college/Fall-2024/CS-3333/Assignments/2/HW2.typ

#set page(margin: (x: .5in, y: .5in))

#let solvein(solution) = {
  let outset = 3pt
  h(outset)
  box(
    outset: outset,
    stroke: blue + .3pt,
    fill: rgb(0, 149, 255, 15%),
    radius: 4pt,
  )[#solution]
}

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

#let notein(content) = {
  let outset = 3pt
  h(outset)
  box(
    outset: outset,
      stroke: luma(20%) + .3pt,
      fill: luma(95%),
    radius: 4pt,
  )[#content]
}

#let note(content) = [
  #align(
    center,
    block(
      inset: 5pt,
      stroke: luma(20%) + .3pt,
      fill: luma(95%),
      radius: 4pt,
    )[#align(left)[#content]],
  )
]

#align(center)[
  = CS 3333 Mathematical Foundations
  Homework 2 (100 points)\
  #underline[Price Hiller] *|* #underline[zfp106]
]

#line(length: 100%, stroke: .25pt)

= Submission:

Same as HW1.

= Questions

+ Answer Yes or No (12 pts).
  #enum(
    numbering: "a.",
    [
      Is $87 ≡ 51 (mod 5)$? #solvein[No]
    ],
    [
      Is $34 ≡ 14 (mod 3)$? #solvein[No]
    ],
    [
      Is $7 ≡ 55 (mod 24)$? #solvein[Yes]
    ],
    [
      Is $29 ≡ 41 (mod 12)$? #solvein[Yes]
    ],
  )
+ List 5 integers that are congruent to 6 modulo 19 (10 points).

  #solvein[
    + $6$, #notein[$∵ (6 - 6) mod 19 = 0$]
    + $25$, #notein[$∵ (25 - 6) mod 19 = 0$]
    + $44$, #notein[$∵ (44 - 6) mod 19 = 0$]
    + $63$, #notein[$∵ (63 - 6) mod 19 = 0$]
    + $82$, #notein[$∵ (82 - 6) mod 19 = 0$]
  ]

+ Calculate the following problems (18 pts).
  #enum(
    numbering: "a.",
    [
      $7+_(11) 34 =$ #solvein[$8$]
      #note[
        + $(7 + 34) mod 11$
        + $(41) mod 11$
        + $8$
      ]
    ],
    [
      $5⋅_(13) 19 =$ #solvein[ $4$ ]
      #note[
        + $(5 ⋅ 19) mod 13$
        + $(95) mod 13$
        + $4$
      ]
    ],
    [
      $17+_(11) 1 =$ #solvein[$7$]
      #note[
        + $(17 + 1) mod 11$
        + $(18) mod 11$
        + $7$
      ]
    ],
    [
      $47+_(13) 0 =$ #solvein[$8$]
      #note[
        + $(47 + 0) mod 13$
        + $(47) mod 13$
        + $8$
      ]
    ],
    [
      $55⋅_(11) 1 =$ #solvein[$0$]
      #note[
        + $(55 ⋅ 1) mod 11$
        + $(55) mod 11$
        + $0$
      ]
    ],
    [
      $55⋅_(11) 0 =$ #solvein[$0$]
      #note[
        + $(55 ⋅ 0) mod 11$
        + $(0) mod 11$
        + $0$
      ]
    ],
  )

+ Find all positive primes $<= 50$ (You can just list the positive prime numbers directly.) (10 points).

  #align(center)[
    #solvein[
      All primes $<= 50$
      #table(
        columns: (auto, auto, auto, auto, auto),
        [2], [3], [5], [7], [11],
        [13], [17], [19], [23], [29],
        [31], [37], [41], [43], [47],
      )]
    #note[
      Some Rust code to generate those primes :)
      ```rust
      fn main() {
          let is_prime = |num: usize| -> bool { num > 1 && !((2..num).any(|n| num % n == 0)) };
          let primes: Vec<_> = (0..50).filter(|x| is_prime(*x)).collect();
          println!(
              "Prime numbers: {:?}\nNumber of prime numbers found: {}",
              primes,
              primes.len()
          );
      }
      ```
    ]
  ]


