \quad = 2(3k + 4) \quad \text{ by algebra} $$ + +Let $t = 3k + 4$. + +Then $3n + 5 = 2(3k + 4) = 2t$, where $t$ is an integer because products and +sums of integers are integers. + +Therefore $3n + 5$ is even by the definition of even integers. + +Q.E.D. + 2. For ever integer $m$, if $m$ is even then $3m + 5$ is odd. +**Theorem:** Suppose $m$ is any even integer. + +**Proof:** + +Since $m$ is even, $m = 2k$ for some integer $k$. + +Then: + +$$ 3m + 5 = 3(2k) + 5 \quad \text{ by substitution} $$ + +$$ \quad = 6k + 5 $$ + +$$ \quad = 6k + 4 + 1 $$ + +$$ \quad = 2(3k + 2) + 1 \quad \text{ by algebra} $$ + +Let $t = 3k + 2$. + +Then $3m + 5 = 2(3k + 2) + 1 = 2t + 1$ where $t$ is an integer because the +product and sum of integers are integers. + +Therefore $3m + 5$ is odd by the definition of odd integers. + +Q.E.D. + 3. For every integer $n$, $2n - 1$ is odd. +**Theorem:** + +Suppose $n$ is any integer. + +**Proof:** + +Then: + +$$ 2n - 1 = 2n - 2 + 1 \quad \text{ by algebra} $$ + +$$ \quad = 2(n - 1) + 1 \quad \text{ by factoring} $$ + +Let $t = n - 1$. + +Then $2n - 1 = 2(n - 1) + 1 = 2t + 1$ where $t$ is an integer because the +difference of integers is an integer. + +Therefore $2n - 1$ is odd by the definition of an odd integer. + +Q.E.D. + 4. **Theorem 4.2.2:** The difference of any even integer minus any odd integer is odd. +**Theorem:** Suppose $m$ is any even integer and $n$ is any odd integer. + +**Proof:** + +Since $m$ is even and $n$ is odd, $m = 2k$ and $n = 2s + 1$ where $k$ is some +integer and $s$ is some integer. + +Then + +$$ m - n = 2k - (2s + 1) \quad \text{ by substitution} $$ + +$$ \quad = 2k - 2s - 1 $$ + +$$ \quad = 2k - 2s - 2 + 1 $$ + +$$ \quad = 2(k - s - 1) + 1 $$ + +Let $t = k - s - 1$. + +Then $m - n = 2(k - s - 1) + 1 = 2t + 1$ where $t$ is an integer because the +difference of integers is an integer. + +Therefore $m - n$ is odd by the definition of odd integers. + +Q.E.D. + 5. If $a$ and $b$ are any odd integers, then $a^2 + b^2$ is even. +**Theorem:** Suppose $a$ is any odd integer and $b$ is any odd integer. + +**Proof:** + +Since $a$ is an odd integer and $b$ is an odd integer, $a = 2k + 1$ and +$b = 2s + 1$ where $k$ is some integer and $s$ is some integer. + +Then: + +$$ a^2 + b^2 = (2k + 1)^2 + (2s + 1)^2 \quad \text{ by substitution} $$ + +$$ \quad = (2k + 1)(2k + 1) + (2s + 1)(2s + 1) \quad \text{ by exponentiation} $$ + +$$ \quad = (4k^2 + 4k + 1) + (4s^2 + 4s + 1) $$ + +$$ \quad = 4k^2 + 4k + 1 + 4s^2 + 4s + 1 $$ + +$$ \quad = 4k^2 + 4k + 4s^2 + 4s + 2 $$ + +$$ \quad = 2(2k^2 + 2k + 2s^2 + 2s + 1) $$ + +Let $t = 2k^2 + 2k + 2s^2 + 2s + 1$. + +Then $a^2 + b^2 = 2(2k^2 + 2k + 2s^2 + 2s + 1) = 2t$ where $t$ is an integer +because the product and sum of integers is an integer. + +Therefore $a^2 + b^2$ is even by the definition of even integers. + +Q.E.D. + 6. If $k$ is any odd integer and $m$ is any even integer, then $k^2 + m^2$ is odd. +**Theorem:** + +Suppose $k$ is any odd integer and $m$ is any even integer. + +**Proof:** + +Since $k$ is an odd integer and $m$ is an even integer, $k = 2a + 1$ and +$m = 2b$ where $a$ is some integer and $b$ is some integer. + +Then: + +$$ k^2 + m^2 = (2a + 1)^2 + (2b)^2 \quad \text{ by substitution} $$ + +$$ \quad = (2a + 1)(2a + 1) + (2b)(2b) \quad \text{ by exponentiation} $$ + +$$ \quad = (4a^2 + 4a + 1) + (4b^2) $$ + +$$ \quad = 4a^2 + 4a + 4b^2 + 1 $$ + +$$ \quad = 2(2a^2 + 2a + 2b^2) + 1 $$ + +Let $t = 2a^2 + 2a + 2b^2$. + +Then $k^2 + m^2 = 2(2a^2 + 2a + 2b^2) + 1 = 2t + 1$ where $t$ is an integer +because the product and sum of integers is an integer. + +Therefore $k^2 + m^2$ is odd by the definition of an odd integer. + +Q.E.D. + 7. The difference between the squares of any