#import "@youwen/zen:0.1.0": *

#show: zen.with(
  title: "Homework 2",
  author: "Youwen Wu",
)

#set heading(
  numbering: (
    num => {
      return "1." + str(num)
    }
  ),
)

#set par(first-line-indent: 0pt, spacing: 1em)

Problems:

1.3: \#1fgjkLmp, 2fgjkLmp, 6, 8, 9, 10, 13

1.4: \#5cdghi, 6bd, 7defghijk, 8, 9a (use any proof method, not necessarily proof by working backward), 11bcd

*1.3*

1f.

Let $H(p)$ be true if a person is honest and false otherwise.

$ ((forall x)H(x)) or ((forall x)(not H(x))) $

1g.

Again let $H(p)$ be true if a person is honest and false otherwise.

$ (exists x)(exists y)(H(x) and not H(y)) $

1j.

$ (forall x)(exists y)(x > y) $

1k.

$ (exists.not x)(forall y)(x > y) $

1L.

$ (x in ZZ)(y in ZZ)(y > x)(exists z in RR)(x < z < y) $

1m.

$ (exists x in ZZ^+)(exists.not y in ZZ^+)(y < x) $

1p.

$ (forall x)(x > 0)(exists y)(2^y = x) $

2f.

Let $H(p)$ be true if a person is honest and false otherwise.

$ (exists x)(exists y)(H(x) and not H(y)) $

In English: Some people are honest and some people are not honest.

2g.

Let $H(p)$ be true if a person is honest and false otherwise.

$ ((forall x)H(x)) or ((forall x)(not H(x))) $

In English: all people are honest or no one is honest.

2j.


$ (exists x)(forall y) not (x > y) $

2k.

$ (forall x)(exists y)(y > x) $

2L.

$ (exists x in ZZ)(exists y in ZZ)(forall z)((z > x) and (z > y)) $

2m.

$ (forall x in ZZ^+)(exists y in ZZ^+)(y < x) $

2p.

$ (exists x)(x > 0)(forall y) not (2^y = x) $

6a.

$T$, $U$, $V$.

6b.

$T$ only.

6c.

$T$, $U$, $V$.

6d.

$T$ only.

8a. False. Consider $x = -10$.

8b. True.

8c. True.

8d. True.

8e. False.

8f. True.

8g. True.

8h. True.

8i. True ($x=0)$.

8j. False.

8k. False.

8L. True.

9a. All natural numbers are at least one.

9b. There is exactly one and only one real number that is both greater than or
equal to 0 and less than or equal to 0.

9c. All natural numbers are prime, and if they are not two then they are odd.

9d. There is one and only one real number that is $e$ raised to the power of
one.

9e. There is no real number such that its square is negative.

9f. There is a unique real number that is the square root of 0.

9g. Any odd natural number is also odd when squared.

10a. True.

10b. False.

10c. False.

10d. False.

10e. True ($x=0$).

10f. False.

10g. True.

10h. False.

10i. False.

10j. True.

10k. False.

\13. (b), (c)

*1.4*

5c. If $x$ and $y$ are even, then $x y$ is divisible by 4.

#proof[
  $
    exists j,k in ZZ, x = 2j, y = 2k \
    x y = 4 j k
  $

  Clearly $x y$ has $4$ in its factors and so $x y | 4$.
]

5d.

#proof[
  $
    exists j,k in ZZ, x = 2j, y = 2k \
    3x - 5y = 6j - 10k = 2(3j - 5k) = 2n, n in ZZ
  $
]

5e.

#proof[
  $
    exists j,k in ZZ, x = (2j + 1), y = (2k + 1) \
    x + y = 2j + 1 + 2k + 1 = 2(j + k + 1) = 2n, n in ZZ
  $
]

5f.

#proof[
  $
    exists j,k in ZZ, x = (2j + 1), y = (2k + 1) \
    3x - 5y = 6j + 3 - 10k - 5 = 2(3j-5k-1) = 2n, n in ZZ
  $
]

5g.

#proof[
  $
    exists j,k in ZZ, x = (2j + 1), y = (2k + 1) \
    x y = (2j + 1)(2k + 1) &= 4 j k + 2j + 2k + 1 \
    = 2(2 j k + j + k) + 1 &= 2n + 1, n in ZZ
  $
]

5i.

#proof[
  If one of $x$, $y$, and $z$ are odd, then let there be $i,j,k in ZZ$, such
  that for the two even terms are represented as $2i$ and $2j$, and the single
  odd term represented as $2k + 1$. Then clearly
  $
    x + y + z = 2i + 2j + 2k + 1 = 2(i + j + k) + 1 = 2n + 1, n in ZZ
  $
]

6b.

#proof[
  Start by rewriting
  $
    |b - a| = |-(a - b)|
  $

  But by the definition of the absolute value,
  $ |-(a - b)| = |a-b| $
  So indeed
  $ |a - b| = |b - a| $
]

6d.

#proof[
  In the case where $a$ and $b$ are positive or zero the two sides are equal.
  $
    |a + b| <= |a| + |b|
  $
  Otherwise, consider if $a$ is positive, $b$ is negative. We replace all
  occurrences of $b$ with $-|b|$.

