#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.*

F. This is a proof that $not A => B$ is a contradiction, which is not
sufficient to show $A => B$.

= 1.6

*1b.*

#proof[
  Consider $m = 1$, $n = -1$. Then $15(1) + 12(-1) = 3$.
]

*1c.*

Rewrite $2m + 4n = 7$ as $2(m + 2n) = 7$. Notice that $2(m + 2n) = 2k, exists k
in nonzero$ and therefore is an even number for any $m$, $n$. But 7 is not
even. So no choices of $m$ and $n$ can create a number equal to 7. So no $m$ or
$n$ exists.

*4a.*

#proof[
  Consider $x = 2$. Then $x^2 + x + 41 = 46$, which is not prime. So it's false.
]

*4b.*

#proof[
  For any $x$, choose $y = -1 dot x$. Then $x + (-x) = x(1 - 1) = 0x = 0$. This
  is also trivially true when considering our usual field of reals because the
  existence of $y$ such that $x + y = 0$ is an axiom.
]

*4c.*

Consider $x = 2$ and $y = 1$. Then $1 > 2$ is false.

*4d.*

Consider $a = 10$, $b = 2$, $c = 5$. Then $a | b c$ because $10 | 2 dot 5$ but
10 does not divide either 2 or 5.

*6a.*

#proof[
  Rewriting the inequality,
  $
    1 / n <= 1\
    1 <= n
  $
  This is true for all $n in NN$ by the definition of $NN$.
]

*6b.*

#proof[
  Rewriting,
  $
    1 / n < 0.13 \
    1 < 0.13n \
    n > 1 / 0.13
  $
  So such an $M$ is $1/0.13$, because all naturals greater than $1/0.13$ are
  greater than $1/0.13$.
]

*6e.*

#proof[
  $
    forall n in NN, n + 1 > n
  $
]

*6f.*

$
  forall k in ZZ, m = -k, k + m = 0 \
  forall n in NN, 0 <= 0 < n
$

*6i.*

First consider $K = 10$, and therefore

$
  forall r > 10, r^2 > 100
$
Then note that our inequality is equivalent to $r^2 > 100$, which is true for
all $r$.
$
  1 / (r^2) < 0.01 \
  r^2 > 100 \
$
So a $K$ exists, namely $K = 10$.

*6k.*

Consider $M = 51$. Then $forall r > 51, 2r > 102$ and $2r > 100$ is always
true.

$
  1 / (2r) < 0.01 \
  2r > 100
$

So such an $M$ exists, namely, $M = 51$.

= 2.1

*4a.*

False.

*4b.*

True.

*4c.*

False.

*4d.*

True.

*4e.*

True.

*4f.*

False.

*4g.*

True.

*4h.*

False.

*4i.*

False.

*4j*.

True.

*5a.*

True.

*5b.*

True.

*5c.*

True.

*5d.*

True.

*5e.*

False.

*5f.*

True.

*5g.*

True.

*5h.*

True.

*5i.*

False.

*5j.*

True.

*5k.*

True.

*5L.*

True.

*6a.*

$
  A = {1,2}, B = {1,2,3}, C = {1,2,4}
$

*6b.*

$
  A = B = C
$

A particular example might be $A = {1}, B = {1}, C = {1}$.

*6c.*

$
  A = {1}, B = {1,2}, C = {3}
$

*6d.*

$
  A = {1,2}, B = {1,2,3}, C = {5,6,7}
$

*8.*

Theorem 2.1.1: $forall A, B, C, A subset.eq B and B subset.eq C => A subset.eq C$.

#proof[
  $
    forall a in A, a in B \
    forall a in A, exists b in B, a = b \
    forall b in B, b in C \
    forall a in A, a in C => A subset.eq C
  $
]

*11a.*

#proof[
  $
    {x in RR : 3 / 4 x - 2 > 10} \
    = {x in RR : x > 16 } \
    = (16, infinity)
  $
]

*11b.*

#proof[
  $
    {x in RR : |x - 4| = 2|x| - 2} \
    = {x in RR : |x - 4| = 2|x| - 2} \
  $

  Now we consider various cases. Consider $x >= 4$. Then
  $
    {x in RR : x - 4 = 2x - 2} \
    = {x in RR : x - 4 = 2x - 2} \
    = {x in RR : x = -2} \
  $
  But in this case $x >= 4$. Since there is no $x >= 4$ such that $x = -2$,
  this is actually $emptyset$.

  Now consider $0 <= x < 4$.
  $
    {x in RR : 4 - x = 2x - 2} \
    = {x in RR : x = 2} \
    = {2}
  $


  Now consider $x < 0$. Then
  $
    {x in RR : 4 - x = -2x - 2} \
    {x in RR : x = -6}
  $
  So the set contains $6$ when $x < 0$.

  And since these inequalities span all $x in RR$, the only members of
  the set are ${2, -6}$.
]

*11c.*

#proof[
  $
    {x in RR : 2|x+3| + x = 0}
  $
  For $x >= -3$, we have
  $
    {x in RR : 2(x + 3) + x = 0} \
    = {x in RR : x = -2} \
    = {-2}
  $
  For $x < -3$, we have
  $
    {x in RR : 2(3 - x) + x = 0} \
    = {x in RR : x = 6 } \
    = {6}
  $
  And since these inequalities partition $RR$ the original set is ${-2, 6}$.
]

*11d.*

$
  {x in RR : |x| = 6 - |2x|}
$

Consider $x >= 0$. Then
$
  {x in RR : x = 6 - 2x} \
  = {x in RR : x = 2}
$
Consider $x < 0$. Then
$
  {x in RR : -x = 6 + 2x} \
  = {x in RR : x = -2}
$
So
$
  {x in RR : |x| = 6 - |2x|} = {-2, 2}
$

*11e.*

$
  {x in RR : |x + 3| <= -4x - 2}
$

Consider $x >= -3$. Then
$
  {x in RR : x <= -1} \
  = (-infinity, -1]
$

Now consider $x < -3$ Then
$
  {x in RR : 3 - x <= -4x - 2} \
  = {x in RR : x <= -5 / 3} \
  = (-infinity, -5 / 3]
$
But $(-infinity, -5/3) union (-infinity, 1] = (-infinity, 1]$ so that is our
answer.

*11f.*