alexandria/documents/by-course/math-8/pset-7/main.typ

#import "@youwen/zen:0.1.0": *
#import "@preview/mitex:0.2.5": *

#show: zen.with(
  title: "Homework 7",
  author: "Youwen Wu",
)

#set heading(numbering: none)
#show heading.where(level: 2): it => [#it.body.]
#show heading.where(level: 3): it => [#it.body.]

#set par(first-line-indent: 0pt, spacing: 1em)

Problems:

3.2: \#9cd, 10ac, 16a

3.4: \#1, 7ad, 11ac

4.1: \#1cef, 2, 3b, 4d, 7, 9, 12b, 13, 16, 17ab, 18ab

#outline()

= 3.2

== 9

=== c

$
  overline(0) = {...,0,1,2,3...} = ZZ
$

=== d

$
  &overline(0) = {...,-7,0,7,14,...} \
  &overline(1) = {...,-8,-1,1,8,15,...} \
  &overline(2) = {...,-9,-2,2,9,16,...} \
  &overline(3) = {...,-10,-3,3,10,17,...} \
  &overline(4) = {...,-11,-4,4,11,18,...} \
  &overline(5) = {...,-12,-5,5,12,19,...} \
  &overline(6) = {...,-13,-6,6,13,20,...} \
$

== 10

=== a

5 and -5.

=== c

14 and -10

== 16

=== a

$ 6 equiv 1 space (mod 5)$, but $6 equiv 6 space (mod 10)$.

= 3.4

== 1

=== a

$6 + 6 equiv 5$

=== b

$5 dot 4 equiv 6$

=== c

$3 dot 3 + 5 dot 2 equiv 5$

=== d

$2 dot 4 + 3 dot 5 equiv 2$

=== e

$5 dot 1 + 3 dot 2 equiv 4$

=== f

$0 dot 3 + 2 dot 4 equiv 1$

== 7

=== a

$(238 + 496 - 44) mod 9 = 6$

=== d

$317 dot 403 mod 9 = 5$

== 11

=== a

No

=== c

Yes

= 4.1

== 1

=== c

It is a function. Domain: ${1,2}$, codomain: ${1,2}$.

=== e

No.

=== f

No.

== 2

It fails the "vertical line test" when graphed.

Consider the input 1 so $f(1) = plus.minus sqrt(1)$. Then $f(1) = 1$ and $f(1) = -1$, so it relates 1 to multiple distinct values. Therefore it is not a function.

== 3

=== b

- Domain: $RR$
- Range: $y >= 5$, $y in RR$
- Another codomain: $RR^+$ (positive real numbers)

== 4

=== d

Domain: ${-2}$

== 7

#proof[
  Now we show the converse. Suppose the conditions hold, that is,
  1. $"Dom"(f) = "Dom"(g)$
  2. $forall x in "Dom"(f)$, $f(x) = g(x)$

  For any $(x,y) in f$, $y = f(x)$. Then by (1) $x$ is in the domain of $g$.
  By (2), $g(x) = f(x) = y$, so $(x,y) in g$. So $f subset.eq g$. Repeat an
  identical reasoning showing $g subset.eq f$. Therefore $f = g$.
]

== 9

=== a

$chi_A (1) = 1$

=== b

$chi_A (3) = 0$

=== c

$chi_A (pi) = 0$

=== d

$chi_A (2) - chi_A (0.2) = 1$

== 12

=== b

- 1st term: 0
- 5th term: 4
- 10th term: 11

== 13

=== a

$f(3) = 3$

=== b

0

=== c

3

=== d

${x : x - 1 | 6 = 0}$

== 16

=== a

$"Rng"(pi_1) = {a in A : exists b in B "such that" (a,b) in S}$

=== b


$"Rng"(pi_1) = {b in B : exists a in A "such that" (a,b) in S}$

== 17

=== a

There are $n^m$ functions.

=== b


There are $m n$ functions.

== 18

=== a

#mitext(`
Let $f: A \to B$ be a function and define the relation $T$ on $A$ by
\[
x \,T\, y \quad \Longleftrightarrow \quad f(x) = f(y).
\]
We must show that $T$ is reflexive, symmetric, and transitive.

For any $x \in A$, we have $f(x) = f(x)$. Hence,
\[
x \,T\, x,
\]
so $T$ is reflexive.

Suppose $x \,T\, y$. By definition, this means $f(x) = f(y)$. Since equality in $B$ is symmetric, we also have $f(y) = f(x)$. Therefore,
\[
y \,T\, x,
\]
so $T$ is symmetric.

Suppose $x \,T\, y$ and $y \,T\, z$. Then $f(x) = f(y)$ and $f(y) = f(z)$. By transitivity of equality in $B$, $f(x) = f(z)$. Hence,
\[
x \,T\, z,
\]
so $T$ is transitive.

Since $T$ is reflexive, symmetric, and transitive, it is an equivalence relation on $A$.
`)

=== b

#mitext(`
We want to describe the equivalence classes of $0$, $2$, and $4$ under the relation $x \,T\, y$ if and only if $x^2 = y^2$.

\[
[0] = \{y \in \mathbb{R}: y^2 = 0^2\} = \{0\}.
\]

\[
[2] = \{y \in \mathbb{R}: y^2 = 2^2\} = \{2, -2\}.
\]

\[
[4] = \{y \in \mathbb{R}: y^2 = 4^2\} = \{4, -4\}.
\]
`)