two consecutive integers is odd. +**Theorem:** + +Suppose $n$ is any integer. + +**Proof:** + +Since $n$ is an integer, $n + 1$ is a consecutive integer of $n$. + +Then: + +$$ n^2 - (n + 1)^2 = n^2 - (n + 1)(n + 1) \quad \text{ by exponentiation} $$ + +$$ \quad = n^2 - (n^2 + 2n + 1) $$ + +$$ \quad = n^2 - n^2 - 2n - 1 \quad \text{ by distribution} $$ + +$$ \quad = -2n - 1 \quad \text{ by distribution} $$ + +$$ \quad = -2n - 2 + 1 $$ + +$$ \quad = 2(-n - 1) + 1 $$ + +Let $t = -n - 1$. + +Then $n^2 - (n + 1)^2 = 2(-n - 1) + 1 = 2t + 1$ where $t$ is an integer because +the product and difference of integers is an integer. + +Therefore $n^2 - (n + 1)^2$ is odd by the definition of an odd integer. + +Q.E.D. + 8. For any integers $m$ and $n$, if $m$ is even and $n$ is odd then $5m + 3n$ is odd. +**Theorem:** + +Suppose $m$ is any even integer and $n$ is any odd integer. + +**Proof:** + +Since $m$ is an even integer and $n$ is an odd integer, $m = 2k$ and +$n = 2s + 1$ where $k$ is some integer and $s$ is some integer. + +Then: + +$$ 5m + 3n = 5(2k) + 3(2s + 1) \quad \text{ by substitution} $$ + +$$ \quad = 10k + 6s + 3 $$ + +$$ \quad = 10k + 6s + 2 + 1 $$ + +$$ \quad = 2(5k + 3s + 1) + 1 $$ + +Let $t = 5k + 3s + 1$. + +Then $5m + 3n = 2(5k + 3s + 1) + 1 = 2t + 1$ where $t$ is an integer because the +products and sums of integers is an integer. + +Therefore $5m + 3n$ is odd by the definition of an odd integer. + +Q.E.D. + 9. If an integer greater than $4$ is a perfect square, then the immediately preceding integer is not prime. +**Theorem:** + +Suppose $n$ is any integer where $n > 4$ and $n$ is a perfect square. + +**Proof:** + +Since $n$ is a perfect square and $n > 4$, then $n = k^2$ for some integer $k$ +where $k > 2$ or $k < -2$. + +Then: + +$$ n - 1 = k^2 - 1 \quad \text{ by substitution} $$ + +$$ \quad = (k + 1)(k - 1) \quad \text{ by algebra} $$ + +In order for $n - 1$ to be prime, either $k + 1$ or $k - 1$ must be equal to +$1$. + +If $k > 2$, then both $k + 1 > 1$ and $k - 1 > 1$ are true. + +If $k < -2$, then both $k + 1 < 1$ and $k - 1 < 1$ are true. + +Therefore neither $k + 1$ nor $k - 1$ can ever be equal to $1$. + +Therefore $n - 1$ is not prime by the definition of a prime number. + +Q.E.D. + 10. If $n$ is any even integer, then $(-1)^n = 1$. +**Theorem:** + +Suppose $n$ is any even integer. + +**Proof:** + +Since $n$ is an even integer, then $n = 2k$ where $k$ is some integer. + +Then: + +$$ (-1)^n = (-1)^{2k} \quad \text{ by substitution} $$ + +$$ \quad = (-1)^{2 \cdot k} $$ + +$$ \quad = ((-1)^2)^k $$ + +$$ \quad = 1^k $$ + +$$ \quad = 1 \quad \text{ by the laws of exponents} $$ + +Therefore $(-1)^n = 1$. + +Q.E.D. + 11. If $n$ is any odd integer, then $(-1)^n = -1$. +**Theorem:** + +Suppose $n$ is any odd integer. + +**Proof:** + +Since $n$ is an odd integer, then $n = 2k + 1$ where $k$ is some integer. + +Then: + +$$ (-1)^n = (-1)^{2k + 1} \quad \text{ by substitution} $$ + +$$ (-1)^n = (-1)^{2 \cdot k} \cdot (-1)^1 $$ + +$$ (-1)^n = ((-1)^2)^k \cdot (-1)^1 $$ + +$$ (-1)^n = 1^k \cdot -1 $$ + +$$ (-1)^n = 1 \cdot -1 $$ + +$$ (-1)^n = -1 \quad \text{ by the laws of exponents} $$ + +Therefore $(-1)^n = -1$. + +Q.E.D. + Prove that the statements in 12-14 are false. 