cs-2233: complete assignment 3

Spring-2023/CS-2233/Assignment-3/Solution.typ

@@ -0,0 +1,161 @@
+#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
+  ]
+]

Spring-2023/CS-2233/TODO.org

@@ -1,4 +1,10 @@
-* DONE Assignment 1
+* DONE Assignment 1                                             :college:cs2233:
 DEADLINE: <2024-01-26 Fri> SCHEDULED: <2024-01-25 Thu>
 
 Complete Zybooks section ~1~ and the first homework assignment
+
+* TODO Assignment 3                                             :college:cs2233:
+DEADLINE: <2024-02-11 Sun> SCHEDULED: <2024-02-11 Sun>
+
+Complete Zybooks section ~2~ and the third homework assignment
+