#let m(math) = align(center)[$#math$]
#let pgbreakmsg = align(center, text(blue, weight: "black", size: 1.5em)[See Next Page\ ↓])

#let solve(work, solution) = align(
  center,
)[
  #let solution = align(center, block(
    inset: 5pt,
    stroke: blue + .3pt,
    fill: rgb(0, 149, 255, 15%),
    radius: 4pt,
  )[#align(left)[#solution]])

  #if work == [] [
    #solution
  ] else [
    #block(inset: 6pt, radius: 4pt, stroke: luma(50%) + .5pt, fill: luma(90%))[
      #align(left, text(font: "Liberation Sans", size: .85em, work))
      #solution
    ]
  ]
]

#let problem-header(number, points) = [== Problem #number. #text(weight: "regular")[[#points
    points]]]

#let problem(number, points, body) = [
  == Problem #number. #text(weight: "regular")[[#points points]]
  #body
]

#set page(margin: (x: .4in, y: .4in))
#set table(align: center)

_*Price Hiller*_
#v(-.8em)
_*zfp106*_
#v(-.8em)
Homework Assignment 3
#v(-.8em)
CS 2233
#v(-.8em)
Section 001

#align(
  center,
  block(
    inset: 6pt,
    radius: 4pt,
    stroke: luma(50%) + .5pt,
    fill: luma(90%),
  )[If you are interested in viewing the source code of this document, you can do so
    by clicking
    #text(
      blue,
      link(
        "https://git.orion-technologies.io/Price/college/src/branch/Development/Spring-2023/CS-2233/Assignment-3/Solution.typ",
        "here",
      ),
    ).],
)
= Problems

#problem(
  1,
  10,
)[
  - Complete all participation activities in zyBook sections $2.1$, $2.2$, and $2.4$-$2.6$.
  #solve[][Done]
]

#problem(
  2,
  10,
)[
  Prove that if $a$, $b$, and $c$ are odd integers, then $a + b + c$ is an odd
  integer.

  #solve[
    An odd integer is expressed as $2k + 1$ where $k$ is some integer.
  ][
    1. Suppose that $a$, $b$, and $c$ are odd integers. I shall prove that $a + b + c$ is an odd
    integer.
    2. Since $a$, $b$, and $c$ are integers, $a + b + c$ is also an integer
    3. Since $a$ is odd, there is an integer $q$ such that $a = 2q + 1$
    4. Since $b$ is odd, there is an integer $w$ such that $b = 2w + 1$
    5. Since $c$ is odd, there is an integer $e$ such that $c = 2e + 1$
    6. $a + b + c = (2q + 1) + (2w + 1) + (2e + 1) = 2(q + w + e + 1) + 1$
    7. Let $m = q + w + e + 1$
    8. Since $q$, $w$, $e$, and $1$ are integers, $m$ must be an integer
    9. Since $a + b + c = 2m + 1$, therefore $a + b + c$ is
      an odd integer
  ]
]
#problem(
  3,
  30,
)[
  Recall that a rational number can be put in the form $p/q$ where $p$ and $q$ are
  integers and $q ≠
  0$. Prove the following for any rational number, $x$:

  a.) If $x$ is rational, then $x - 5$ is rational
  #solve[][
    1. Let $x$ be a rational number. I will show that $x - 5$ is also a rational
      number.
    2. Since $x$ is rational, there exists integers $p$ and $q$ such that $x = p/q$ and $q ≠ 0$.
    3. Thus $x - 5 = p/q - 5$
    4. Since $5$ is an integer, $5$ is also a rational number written as $5/1$
    5. Therefore $p/q - 5 = p/q - 5/1$
    6. Working $p/q - 5/1$ we get $(p - 5q)/(q × 1)$ which is $(p - 5q)/q$
    7. Since $x - 5$ is equal to the ratio of two integers where the deonominator $≠ 0$, then $x - 5$ is a rational number
  ]
  b.) If $x - 5$ is rational, then $x/3$ is rational

  #solve[][
    1. Let $x - 5$ be rational. I will show that $x/3$ is also rational
    2. $x - 5 = p/q$ for some integers $p$, $q$, where $q ≠ 0$
    3. Add $5$ to both sides giving $x = p/q + 5$
    4. Divide both sides by $3$ giving $x/3 = (p/q + 5)/3$
    5. $(p/q + 5)/3$ is also rational
    6. Therefore $x/3$ is rational
  ]
\
  c.) If $x/3$ is rational, then $x$ is rational
  #solve[][
    1. Let $x/3$ be rational. I will show that $x$ is rational
    2. $x/3 = p/q$ for some integers $p$ and $q$ where $q ≠ 0$
    3. Multiply both sides of the equation by $q$ to get $x = 3p/q$
    4. Since $p$, $q$, and $3$ are integers, $x$ is rational
  ]
]

#problem(
  4,
  20,
)[
  Consider the following statement: For all integers $m$ and $n$, if $m - n$ is
  odd, then $m$ is odd or $n$ is odd.

  a.) Prove the statement using a proof by contrapositive
  #solve[][
    1. Let $m$ and $n$ be integers. I will show that if $m$ is even and $n$ is even
      then $m - n$ is even
    2. Since $m$ is even, there exists an integer $k$ such that $m = 2k$
    3. Since $n$ is even, there exists an integer $j$ such that $n = 2j$
    4. Thus, $m - n = 2k - 2j = 2(k - j )$
    5. Since $k - j$ is an integer, $2(k - j)$ is even
    6. Therefore, if $m$ is even and $n$ is even, $m - n$ is even
  ]
  b.) Prove the statement by using a proof by contradiction
  #solve[][
    1. Let $m$ and $n$ be integers. Assume then that $m - n$ is odd, but both $m$ and $n$ are
      even
    2. If $m$ is even, then $m = 2k$ for some integer $k$
    3. If $n$ is even, then $n = 2j$ for some integer $j$
    4. Then $m - n = 2k - 2j = 2(k - j)$
    5. But $2(k - j)$ is even, which contradicts that $m - n$ is odd
    6. Thus, if $m - n$ is odd, then $m$ must be odd or $n$ must be odd
  ]
]