12. There exists an integer $m \geq 3$ such that $m^2 - 1$ is prime. +Take the negation first: + +For all integers $m \geq 3$, $m^2 - 1$ is not prime. + +**Theorem:** + +There is no integer $m \geq 3$ such that $m^2 - 1$ is prime. + +**Proof:** + +By algebra, we know that: + +$$ m^2 - 1 = (m + 1)(m - 1) $$ + +We also know that for $m^2 - 1$ to be prime, either $m + 1$ or $m - 1$ must be +equal to $1$. + +Since $m \geq 3$, we know that both $m + 1 \geq 4$ and $m - 1 \geq 2$ are both +true. Thus both factors are greater than 1. + +Therefore $m^2 - 1$ is a product of two integers greater than $1$, so it is not +prime. + +Therefore $m^2 - 1$ is not prime by the definition of prime numbers. + +Q.E.D. + 13. There exists an integer $n$ such that $6n^2 + 27$ is prime. +Take the negation first: + +For all integers $n$, $6n^2 + 27$ is not prime. + +**Theorem:** + +There is no integer $n$ such that $6n^2 + 27$ is prime. + +**Proof:** + +By algebra we know that: + +$$ 6n^2 + 27 = 3(2n^2 + 9) $$ + +Since $n^2$ is always positive or $0$, by the laws of exponentiation and by +algebra, we can conclude that $2n^2 + 9 \geq 9$ is true. + +Since $3 > 1$ and $2n^2 + 9 > 1$, we then know that $6n^2 + 27$ is a product of +two integers greater than $1$, so it is not prime. + +$6n^2 + 27$ is not prime by the definition of prime numbers. + +Q.E.D. + 14. There exists an integer $k \geq 4$ such that $2k^2 - 5k + 2$ is prime. +Take the negation first: + +For all integers $k \geq 4$, $2k^2 - 5k + 2$ is not prime. + +**Theorem:** + +There is no integer $k \geq 4$ such that $2k^2 - 5k + 2$ is prime. + +**Proof:** + +By algebra we know: + +$$ 2k^2 - 5k + 2 = (k - 2)(2k - 1) $$ + +Since we know that $k \geq 4$, we know that $k - 2 \geq 2$ and $2k - 1 \geq 7$. + +Since $k - 2 > 1$ and $2k - 1 > 1$, we then know that $2k^2 - 5k + 2$ is a +product of two integers greater than $1$, so it is not prime. + +Therefore $2k^2 - 5k + 2$ is not prime by definition of prime numbers. + +Q.E.D. + Find the mistakes in the "proofs" shown in 15-19. 15. @@ -692,6 +1060,10 @@ Find the mistakes in the "proofs" shown in 15-19. **"Proof:** For $k = 2$, $k > 0$ and $k^2 + 2k + 1 = 2^2 + 2 \cdot 2 + 1 = 9$. And since $9 = 3 \cdot 3$, then $9$ is composite. Hence the theorem is true." +Answer: This proof just shows that the theorem is true for a single case, +$k = 2$, in order to prove a universal claim as the theorem presents, the proof +must prove the conclusion true for every integer $k$ where $k > 0$. + 16. **Theorem:** The difference between any odd integer and any even integer is odd. @@ -702,8 +1074,14 @@ even, $m = 2k$ where $k$ is an integer. Then $$ n - m = (2k + 1) - 2k = 1 $$ -and $1$ is odd. Therefore, the difference between any odd integer and any even -integer is odd." +Answer: This proof makes the mistake of using $k$ to represent two different +quantities. By setting $n = 2k + 1$ and $m = 2k$, the proof implies that +$n = m + 1$, and thus deduces the conclusion for only this situation. This proof +falsely then "proves" that the difference between _any_ even and odd integer +will always equal $1$, but taking most examples of even and odd integers as +cases for this would show that this is false. In essence, this proof makes the +mistake of assigning the same variable name to represent two different integers, +and then by algebra comes to a false conclusion. 17. @@ -727,6 +1105,20 @@ and both $r$ and $s$ are strictly between $1$ and $k^2 + 2k + 1$, then $k^2 + 2k + 1$ is not prime. Hence $k^2 + 2k + 1$ is composite as was to be shown." +Answer: This proof makes the mistake of assuming what is to be proved. Instead +of proving that $k^2 + 2k + 1$ is composite, it assumes the definition of +composite numbers applies to the expression and then extrapolates logic about +$r$ and $s$ that cannot be known because it has not yet been proven that +$k^2 + 2k +1$ is composite. This starts at the line starting with "Since", which +cannot be asserted as that is an assertion of the conclusion, not the +hypothesis. + +Teacher's answer: This incorrect proof assumes what is to be proved. The word +_since_ in the third sentence is completely unjustified. The second sentence +tells only