+ Find all positive integers less than $50$ that are relatively prime to $50$ (showing the $\g\cd$ is optional. You can list all the numbers.) (10 pts).
  #align(center)[
    #solvein[
      All relative primes $< 50$
      #table(
        columns: (auto, auto, auto, auto, auto),
        [2], [3], [5], [7], [ 11],
        [13], [17], [19], [23], [29],
        [31], [37], [41], [43], [47],
      )]
    #note[
      Some Rust code to generate those co-primes :)
      ```rust
      fn gcd(a: &usize, b: &usize) -> usize {
          let mut a = *a; // Deref to avoid modifying the passed value
          let mut b = *b; // Deref to avoid modifying the passed value
          assert!(a != 0 && b != 0);
          while b != 0 {
              (a, b) = (b, a % b)
          }
          a
      }

      fn main() {
          let relative_primes: Vec<_> = (1..51).filter(|x| gcd(x, &50) == 1).collect();
          println!(
              "Relatively prime numbers: {:?}\nNumber of relative prime numbers found: {}",
              relative_primes,
              relative_primes.len()
          );
      }
      ```
    ]
  ]
+ Express $\g\cd(990, 502) = 2$ as a linear combination of $990$ and $502$ by working backwards through the steps of the Euclidean algorithm (10pts). [Hint: refer to Example 17 in the textbook Section 4.3.8]
  #note[
    Working it fowards:
    + $990 = 1 ⋅ 502 + 488$
    + $502 = 1 ⋅ 488 + 14$
    + $488 = 34 ⋅ 14 + 12$
    + $14 = 1 ⋅ 12 + 2$
    + $12 = 6 ⋅ 2 + 0$
    + $\g\cd(990, 502) = 2$, from step $4$

    Now backwards:
    + $2 = 14 - 1 ⋅ 12$
    + $2 = 14 - 1 ⋅ (488 - 34 ⋅ 14)$
    + $2 = -1 ⋅ 488 + 35 ⋅ 14$
    + $2 = -1 ⋅ 488 + 35 ⋅ (502 - 488)$
    + $2 = -36 ⋅ 488 + 35 ⋅ 502$
    + $2 = -36 ⋅ (990 - 488) + 35 ⋅ 502$
    + #solvein[$2 = - 36 ⋅ 990 + 71 ⋅ 502$]
    + #solvein[$2 = 71 ⋅ 502 - 36 ⋅ 990$]
  ]

+ Show that if $a ≡ b (mod m)$ and $c ≡ d (mod m)$, where $a$, $b$, $c$, $d$, and $m$, are integers with $m >= 2$, then $a - c ≡ b - d (mod m)$ (10 pts).

  #solve[
    + $a ≡ b (mod m) = m | (a - b)$
    + $a = b + m k$ where $k$ is an integer
    + $c ≡ d (mod m) = m | (c - d)$
    + $c = d + m j$ where $j$ is an integer
    + $a - c = (b + m k) - (d + m j)$
    + $a - c = (b - d) + (m k - m s)$
    + $a - c = (b - d) + m(k - s)$
    + $(a - c) - (b - d) = m (k - s)$
    + $(a - c) - (b - d) = m i$, where $i = k - s$ is an integer
    + $m | (a - c) - (b - d)$
    + $∴ a - c ≡ b - d (mod m)$
  ]

