451 lines
8 KiB
Text
451 lines
8 KiB
Text
#import "@youwen/zen:0.1.0": *
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#show: zen.with(
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title: "Homework 2",
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author: "Youwen Wu",
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)
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#set heading(
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numbering: (
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num => {
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return "1." + str(num)
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}
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),
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)
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#set par(first-line-indent: 0pt, spacing: 1em)
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Problems:
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1.3: \#1fgjkLmp, 2fgjkLmp, 6, 8, 9, 10, 13
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1.4: \#5cdghi, 6bd, 7defghijk, 8, 9a (use any proof method, not necessarily proof by working backward), 11bcd
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*1.3*
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1f.
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Let $H(p)$ be true if a person is honest and false otherwise.
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$ ((forall x)H(x)) or ((forall x)(not H(x))) $
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1g.
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Again let $H(p)$ be true if a person is honest and false otherwise.
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$ (exists x)(H(x)) and ((exists y) not H(y)) $
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1j.
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$ (forall x)(exists y)(x > y) $
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1k.
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$ (forall x)(exists y)(not (x > y)) $
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1L.
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$ (forall x in ZZ)(forall y in ZZ)(y > x)(exists z in RR)(x < z < y) $
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1m.
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$ (exists x in ZZ^+)(forall y in ZZ^+)(x <= y) $
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1p.
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$ (forall x)(x > 0)(exists! y)(2^y = x) $
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2f.
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Let $H(p)$ be true if a person is honest and false otherwise.
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$ (exists x)(H(x) and ((exists y) not H(y)) $
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In English: Some people are honest and some people are not honest.
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2g.
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Let $H(p)$ be true if a person is honest and false otherwise.
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$ ((forall x)H(x)) or ((forall x)(not H(x))) $
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In English: all people are honest or no one is honest.
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2j.
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$ (exists x)(forall y) (x <= y) $
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There is an integer such that it is smaller than every other integer y.
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2k.
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$ (exists y)(forall x)(y >= x) $
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There is an integer that is greater than all other integers.
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2L.
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$ (exists x in ZZ)(exists y in ZZ)(x < y)(forall z)(not (x < z < y)) $
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There exists an integer and a larger integer such that there is no real number
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between them.
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2m.
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$ (forall x in ZZ^+)(exists y in ZZ^+)(y < x) $
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Any positive integer has an integer less than itself.
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2p.
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$ (exists x)(x > 0)(forall y) not (2^y = x) $
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There is a positive real number such that there is no real number $y$ that
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satisfies $2^y = x$.
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6a.
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$T$, $U$, $V$.
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6b.
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$T$ only.
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6c.
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$T$, $U$, $V$.
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6d.
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$T$ only.
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8a. False. Consider $x = -10$.
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8b. True.
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8c. True.
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8d. True.
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8e. False.
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8f. True.
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8g. True.
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8h. True.
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8i. True ($x=0)$.
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8j. False.
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8k. False.
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8L. True.
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9a. All natural numbers are at least one.
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9b. There is exactly one and only one real number that is both greater than or
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equal to 0 and less than or equal to 0.
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9c. All natural numbers are prime, and if they are not two then they are odd.
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9d. There is one and only one real number that is $e$ raised to the power of
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one.
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9e. There is no real number such that its square is negative.
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9f. There is a unique real number that is the square root of 0.
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9g. Any odd natural number is also odd when squared.
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10a. True.
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10b. False.
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10c. False.
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10d. False.
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10e. True ($x=0$).
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10f. False.
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10g. True.
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10h. False.
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10i. False.
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10j. True.
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10k. False.
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\13. (b), (c)
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*1.4*
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5c. If $x$ and $y$ are even, then $x y$ is divisible by 4.
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#proof[
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$
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exists j,k in ZZ, x = 2j, y = 2k \
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x y = 4 j k
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$
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$x y$ has $4$ in its factors and so $4 | x y $.
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]
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5d.
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#proof[
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$
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exists j,k in ZZ, x = 2j, y = 2k \
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3x - 5y = 6j - 10k = 2(3j - 5k) = 2n, n in ZZ
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$
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]
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5e.
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#proof[
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$
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exists j,k in ZZ, x = (2j + 1), y = (2k + 1) \
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x + y = 2j + 1 + 2k + 1 = 2(j + k + 1) = 2n, n in ZZ
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$
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]
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5f.
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#proof[
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$
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exists j,k in ZZ, x = (2j + 1), y = (2k + 1) \
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3x - 5y = 6j + 3 - 10k - 5 = 2(3j-5k-1) = 2n, n in ZZ
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$
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]
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5g.
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#proof[
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$
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exists j,k in ZZ, x = (2j + 1), y = (2k + 1) \
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x y = (2j + 1)(2k + 1) &= 4 j k + 2j + 2k + 1 \
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= 2(2 j k + j + k) + 1 &= 2n + 1, n in ZZ
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$
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]
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5i.
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#proof[
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If one of $x$, $y$, and $z$ are odd, then let there be $i,j,k in ZZ$, such
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that for the two even terms are represented as $2i$ and $2j$, and the single
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odd term represented as $2k + 1$. Then clearly
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$
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x + y + z = 2i + 2j + 2k + 1 = 2(i + j + k) + 1 = 2n + 1, n in ZZ
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$
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]
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6b.
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#proof[
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Start by rewriting
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$
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|b - a| = |-(a - b)|
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$
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But by the definition of the absolute value,
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$ |-(a - b)| = |a-b| $
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So indeed
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$ |a - b| = |b - a| $
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]
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6d.
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#proof[
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In the case where $a$ and $b$ are positive or zero the two sides are equal.