what happens _if_ $k^2 + 2k + 1$ is composite. But at that point in +the proof, it has not been established that $k^2 + 2k + 1$ _is_ composite. In +fact, that is exactly what is to be proved. + 18. **Theorem:** The product of any even integer and any odd integer is even. @@ -742,6 +1134,12 @@ $$ mn = (2p)(2q + 1) = 2r $$ where $r$ is an integer. By definition of even, then, $m \cdot n$ is even, as was to be shown." +Answer: This incorrect proof exhibits confusion between what is known and what +is still to be shown. The writer correctly uses the definitions of even and odd +integers to express $m$ and $n$ as $2p$ and $2q + 1$, but assumes the conclusion +that $mn$ must be an expression of $2r$, which is exactly what is to be shown, +but has not yet been proven. In essence, they have jumped to the conclusion. + 19. **Theorem:** The sum of any two even integers equals $4k$ for some integer $k$. @@ -754,6 +1152,13 @@ $$ m + n = 2k + 2k = 4k $$ That is what was to be shown." +Answer: This incorrect proof suffers from multiple problems. One is that it uses +the same variable name $k$ to represent two potentially different integers when +expressing both $m$ and $n$ as even integers. The writer then incorrectly sums +them to $4k$ and concludes they have proven the conclusion, but the form of $4k$ +does not explicitly show that $m + n$ is even by the definition of even +integers. + In 20-38 determine whether the statement is true or false. Justify your answer with a proof or a counterexample, as appropriate. In each case use only the definitions of the terms and the Assumptions listed on page 161, not any @@ -761,58 +1166,694 @@ previously established properties. 20. The product of any two odd integers is odd. +**Theorem:** + +Suppose $n$ is any odd integer and $m$ is any odd integer. + +**Proof:** + +Since $n$ and $m$ are odd integers, $n = 2k + 1$ and $m = 2s + 1$ where $k$ is +some integer and $s$ is some integer. + +Then: + +$$ n \cdot m = (2k + 1)(2s + 1) \quad \text{ by substitution} $$ + +$$ \quad = 4ks + 2s + 2k + 1 $$ + +$$ \quad = 2(2ks + s + k) + 1 \quad \text{ by algebra} $$ + +Let $t = 2ks + s + k$. + +Then $n \cdot m = 2(2ks + s + k) + 1 = 2t + 1$ where $t$ is an integer because +the products and sums of integers is an integer. + +Therefore $n \cdot m$ is odd by the definition of odd integers. + +Q.E.D. + 21. The negative of any odd integer is odd. +**Theorem:** + +Suppose $n$ is any odd integer. + +**Proof:** + +Since $n$ is odd, $n = 2k + 1$ where $k$ is some integer. + +Then: + +$$ -n = -(2k + 1) \quad \text{ by substitution} $$ + +$$ \quad = -2k - 1 $$ + +$$ \quad = -2k - 2 + 1 $$ + +$$ \quad = 2(-k - 1) + 1 $$ + +Let $t = -k - 1$. + +Then $-n = 2(-k - 1) + 1 = 2t + 1$ where $t$ is an integer because the products +and differences of integers is an integer. + +Therefore $-n$ is odd by definition of an odd integer. + +Q.E.D. + 22. For all integers $a$ and $b$, $4a + 5b + 3$ is even. +False. Intuition says if $a = b = 0$ then $4a + 5b + 3 = 3$ which is not even. +Let's prove this more formally. + +Take the negation: + +There exists some integer $a$ and some integer $b$ such that $4a + 5b + 3$ is +not even. + +**Theorem:** There is some integer $a$ and some integer $b$ such that +$4a + 5b + 3$ is not even. + +Let $a = 0$ and let $b = 0$ + +Then: + +$$ 4a + 5b + 3 = 4(0) + 5(0) + 3 \quad \text{ by substitution} $$ + +$$ \quad = 0 + 0 + 3 $$ + +$$ \quad = 3 $$ + +Since $3$ is not even, $4a + 5b + 3$ is not even for the given $a$ and $b$. + +Therefore there exists integers $a$ and $b$ such that $4a + 5b + 3$ is not even +and the original given statement is false. + +Q.E.D. + 23. The product of any even integer and any integer is even. +**Theorem:** + +Suppose $n$ is any even integer and $m$ is any integer. + +**Proof:** + +Since $n$ is even, $n = 2k$ for some integer $k$. + +Then: + +$$ n \cdot m = (2k)(m) \quad \text{ by substitution} $$ + +$$ \quad = 2km $$ + +$$ \quad = 2(km) $$ + +Let $t = km$. + +Then $n \cdot m = 2(km) = 2t$ where $t$ is an integer because the product of +integers is an integer. + +Therefore $n \cdot m$ is even by definition of even integers. + +Q.E.D. + 24. If a sum of two integers is even, then one of the summands is even. (In the expression $a + b$, $a$ and $b$ are called **summands**.) +This is false, quickly consider $1 + 3 = 4$ where both the summands are odd, but +the sum is even. + +Take the negation first: + +There exists two integers whose sum is even but neither integer is even. + +**Claim:** + +There is some integer $a$ and there is some integer $b$ such that $a + b$ is +even and neither $a$ nor $b$ is even. + +**Proof:** + +Let $a = 1$ and $b = 3$. + +Then: + +$$ a + b = 4 $$ + +$$ \quad = 2(2) $$ + +Then $a + b$ is even by the definition of even integers. + +Then: + +$$ a = 1 $$ + +$$ \quad = 2(0) + 1 $$ + +And: + +$$ b = 3 $$ + +$$ \quad = 2(1) + 1 $$ + +Then both $a$ and $b$ are odd by the definition of odd integers. + +Therefore $a + b$ is even, but $a$ and $b$ are not even for the given $a$ and +$b$, therefore the statement is false. + +Q.E.D. + 25. The difference of any two even integers is even. +**Theorem:** + +Suppose $m$ is an even integer and $n$ is an even integer. + +**Proof:** + +Since $m$ and $n$ are even integers, $m = 2k$ and $n = 2s$ where $k$ is some +integer and $s$ is some integer. + +Then: + +$$ n - m = (2s) - (2k) \quad \text{ by substitution} $$ + +$$ \quad = 2s - 2k $$ + +$$ \quad = 2(s - k) \quad \text{ by algebra} $$ + +Let $t = s - k$. + +Then $n -m = 2(s - k) = 2t$ where $t$ is an integer because the difference of +integers is an integer. + +Therefore $n - m$ is even by the definition of even integers. + +Q.E.D. + 26. For all integers $a$, $b$, and $c$, if $a$, $b$, and $c$ are consecutive, then $a + b + c$ is even. +This is false. Take the negation for the claim. + +**Claim:** + +There exists some integer $a$, some integer $b$, and some integer $c$ such that +$a$, $b$, and $c$ are consecutive and $a + b + c$ is not even. + +**Proof:** + +Let $a = 2$, $b = 3$, $c = 4$. + +Then: + +$$ a + b + c = 2 + 3 + 4 \quad \text{ by substitution} $$ + +$$ \quad = 9 $$ + +$$ \quad = 8 + 1 $$ + +$$ \quad = 2(4) + 1 $$ + +Therefore for the given $a$, $b$, and $c$, $a + b + c$ is not even, by the +definition of an odd number. + +Therefore the given $a$, $b$, and $c$ are consecutive numbers, but their sum is +not even. The statement is false. + +Q.E.D. + 27. The difference of any two odd integers is even. +**Theorem:** + +Suppose $n$ is any odd integer and $m$ is any odd integer. + +**Proof:** + +Since $n$ and $m$ are odd integers, $n = 2k + 1$ and $m = 2s + 1$ where $k$ is +some integer and $s$ is some integer. + +Then: + +$$ n - m = (2k + 1) - (2s + 1) \quad \text{ by substitution} $$ + +$$ \quad = 2k + 1 - 2s - 1 $$ + +$$ \quad = 2k - 2s $$ + +$$ \quad = 2(k - s) $$ + +Let $t = k - s$. + +Then $n - m = 2(k - s) = 2t$ where $t$ is an integer because the difference of +integers is an integer. + +Therefore $n - m$ is even by definition of an even integer. + +Q.E.D. + 28. For all integers $n$ and $m$, if $n - m$ is even then $n^3 - m^3$ is even. +**Theorem:** + +Suppose $n$ is any integer and $m$ is any integer and $n - m$ is even. + +**Proof:** + +Since we know that $n - m$ is even, $n - m = 2k$ where $k$ is some integer. + +Then: + +$$ n^3 - m^3 = (n - m)(n^2 + nm + m^2) \quad \text{ by factoring} $$ + +$$ \quad = 2k(n^2 + nm + m^2) \quad \text{ by