#import "@youwen/zen:0.1.0": *
#import "@preview/mitex:0.2.5": *

#show: zen.with(
  title: "Homework 3",
  author: "Youwen Wu",
)

#set heading(numbering: none)

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

#let nonzero = $ZZ_(!=0)$

Problems:

1.5: $hash$ 3cdefgh, 4de, 6de, 7ab, 9, 10, 11, 12a

1.6: $hash$ 1bd, 4abcd, 6abefik

2.1: $hash$ 4, 5, 6abcd, 8, 11ab, 14ab, 15abcd

= 1.5

*3c.*

If $x^2$ is not divisible by 4, then $x$ is odd.

#proof[
  Suppose $x$ is not odd. We seek to show that $x^2$ is divisible by $4$.

  Then $x$ is even and $exists k in ZZ_(!=0), x = 2k$. Thus
  $x^2 = 4k^2$, which implies $4 | x^2$. So $x$ is not odd implies $x^2$ is
  divisible by 4, and therefore the contrapositive (which is our original
  statement) is also true.
]

*3d.*

If $x y$ is even, then either $x$ or $y$ is even.

#proof[
  Suppose that $not (x "or" y" is even")$. In other words, $x$ and $y$ are both
  odd. We seek to show that $x y$ is odd.

  Then $exists j,k in ZZ_(n!=0), x = 2j + 1, y = 2k + 1$. So $x y = 4 j k + 2j +
  2k + 1 = 2 (2 j k + j + k) + 1$ where $2 j k + j + k$ is an integer so $x y$
  is odd. This is the contrapositive, so the original statement is also true.
]

*3e.*

If $x + y$ is even, then $x$ and $y$ have the same parity.

#proof[
  Suppose $x$ and $y$ had different parities. We seek to show that $x + y$ is
  odd.

  Without loss of generality, let us inspect the case where $x$ is odd. Then
  $y$ must be even and
  $
    exists j,k in ZZ_(!=0), x = 2j + 1, y = 2k \
    x + y = 2j + 1 + 2k = 2(j + k) + 1 = 2n + 1, n in ZZ_(n!=0)
  $
  Repeat the same reasoning for when $x$ is even and $y$ is odd. We've shown
  that $x$ and $y$ having different parities implies that $x + y$ is odd.
  Therefore the contrapositive is also true.
]

*3f.*

If $x y$ is odd, then both $x$ and $y$ are odd.

#proof[
  Suppose that both $x$ and $y$ are odd was not true, in other words $x$ or $y$
  are even (here or is the logical $or$). We seek to show that this implies $x y
$ is even. Then we have two cases: either $x$ or
  $y$, but not both, are even, and the case where $x$ and $y$ are both even.

  We look at the first case, without loss of generality, assume $x$ is even and
  $y$ is odd. Then

  $
    exists j,k in ZZ_(!= 0), x = 2j, y = 2k + 1 \
    x y = 2j(2k + 1) = 4 j k + 2j = 2(2 j k + j) = 2n, n in ZZ_(!=0)
  $

  So $x$ and $y$ having different parities implies $x y$ is even. The argument holds identically when instead $y$ is even and $x$ is odd. Now we turn our attention to the case when $x$ and $y$ are both even. Then

  $
    exists j,k in ZZ_(!=0), x = 2j, y = 2k \
    x y = 2j dot 2k = 4 j k = 2 (2 j k) = 2n, n in ZZ_(!=0)
  $

  So this also implies that $x y$ is even. Therefore $x$ or $y$ being even indeed
  implies $x y$ is also even, and the contrapositive is also true.
]

*3g.*

If 8 does not divide $x^2 - 1$, then $x$ is even.

#proof[
  Suppose that $x$ is odd. We seek to show that 8 does in fact divide $x^2 - 1$.

  $
    exists k in nonzero, x = 2k + 1 \
    x^2 - 1 = 4k^2 + 4k = 4(k^2 + k)
  $
  Now we need to see if it's divisible by 8. First consider the unique case $k = 0$, where $x^2 - 1 = 0$. Clearly $8 | 0$ so 8 does divide $x^2-1$.

  Now we consider all other values. Notice that for all other possible values
  of $k$, $k^2 + k$ is greater than 1. We see this by noting that $k^2 + k$ is
  a quadratic with its absolute minima at $k = 1/2$, therefore we can check the
  two non-zero integers closest to this value. For $k = 1$, $k^2 + k = 2$.
  Since we already checked $k=-1$, let's check $k=-2$, which gives $k^2 + k =
  2$. For all values of $k$ greater than 1 or less than $-2$, $k^2 + k$ must be
  greater than 2 (because it's a quadratic).

  Therefore $4(k^2 + k)$ can be written as $8n$ for some integer $n$, so 8
  indeed divides $x^2 - 1$ for all possible $k$, and the contrapositive is also
  true.
]

*3h.*

If $x$ does not divide $y z$, then $x$ does not divide $z$.

#proof[
  Assume $x$ does divide $z$. We seek to show that $x$ does divide $y z$. If $x
  | z$ then $exists k in nonzero, k x = z$. So $y z$ can be written as $y k x$.
  But this shows $exists j in nonzero, j x = y z$, namely $j = y k$, which
  means $x | y z$, and the contrapositive is also true.
]

*4d.*

If $(x+1)(x-1) < 0$, then $x < 1$.

