162 lines
4.9 KiB
Plaintext
162 lines
4.9 KiB
Plaintext
#let m(math) = align(center)[$#math$]
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#let pgbreakmsg = align(center, text(blue, weight: "black", size: 1.5em)[See Next Page\ ↓])
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#let solve(work, solution) = align(
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center,
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)[
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#let solution = align(center, block(
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inset: 5pt,
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stroke: blue + .3pt,
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fill: rgb(0, 149, 255, 15%),
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radius: 4pt,
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)[#align(left)[#solution]])
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#if work == [] [
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#solution
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] else [
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#block(inset: 6pt, radius: 4pt, stroke: luma(50%) + .5pt, fill: luma(90%))[
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#align(left, text(font: "Liberation Sans", size: .85em, work))
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#solution
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]
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]
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]
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#let problem-header(number, points) = [== Problem #number. #text(weight: "regular")[[#points
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points]]]
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#let problem(number, points, body) = [
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== Problem #number. #text(weight: "regular")[[#points points]]
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#body
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]
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#set page(margin: (x: .4in, y: .4in))
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#set table(align: center)
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_*Price Hiller*_
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#v(-.8em)
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_*zfp106*_
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#v(-.8em)
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Homework Assignment 3
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#v(-.8em)
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CS 2233
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#v(-.8em)
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Section 001
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#align(
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center,
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block(
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inset: 6pt,
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radius: 4pt,
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stroke: luma(50%) + .5pt,
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fill: luma(90%),
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)[If you are interested in viewing the source code of this document, you can do so
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by clicking
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#text(
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blue,
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link(
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"https://git.orion-technologies.io/Price/college/src/branch/Development/Spring-2023/CS-2233/Assignment-3/Solution.typ",
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"here",
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),
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).],
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)
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= Problems
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#problem(
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1,
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10,
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)[
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- Complete all participation activities in zyBook sections $2.1$, $2.2$, and $2.4$-$2.6$.
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#solve[][Done]
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]
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#problem(
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2,
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10,
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)[
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Prove that if $a$, $b$, and $c$ are odd integers, then $a + b + c$ is an odd
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integer.
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#solve[
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An odd integer is expressed as $2k + 1$ where $k$ is some integer.
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][
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1. Suppose that $a$, $b$, and $c$ are odd integers. I shall prove that $a + b + c$ is an odd
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integer.
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2. Since $a$, $b$, and $c$ are integers, $a + b + c$ is also an integer
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3. Since $a$ is odd, there is an integer $q$ such that $a = 2q + 1$
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4. Since $b$ is odd, there is an integer $w$ such that $b = 2w + 1$
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5. Since $c$ is odd, there is an integer $e$ such that $c = 2e + 1$
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6. $a + b + c = (2q + 1) + (2w + 1) + (2e + 1) = 2(q + w + e + 1) + 1$
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7. Let $m = q + w + e + 1$
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8. Since $q$, $w$, $e$, and $1$ are integers, $m$ must be an integer
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9. Since $a + b + c = 2m + 1$, therefore $a + b + c$ is
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an odd integer
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]
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]
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#problem(
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3,
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30,
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)[
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Recall that a rational number can be put in the form $p/q$ where $p$ and $q$ are
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integers and $q ≠
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0$. Prove the following for any rational number, $x$:
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a.) If $x$ is rational, then $x - 5$ is rational
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#solve[][
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1. Let $x$ be a rational number. I will show that $x - 5$ is also a rational
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number.
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2. Since $x$ is rational, there exists integers $p$ and $q$ such that $x = p/q$ and $q ≠ 0$.
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3. Thus $x - 5 = p/q - 5$
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4. Since $5$ is an integer, $5$ is also a rational number written as $5/1$
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5. Therefore $p/q - 5 = p/q - 5/1$
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6. Working $p/q - 5/1$ we get $(p - 5q)/(q × 1)$ which is $(p - 5q)/q$
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7. Since $x - 5$ is equal to the ratio of two integers where the deonominator $≠ 0$, then $x - 5$ is a rational number
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]
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b.) If $x - 5$ is rational, then $x/3$ is rational
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#solve[][
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1. Let $x - 5$ be rational. I will show that $x/3$ is also rational
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2. $x - 5 = p/q$ for some integers $p$, $q$, where $q ≠ 0$
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3. Add $5$ to both sides giving $x = p/q + 5$
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4. Divide both sides by $3$ giving $x/3 = (p/q + 5)/3$
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5. $(p/q + 5)/3$ is also rational
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6. Therefore $x/3$ is rational
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]
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\
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c.) If $x/3$ is rational, then $x$ is rational
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#solve[][
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1. Let $x/3$ be rational. I will show that $x$ is rational
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2. $x/3 = p/q$ for some integers $p$ and $q$ where $q ≠ 0$
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3. Multiply both sides of the equation by $q$ to get $x = 3p/q$
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4. Since $p$, $q$, and $3$ are integers, $x$ is rational
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]
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]
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#problem(
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4,
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20,
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)[
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Consider the following statement: For all integers $m$ and $n$, if $m - n$ is
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odd, then $m$ is odd or $n$ is odd.
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a.) Prove the statement using a proof by contrapositive
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#solve[][
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1. Let $m$ and $n$ be integers. I will show that if $m$ is even and $n$ is even
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then $m - n$ is even
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2. Since $m$ is even, there exists an integer $k$ such that $m = 2k$
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3. Since $n$ is even, there exists an integer $j$ such that $n = 2j$
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4. Thus, $m - n = 2k - 2j = 2(k - j )$
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5. Since $k - j$ is an integer, $2(k - j)$ is even
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6. Therefore, if $m$ is even and $n$ is even, $m - n$ is even
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]
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b.) Prove the statement by using a proof by contradiction
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#solve[][
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1. Let $m$ and $n$ be integers. Assume then that $m - n$ is odd, but both $m$ and $n$ are
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even
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2. If $m$ is even, then $m = 2k$ for some integer $k$
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3. If $n$ is even, then $n = 2j$ for some integer $j$
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4. Then $m - n = 2k - 2j = 2(k - j)$
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5. But $2(k - j)$ is even, which contradicts that $m - n$ is odd
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6. Thus, if $m - n$ is odd, then $m$ must be odd or $n$ must be odd
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]
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]
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