substitution} $$ + +$$ \quad = 2[k(n^2 + nm + m^2)] $$ + +Let $t = k(n^2 + nm + m^2)$. + +Then $n^3 - m^3 = 2t$ where $t$ is an integer because products and sums of +integers is an integer. + +Therefore $n^3 - m^3$ is even by the definition of even integers. + +Q.E.D. + 29. For every integer $n$, if $n$ is prime then $(-1)^n = -1$. +This is false when $n = 2$. Let's prove our claim. + +**Claim:** + +There exists some integer $n$ such that $n$ is prime and $(-1)^n \neq -1$. + +**Proof:** + +Let $n = 2$. + +Then: + +$$ (-1)^n = (-1)^2 \quad \text{ by substitution} $$ + +$$ \quad = 1 $$ + +$$ 1 \neq -1 $$ + +Therefore since there is a prime number for $n$ such that $(-1)^n \neq -1$, the +given statement is false. + +Q.E.D. + 30. For every integer $m$, if $m > 2$ then $m^2 - 4$ is composite. +This is false. If $m = 3$, then $m^2 - 4 = 9 - 4 = 5$ which is not composite. +Let's prove our claim. + +**Claim:** + +There exists some integer $m$ such that $m > 2$ and $m^2 - 4$ is not composite. + +**Proof:** + +Let $m = 3$. + +Then: + +$$ m^2 - 4 = (3)^2 - 4 \quad \text{ by substitution} $$ + +$$ \quad = 9 - 4 $$ + +$$ \quad = 5 $$ + +$$ \quad = (1)(5) $$ + +Since $m^2 - 4$ cannot be written as the product of 2 factors where both factors +are greater than $1$, $m^2 - 4$ is not composite. + +Therefore since there is some integer $m$ such that $m > 2$ and $m^2 - 4$ is not +composite, this statement is false. + +Q.E.D. + 31. For every integer $n$, $n^2 - n + 11$ is a prime number. +This is false for when $n = 11$, let's formalize our claim. + +**Claim:** + +There exists some integer $n$ such that $n^2 - n + 11$ is not a prime number. + +**Proof:** + +Let $n = 11$. + +Then: + +$$ n^2 - n + 11 = (11)^2 - (11) + 11 \quad \text{ by substitution} $$ + +$$ \quad = 121 - 11 + 11 $$ + +$$ \quad = 121 $$ + +$$ \quad = (11)(11) $$ + +Therefore $n^2 - n + 11$ is not a prime number since it is divisible by a number +other than $1$ and itself for this given $n$. + +Thus there exists some integer $n$ such that $n^2 - n + 11$ is not a prime +number, and therefore the given statement is false. + +Q.E.D. + 32. For every integer $n$, $4(n^2 + n + 1) - 3n^2$ is a perfect square. -33.. Every positive integer can be expressed as a sum of three or fewer perfect -squares. +**Theorem:** + +Suppose $n$ is any integer. + +**Proof:** + +Then: + +$$ 4(n^2 + n + 1) - 3n^2 = 4n^2 + 4n + 4 - 3n^2 \quad \text{ by distribution} $$ + +$$ \quad = n^2 + 4n + 4 $$ + +$$ \quad = (n + 2)(n + 2) $$ + +$$ \quad = (n + 2)^2 $$ + +Let $t = n + 2$. + +Then $4(n^2 + n + 1) - 3n^2 = t^2$ where $t$ is an integer because the sum of +integers is an integer. + +Therefore $4(n^2 + n + 1) - 3n^2$ is a perfect square by the definition of +perfect squares. + +Q.E.D. + +33. Every positive integer can be expressed as a sum of three or fewer perfect + squares. + +This is false. + +**Claim:** + +There exists some positive integer $x$ such that $x$ cannot be expressed as the +sum of three or fewer perfect squares. + +**Proof:** + +Let $x = 7$. We check all sums of three nonnegative perfect squares +$a^2 + b^2 + c^2$, where $a, b, c \in \{0, 1, 2\}$ because $3^2 = 9 > 7$. + +Possible squares: $0^2 = 0$, $1^2 = 1$, $2^2 = 4$. + +Now we check all sums + +1. Using only $0$ and $1$: + +- $0 + 0 + 1 = 1$, $0 + 1 + 1 = 2$, $1 + 1 + 1 = 3$ + +All of these are too small and do not add up to $7$. + +2. Using a $4$ ($2^2$) with $0$ and $1$: + +- $4 + 0 + 0 = 4$, $4 + 0 + 1 = 5$, $4 + 1 + 1 = 6$, $4 + 4 + 0 = 8$ + +All of these do not equal $7$. + +No combination sums to $7$. + +Therefore, since all possible combinations from the given set of numbers that +could potentially sum to $7$ when each individual number is squared have been +exhausted, it can be concluded that $x = 7$ cannot be expressed as the sum of +three or fewer perfect squares. + +Therefore there exists at least one integer $x$ such that $x$ cannot be +expressed as a sum of three or fewer perfect squares, and this statement is +false. + +Q.E.D. 34. (Two integers are **consecutive** if, and only if, one is one more than the other.) Any product of four consecutive integers is one less than a perfect square. +**Theorem:** + +Suppose $n$ is any integer. + +**Proof:** + +Then: + +$$ (n)(n + 1)(n + 2)(n + 3) = (n(n + 3))(n + 1)(n + 2) $$ + +$$ \quad = (n^2 + 3n)(n^2 + 3n + 2)$$ + +$$ \quad = (n^2 + 3n)((n^2 + 3n) + 2)$$ + +Let $x = n^2 + 3n$. + +Then: + +$$ \quad = (x)((x) + 2) $$ + +$$ \quad = x^2 + 2x $$ + +$$ \quad = x^2 + 2x + 1 - 1 $$ + +$$ \quad = (x^2 + 2x + 1) - 1 $$ + +$$ \quad = (x + 1)(x + 1) - 1 $$ + +$$ \quad = (x + 1)^2 - 1 $$ + +Then remove the substitution: + +$$ \quad = ((n^2 + 3n) + 1)^2 - 1 $$ + +$$ \quad = (n^2 + 3n + 1)^2 - 1 $$ + +Since $n^2 + 3n + 1$ is an integer because the products and sum of integers is +an integer, this means that $(n^2 + 3n + 1)^2$ is a perfect square and +$(n^2 + 3n + 1)^2 - 1$ is one less than a perfect square. + +Therefore the product of any four consecutive integers is one less than a +perfect square. + +Q.E.D. + 35. If $m$ and $n$ are any positive integers and $mn$ is a perfect square, then $m$ and $n$ are perfect squares. +This is false. + +**Claim:** + +There is a positive integer $m$ and there is a positive integer $n$ such that +$mn$ is a perfect square and $m$ and $n$ are not perfect squares. + +**Proof:** + +Let $m = 2$ and $n = 8$. + +Then: + +$$ mn = (2)(8) \quad \text{ by substitution} $$ + +$$ \quad = 16 $$ + +$$ \quad = 4^2 $$ + +Then $mn$ is a perfect square, but $m$ and $n$ are not perfect squares. + +Therefore there exists some $m$ and there exists some $n$ such that $mn$ is a +perfect square and $m$ and $n$ are not perfect squares, proving the statement +false. + +Q.E.D. + 36. The difference of the squares of any two consecutive integers is odd. +**Theorem:** + +Suppose $n$ is any integer. + +**Proof:** + +Then: + +$$ (n + 1)^2 - n^2 = (n + 1)(n + 1) - n^2 \quad \text{ by substitution} $$ + +$$ \quad = n^2 + 2n + 1 - n^2 $$ + +$$ \quad = 2n + 1 $$ + +Therefore $(n + 1)^2 - n^2$ is odd by the definition of an odd integer. + +Q.E.D. + 37. For all nonnegative real numbers $a$ and $b$, $\sqrt{ab} = \sqrt{a}\sqrt{b}$. (Note that if $x$ is a nonnegative real number, then there is a unique nonnegative real number $y$, denoted $\sqrt{x}$, such that $y^2 = x$.) +**Theorem:** + +Suppose $a$ is any nonnegative real number and $b$ is any nonnegative real +number. + +**Proof:** + +Since $a \geq 0$ and $b \geq 0$, we know that $\sqrt{a}$ and $\sqrt{b}$ are +defined nonnegative real numbers such that: + +$$ (\sqrt{a})^2 = a $$ + +and + +$$ (\sqrt{b})^2 = b $$ + +Then: + +$$ (\sqrt{a}\sqrt{b})^2 = (\sqrt{a})^2(\sqrt{b})^2 = ab \quad \text{ by the product of powers property} $$ + +We then know that $\sqrt{a}\sqrt{b} \geq 0$ because both factors are +nonnegative. + +So $\sqrt{a}\sqrt{b}$ is a nonnegative real number whose square is $ab$. + +Therefore $\sqrt{ab} = \sqrt{a}\sqrt{b}$ by the definition of square root +(uniqueness of the nonnegative number whose square is $ab$). + +Q.E.D. + 38. For all nonnegative real numbers $a$ and $b$, $$ \sqrt{a + b} = \sqrt{a} + \sqrt{b} $$ +This is false. + +**Claim:** + +There is some nonnegative real number $a$ and some nonnegative real number $b$ +such that $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$. + +**Proof:** + +Let $a = 9$ and $b = 16$. + +Then: + +$$ \sqrt{a + b} = \sqrt{9 + 16} \quad \text{ by substitution} $$ + +$$ \quad = \sqrt{25} $$ + +$$ \quad = 5 $$ + +Then: + +$$ 5 \stackrel{?