#proof[
  Suppose that $x >= 1$. We seek to show that $(x+1)(x-1) >= 0$. Expanding out
  factors,
  $
    (x+1)(x-1) = x^2 - 1>= 0
  $
  This quadratic is zero at exactly $x = 1$ and positive for all $x > 1$. So
  it's true for all possible values of $x$. Therefore our original statement is
  also true.
]

*4e.*

If $x(x-4) > -3$, then $x < 1$ or $x > 3$.

#proof[
  Suppose that $x >= 1$ and $x <= 3$. We seek to show $x(x-4) <= -3$. Expanding
  out factors,

  $
    x(x-4) = x^2 - 4x > -3
  $

  This quadratic has its stationary point at $x = 2$. Let's check its value at $x
= 1$ and $x = 3$.

  At $x = 1$, $x(x-4) = -3$. So for all values greater than 1 until $x = 2$, $x^2 - 4x$ is less than $-3$. Our inequality is satisfied.

  At $x = 3$, $x(x-4) = -3$ again. So for all values less than 3 until $x = 2$,
  $x^2 - 4x$ is less than $-3$. Our inequality is satisfied for both $x >=1$
  and $x <=3$, so $x^2 -4x > -3$ is true for all possible $x$ and the
  contrapositive is also true.
]

*6d.*

If $a - b$ is odd, then $a + b$ is odd.

#proof[
  Suppose, seeking a contradiction, that if $a - b$ is even, then $a + b$ is odd.

  Then $exists k in nonzero,  a - b = 2k$. Which means we can write $a + b = 2k
  + 2b$. But we can factor this as $2(k + b)$ and so $a + b = 2n, n in
  nonzero$, implying it is even. However we assumed that $a + b$ should be odd,
  a contradiction. Therefore $a - b$ must be odd.
]

*6e.*

If $a < b$ and $a b < 3$, then $a = 1$.

#proof[
  Suppose, seeking a contradiction, that $a >= b$ and $a b >= 3$ implies $a =
  1$. Consider the specific cases $a = b$, $a b = 3$. Then $a = 3/a$, and $a =
  plus.minus sqrt(3)$. However we assumed $a = 1$ is implied, a contradiction.
  Therefore we must have $a < b$ and $a b < 3$.
]

*7a.*

$a c$ divides $b$ and $b$ divides $b + 3$ if and only if $a = 2$ and $b = 3$.

We first show the result left to right, namely, $a c | b c => a | b$.

$
  exists k in nonzero, k a c = b c \
  k a = b
$

which is the definition of $a | b$.

Now we show the right to left direction, namely, $a | b => a c | b c$.

$
  exists k in nonzero, k a = b \
  k a dot c = b dot c
$

So $a c | b c$ by the definition of divisibility. Therefore the biconditional
is true, as we have shown both directions.

*7b.*

The right to left direction is very easy in this case. By directly plugging in $a = 2$, $b = 3$, we see that $2 + 1 | 3$ and $3 | 3 + 3$. To show the left to right case,

*9.*

#proof[
  Suppose that instead $n/(n+1) <= n/(n+2)$. We can be assured the following
  operations do not flip the inequality as $n$ cannot be negative.
  $
    n(n+2) <= n(n+1) \
    n + 2 <= n + 1 \
    2 <= 1
  $
  So $n/(n+1) > n/(n+2)$.
]

*10.*

#proof[
  Suppose that $sqrt(5)$ was rational. That is, $sqrt(5) = p/q$ for nonzero
  integers $p$ and $q$. Additionally, assume that $p/q$ is in its most reduced
  form, that is, $p$ and $q$ share no common factors besides 1. Then

  $
    p^2 = 5q \
  $
  implies that
  $ 5 | p^2 $
  We need to show that $5 | p^2 => 5 | p$.

  By the fundamental theorem of arithmetic, $p^2$ has 5 as one of its unique
  prime factors. If 5 was not a factor of $p$, then $p^2 = p dot p$ would not
  have 5 in its factors either. So 5 is a factor of $p$ and thus $5 | p$. Note
  that 5 appears at least twice amongst the prime factors of $p^2$. Then $5q$
  should also have at least two 5s in its prime factorization. Then $q$ has at
  least one 5 in its prime factorization. However we assumed that $p$ and $q$
  share no common factors besides 1, so this is a contradiction. Therefore
  $sqrt(5)$ is not rational.
]

*11.*

#proof[
  We say that two numbers $x$ and $y$ are within $1/2$ unit from one another if
  $|x - y| < 1/2$. Consider the distance between $z$, and $y$, if it is within
  $1/2$, then we are done. Otherwise $z - y >= 1/2$.

  We know that
  $ (1 - z) + x + (y-x) + (z-y) = 1 $
  because this is the length of all the line segments partitioned by $x,y,z$,
  which is the interval 1. If $(z - y) >= 1/2$, then everything else must be
  less than $1/2$. So the maximum value of $y-x$, the distance between $x$ and
  $y$, is less than $1/2$. Therefore either $y$ and $z$ are within $1/2$ unit
  of each other or $y$ and $x$ are.
]

*12a.*