  $
    |a - |b|| <= |a| + |-|b|| \
    |a - |b|| <= |a| + |b|
  $

  Clearly for any $b$ the left side is strictly lower than the right. Repeat
  this exact for $a$ is negative.
]

7d.

#proof[
  Consider two cases, $a$ is even and $a$ is odd. First assume $a$ is even:
  $
    exists k in ZZ, a = 2k \
    2k(2k+1) = 4k^2 + 2k = 2(k^2 + 1) = 2n, n in ZZ
  $
  Then assume $a$ is odd:
  $
    exists k in ZZ, a = 2k + 1 \
    2k(2k+1+1) = 4k^2 + 4k = 2(k^2 + 2) = 2n, n in ZZ
  $
]

7e.

#proof[
  If $1 | a$, then we can find some $k in ZZ$ such that $1k = a$. Such a $k$ is
  $a$, because $1 dot a = a$. Therefore $1 | a$.
]

7f.

#proof[
  If $a | a$, we can find some $k in ZZ$ such that $a k = a$. Such a $k$ is $1$,
  because $a dot 1 = a$. Therefore $a | a$.
]

7g.

#proof[
  If $a | b$, then there is a $k in ZZ$ such that $a k = b$. So $a = b/k$. If
  $k = 1$, then $b/k = b$. Otherwise for $k > 0$ or $k < 0$ $b/k$ is less than
  $b$. So $b/k = a <= b$.
]

7h.

#proof[
  Since $a | b$, we know $exists k in ZZ, k a = b$. We also know
  that
  $ (exists j in ZZ) (j a = b c) => (a | b c) $

  We can find $j$:
  $
    j a = b c \
    j cancel(a) = k cancel(a) c \
    j = k c
  $

  Therefore $a | b c$.
]

7i.

#proof[
  Suppose that $a = b = 1$ was not the case. Then either $a$ or $b$ must be
  greater than 1, or both $a$ and $b$ are greater than 1.

  If $a > 1$ and $b = 1$, then $a b = a$. But is not 1, and $a b = 1$, so there
  is a contradiction.

  Repeat the same argument for $b >1$, $a = 1$.

  In the case that both $a > 1$ and $b > 1$, $a b > 1$. Again, $a b = 1$ so
  this cannot be true.

  There are no cases where $a$ and $b$ can be anything other than 1,
  so $a = b = 1$.
]

7j.

#proof[
  If $a | b$ then $exists k in ZZ, k a = b$. If $b | a$ then $exists j in ZZ, j
  b = a$.

  $
    k j cancel(b) = cancel(b) \
    k j = 1
  $

  By the fact proven above in Problem 7i, this implies $k = j = 1$.

  So $1 a = b => a = b$.
]

7k.

#proof[
  $
    a | b => exists k in ZZ, k a = b \
    c | d => exists j in ZZ, j c = d \
  $

  Multiply the equations together:

  $ k j a c = b d $

  If $exists n in ZZ, n a c = b d$, then $a c | b d$. We see that $n = k j$.
  Therefore $a c$ indeed divides $b d$.
]

8a.

#proof[
  Consider the case where $n$ is even. Then $exists k in ZZ, n = 2k$.

  $
    n^2 + n + 3 &= (2k)^2 + 2k + 3 \
    = 4k^2 + 2k + 3 &= 2(2k^2 + k + 1) + 1 = 2m + 1, m in ZZ
  $

  Consider the case where $n$ is odd. Then $exists k in ZZ, n = 2k + 1$.

  $
    n^2 + n + 3 &= (2k+1)^2 + 2k+1 + 3 \
    = 4k^2 + 6k + 5 &= 2(2k^2 + 3k + 2) + 1 = 2m + 1, m in ZZ
  $

  So in both cases $n^2 + n + 3$ is odd.
]

8b.

#proof[
  First we rewrite $n^2 + n + 3$.

  $
    n^2 + n + 3 = n(n + 1) + 3
  $

  By 7(d), we know that $n(n+1)$ is even $forall n$. Then by 5(h) we can take
  $x := n(n+1)$ and $y := 3$. Since $x$ is even, $y$ is odd, $x + y$ is odd. So
  $n^2 + n + 3$ is odd.
]

9a.

#proof[
  In the case of $x=0$ and $y=0$ we trivially have $0 >= 0$. Otherwise,

  $
    (x+y) / 2 &>= sqrt(x y) \
    x+y &>= 2sqrt(x y) \
    (x+y)^2 &>= 4x y \
    x^2 + 2 x y + y^2 &>= 4 x y \
    x^2 - 2 x y + y^2 &>= 0
  $

  THIS IS WRONG FIX IT!!!

  You can think of $x^2 - 2 x y + y^2$ as quadratic with $y$ as a constant. For
  all possible values of $y$ the equation is nonnegative (since the absolute
  minimum occurs at the vertex). Therefore the inequality holds true.
]

11b.

Grade: C.

The assertions that $exists q, b = a q$ and $exists q, c = a q$ are essentially
correct but these $q$ are not the same. This can be corrected fairly
straightforwardly by replacing one of the $q$ with another variable serving the
same purpose, then proceeding.

11c.

Grade: A.

11d.

Grade: F.

This only shows $m "is odd" => m^2 "is odd"$, when the claim is the other way around. The converse is not automatically true.