}{=} \sqrt{a} + \sqrt{b} $$ + +$$ 5 \stackrel{?}{=} \sqrt{9} + \sqrt{16} \quad \text{ by substitution} $$ + +$$ 5 \stackrel{?}{=} 3 + 4 $$ + +$$ 5 \neq 7 $$ + +Therefore for the given $a$ and $b$, we have shown that +$\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$, thus proving the statement false. + +Q.E.D. + 39. Suppose that integers $m$ and $n$ are perfect squares. Then $m + n + 2\sqrt{mn}$ is also a perfect square. Why? +**Theorem:** + +Suppose $m$ is any perfect square and $n$ is any perfect square. + +**Proof:** + +Since $m$ and $n$ are perfect squares, $m = k^2$ and $n = s^2$ for some integer +$k$ and some integer $s$. + +Then: + +$$ m + n + 2\sqrt{mn} = (k^2) + (s^2) + 2\sqrt{(k^2)(s^2)} \quad \text{ by substitution} $$ + +$$ \quad = k^2 + s^2 + 2\sqrt{k^2}\sqrt{s^2} \quad \text{ by the product of powers property} $$ + +$$ \quad = k^2 + s^2 + 2ks $$ + +$$ \quad = k^2 + 2ks + s^2 \quad \text{ by the commutative property} $$ + +$$ \quad = (k + s)(k + s) $$ + +$$ \quad = (k + s)^2 $$ + +Therefore $m + n + 2\sqrt{mn}$ is a perfect square by the definition of a +perfect square. + +Q.E.D. + 40. If $p$ is a prime number, must $2^p - 1$ also be prime? Prove or give a counterexample. +False. + +**Claim:** + +There is some prime number $p$ such that $2^p - 1$ is not prime. + +Let $p = 11$. + +Then: + +$$ 2^p - 1 = 2^{11} - 1 \quad \text{ by substitution} $$ + +$$ \quad = 2048 - 1 $$ + +$$ \quad = 2047 $$ + +$$ \quad = (23)(89) $$ + +Thus there is a case where $p$ is a prime number and $2^p - 1$ is not prime, and +therefore the given statement is false. + +Q.E.D. + 41. If $n$ is a nonnegative integer, must $2^{2n} + 1$ be prime? Prove or give a counterexample. + +False. + +**Claim:** + +There exists a nonnegative integer $n$ such that $2^{2n} + 1$ is not prime. + +**Proof:** + +Let $n = 5$. + +Then: + +$$ 2^{2n} + 1 = 2^{2 \cdot 5} + 1 \quad \text{ by substitution} $$ + +$$ \quad = 2^{10} + 1 $$ + +$$ \quad = 1024 + 1 $$ + +$$ \quad = 1025 $$ + +$$ \quad = (25)(41) $$ + +Thus there exists an nonnegative integer $n$ such that $2^{2n} + 1$ is not +prime, and therefore the given statement is false. + +Q.E.D. diff --git a/chapter_4/test_yourself.md b/chapter_4/test_yourself.md index 48a010f..7f96957 100644 --- a/chapter_4/test_yourself.md +++ b/chapter_4/test_yourself.md @@ -42,23 +42,43 @@ Page 204 1. The meaning of every variable used in a proof should be explained with ______. +The body of the proof. + 2. Proofs should be written in sentences that are ______ and ______. +complete; grammatically correct + 3. Every assertion in a proof should be supported by a ______. +reason + 4. The following are some useful "little words and phrases" that clarify the reasoning in a proof: ______, ______, ______, ______, and ______. +because; since; then; thus; so; hence; therefore; consequently; it follows that; +by substitution + 5. A new thought or fact that does not follow as an immediate consequence of the preceding statement can be introduced by writing ______, ______, ______, ______, or ______. +observe that; note that; recall that; but; now + 6. To introduce a new variable that is defined in terms of previous variables, use the word ______. +let + 7. Displaying equations and inequalities increases the ______ of a proof. +readability + 8. Some proof-writing mistakes are ______, ______, ______, ______, ______, ______, and ______. + +arguing from examples; using the same letter to mean two different things; +jumping to a conclusion; assuming what is to be proved; confusion between what +is known and what is still to be shown; use of _any_ when the correct word is +_some_; misuse of the word _if_