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$
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|a + b| <= |a| + |b|
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$
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Otherwise, consider if $a$ is positive, $b$ is negative. We replace all
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occurrences of $b$ with $-|b|$.
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$
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|a - |b|| <= |a| + |-|b|| \
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|a - |b|| <= |a| + |b|
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$
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Clearly for any $b$ the left side is strictly lower than the right. Repeat
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this exact reasoning for when $a$ is negative.
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]
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7d.
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#proof[
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Consider two cases, $a$ is even and $a$ is odd. First assume $a$ is even:
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$
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exists k in ZZ, a = 2k \
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2k(2k+1) = 4k^2 + 2k = 2(k^2 + 1) = 2n, n in ZZ
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$
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Then assume $a$ is odd:
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$
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exists k in ZZ, a = 2k + 1 \
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2k(2k+1+1) = 4k^2 + 4k = 2(k^2 + 2) = 2n, n in ZZ
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$
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]
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7e.
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#proof[
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$1 | a$ if and only if we can find some $k in ZZ$ such that $1k = a$. We can
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always find such a $k$, which is $a$, because $1 dot a = a$. Therefore $1 | a$ is always true.
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]
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7f.
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#proof[
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$a | a$ if and only if we can find some $k in ZZ$ such that $a k = a$. Such a
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$k$ is $1$, because $a dot 1 = a$. Therefore $a | a$.
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]
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7g.
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#proof[
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If $a | b$, then there is a $k in ZZ$ such that $a k = b$. So $a = b/k$. If
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$k = 1$, then $b/k = b$. Otherwise for $k > 0$ or $k < 0$ $b/k$ is less than
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$b$. So $b/k = a <= b$.
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]
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7h.
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#proof[
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Since $a | b$, we know $exists k in ZZ, k a = b$. We also know
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that
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$ (exists j in ZZ) (j a = b c) => (a | b c) $
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We can find $j$:
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$
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j a = b c \
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j cancel(a) = k cancel(a) c \
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j = k c
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$
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Therefore $a | b c$.
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]
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7i.
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#proof[
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Suppose that $a = b = 1$ was not the case. Then either $a$ or $b$ must be
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greater than 1, or both $a$ and $b$ are greater than 1.
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If $a > 1$ and $b = 1$, then $a b = a$. But is not 1, and $a b = 1$, so there
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is a contradiction.
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Repeat the same argument for $b >1$, $a = 1$.
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In the case that both $a > 1$ and $b > 1$, $a b > 1$. Again, $a b = 1$ so
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this cannot be true.
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There are no cases where $a$ and $b$ can be anything other than 1,
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so $a = b = 1$.
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]
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7j.
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#proof[
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If $a | b$ then $exists k in ZZ, k a = b$. If $b | a$ then $exists j in ZZ, j
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b = a$.
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$
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k j cancel(b) = cancel(b) \
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k j = 1
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$
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By the fact proven above in Problem 7i, this implies $k = j = 1$.
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So $1 a = b => a = b$.
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]
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7k.
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#proof[
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$
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a | b => exists k in ZZ, k a = b \
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c | d => exists j in ZZ, j c = d \
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$
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Multiply the equations together:
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$ k j a c = b d $
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If $exists n in ZZ, n a c = b d$, then $a c | b d$. We have such an $n = k
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j$. Therefore $a c$ indeed divides $b d$.
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]
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8a.
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#proof[
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Consider the case where $n$ is even. Then $exists k in ZZ, n = 2k$.
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$
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n^2 + n + 3 &= (2k)^2 + 2k + 3 \
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= 4k^2 + 2k + 3 &= 2(2k^2 + k + 1) + 1 = 2m + 1, m in ZZ
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$
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Consider the case where $n$ is odd. Then $exists k in ZZ, n = 2k + 1$.
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$
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n^2 + n + 3 &= (2k+1)^2 + 2k+1 + 3 \
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= 4k^2 + 6k + 5 &= 2(2k^2 + 3k + 2) + 1 = 2m + 1, m in ZZ
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$
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So in both cases $n^2 + n + 3$ is odd.
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]
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8b.
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#proof[
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First we rewrite $n^2 + n + 3$.
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$
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n^2 + n + 3 = n(n + 1) + 3
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$
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By the proof in 7(d), we know that $forall n, n(n+1)$ is even. Then by 5(h)
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we can take $x := n(n+1)$ and $y := 3$. Since $x$ is even, $y$ is odd, $x +
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y$ is odd. So $n^2 + n + 3$ is odd.
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]
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9a.
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#proof[
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In the case of $x=0$ and $y=0$ we trivially have $0 >= 0$. Otherwise,
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$
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&(x+y) / 2 >= sqrt(x y) \
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&<=> x+y >= 2sqrt(x y) \
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&<=> (x+y)^2 >= 4x y \
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&<=> x^2 + 2 x y + y^2 >= 4 x y \
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&<=> x^2 - 2 x y + y^2 >= 0 \
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&<=> (x - y)^2 >= 0
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$
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We know that $forall x in RR, x^2 >= 0$, so this statement is always true, and thus the original is always true as well.
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]
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11b.
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Grade: C.
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The assertions that $exists q, b = a q$ and $exists q, c = a q$ are essentially
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correct but these $q$ are not the same. This can be corrected fairly
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straightforwardly by replacing one of the $q$ with another variable serving the
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same purpose, then proceeding in a similar fashion.
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11c.
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Grade: A.
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11d.
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Grade: F.
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This only shows $m "is odd" => m^2 "is odd"$, when the claim is the other way around. The converse is not automatically true.
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