+ Let $a$ and $b$ be positive integers. Then $a b = \g\cd(a, b) ⋅ \l\cm(a,b)$.
  #enum(
    numbering: "a.",
    [
      Please prove it and show the intermediate steps (15 pts). [Hint: Use the prime factorizations of $a$ and $b$ and the formulae for $\g\cd(a, b)$ and $\l\cm(a, b)$ in terms of these factorizations. You also can use any other methods.]
      #solve[
        + $a = p^(a_1)_(1) ⋅ p^(a_2)_(2) ⋅⋅⋅ p^(a_n)_(n)$, where $p_n$ are prime numbers and $a_n$ are non-negative integers
        + $b = p^(b_1)_(1) ⋅ p^(b_2)_(2) ⋅⋅⋅ p^(b_n)_(n)$, where $p_n$ are prime numbers and $b_n$ are non-negative integers
        + $\g\cd(
              a, b
            ) = p^(min(a_1,b_1))_1 ⋅ p^(min(a_2,b_2))_2 ⋅⋅⋅ p^(min(a_n,b_n))_n$
        + $\l\cm(
              a, b
            ) = p^(max(a_1,b_1))_1 ⋅ p^(max(a_2,b_2))_2 ⋅⋅⋅ p^(max(a_n,b_n))_n$
        + $g\c\d(a,b) ⋅ \l\cm(a, b) = (
              p^(min(a_1,b_1))_1 ⋅ p^(min(a_2,b_2))_2 ⋅⋅⋅ p^(min(a_n,b_n))_n
            ) ⋅ (
              p^(max(a_1,b_1))_1 ⋅ p^(max(a_2,b_2))_2 ⋅⋅⋅ p^(max(a_n,b_n))_n
            )$

        + $g\c\d(a,b) ⋅ \l\cm(a, b) = (
              p^(min(a_1,b_1) + max(a_1,b_1))_1 ⋅ p^(min(a_2,b_2) + max(a_2,b_2))_2 ⋅⋅⋅ p^(min(a_n,b_n) + max(a_n,b_n))_n
            )$
        + $min(a_n,b_n) + max(a_n,b_n) = a_n + b_n$
        + $g\c\d(a,b) ⋅ \l\cm(
              a,b
            ) = p^(a_1 + b_1)_1 ⋅ p^(a_2 + b_2)_2 ⋅⋅⋅ p^(a_n + b_n)_n = a b$
        + $∴ a b = g\c\d(a,b) ⋅ \l\cm(a,b)$
        #note[If I have to type another `p^(min(a_n,b_n) + max(a_n,b_n))_n`, I _will_ ask for #text(red)[ Ǵ̶̱̝̅̓ȯ̸͙̯͝d̵̛͇͓ͅ'̵̺̑̀͆ͅͅs̶̖̏ ̶̫͔̲͂m̶̱̗̤͒́̏a̵̛̝̳̒n̶̼̱̆ä̷̤ǧ̸̢͜͜e̶̡͂r̵̞̯̺̄].]
      ]
    ],
    [
      Verify that $\g\cd(860, 516)⋅\l\cm(860,516) = 860 * 516$ (5 pts).
      #note[
        + $860=2^2 ⋅ 3^0 ⋅ 5^1 ⋅ 43^1$
        + $516=2^2 ⋅ 3^1 ⋅ 5^0 ⋅ 43^1$
        #underline[GCD]
        + $\g\cd(
              860, 516
            ) = 2^(min(2,2)) ⋅ 3^(min(0,1)) ⋅ 5^(min(1,0)) ⋅ 43^(min(1,1))$
        + $\g\cd(860, 516) = 2^2 ⋅ 3^0 ⋅ 5^0 ⋅ 43^1$
        + $\g\cd(860, 516) = 4 ⋅ 1 ⋅ 1 ⋅ 43$
        + $\g\cd(860, 516) = 172$
        #underline[LCM]
        + $\l\cm(
              860, 516
            ) = 2^(max(2,2)) ⋅ 3^(max(0,1)) ⋅ 5^(max(1,0)) ⋅ 43^(max(1,1))$
        + $\l\cm(860, 516) = 2^2 ⋅ 3^1 ⋅ 5^1 ⋅ 43^1$
        + $\l\cm(860, 516) = 4 ⋅ 3 ⋅ 5 ⋅ 43$
        + $\l\cm(860, 516) = 2580$
        #underline[Check if gcd \* lcm = 860 \* 516]
        + $172 * 2580 = 443760$
        + $860 * 516 = 443760$
        + #solvein[$(172 ⋅ 2580 = 443760) = (443760 = 860 ⋅ 516)$]
      ]
    ],
  )