alexandria/documents/by-course/math-8/pset-6/main.typ

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

#show: zen.with(
  title: "Homework 6",
  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:

3.1: \#2, 3bd, 5ab, 6b, 7ae, 8ef, 11a

3.2: \#1dfg, 6bcdi, 7

3.3: \#2ace, 3a, 7ac, 9d

= 3.1

== 2

=== a

$"Dom"(T) = {3,2,1}$

=== b

$"Rng"(T) = {1,2,3,5,6}$

=== c

$T^(-1) = {(1,3), (3,2), (5,3), (2,2), (6,1), (6,2), (2,1)}$

=== d

$T = {(3,1), (2,3), (3,5), (2,2), (1,6), (2,6), (1,2)}$

== 3

=== b

$
  "Dom"(W) = RR \
  "Rng"(W) = RR
$

=== d

$
  "Dom"(W) = (-infinity,0)union(0,infinity) \
  "Rng"(W) = (-infinity,0)union(0,infinity)
$

== 5

=== a

Suppose $x in RR$. Then there is a $y in RR$ such that $x R y$, namely $y =
6x$. Now consider all $y in RR$. Then there is an $x in RR$ such that $x R y$,
namely $x = y/6$.

=== b

Suppose $x in RR$. Then $x^2$ is either positive or zero. So all $y$ such that
$x R y$ are in $[0,infinity)$. So the range is $[0,infinity)$. Now consider all
$y in [0,infinity)$ such that $x R y$. Then we can always find an $x in RR$
such that $x = sqrt(y)$ as $y$ is not negative. So the domain is $RR$.

== 6

=== b

$
  R_2^(-1) = {(x,y) in RR times RR : y = -1 / 5 (x-2)}
$

== 7

=== a

$
  R compose S = {(3,5),(5,2)}
$

=== e

$
  S compose R = {(1,5),(2,4),(5,4)}
$

== 8

=== e

$
  R_2 compose R_4 &= {(x,y) in RR times RR : y = -5(x^2 + 2) + 2} \
  &= {(x,y) in RR times RR : y = -5x^2 - 8}
$

=== f

$
  R_4 compose R_2 &= {(x,y) in RR times RR : (-5x+2)^2 + 2} \
  &= {(x,y) in RR times RR : 25x^2 - 10x + 6}
$

== 11

=== a

The domain of $R$ is simply all of the first coordinates and the range of
$R^(-1)$ is all of the second coordinates of $R^(-1)$. But by the definition of
$R^(-1)$ all of its second coordinates are precisely the first coordinates of
$R$, so they're the same set.

More formally, $a in "Dom"(R)$ iff. for some $b in B$ there exists $(a,b) in R$
iff. for some $b in B$ there exists $(b,a) in R^(-1)$ iff. $a in
"Rng"(R^(-1))$. Since $a in "Dom"(R) <=> a in "Rng"(R^(-1))$ they are equal
(definition of set equality).

= 3.2

== 1

=== d

Transitive, but not reflexive or symmetric.

=== f

Symmetric only.

=== g

Transitive and reflexive but not symmetric.

== 6

=== b

For natural numbers $m$ and $m$, both have the same digit in the tens place. So
$R$ is reflexive. If natural numbers $m$ and $n$ have the same digit in the
tens place, then $n$ and $m$ have the same digit in the tens place. So $R$ is
symmetric. Finally, if naturals $n$ and $m$ have the same digit in the tens
place, and naturals $m$ and $p$ have the same digit in the tens place, than $n$
and $p$ have the same digit in the tens place. So the relation $R$ is
reflexive, symmetric, and transitive, so it's an equivalence relation on $NN$.

An element of $overline(106)$ less than 50 is 05. Between 150 and 300 is 250.
Greater than 1000 is 2000. Three such elements in the equivalence class
$overline(635)$ are 35, 235, and 1035.

=== c

For $x in RR$, x = x. So $V$ is reflexive. For $x,y in RR$, if $x = y$, then $y
= x$, or if $x y = 1$, then $y x = 1$ by the commutativity of multiplication.
So $V$ is symmetric. For $x,y,z in RR$, if $x = y$ and $y = z$, then $x = z$.

If $x y = 1$ and $y z = 1$, then either

1. $y = 1$ which implies $x = 1$ and $z = 1$ so $x z = 1$ and $x V z$.
2. $y != 1$ and $x = 1/y$ and $z = 1/y$ and we have $x =
  z$ so $x V z$.

So the relation is reflexive, symmetric, and transitive, and it's an
equivalence relation.

- the equivalence class of 3, $overline(3)$ is ${3, 1/3}$
- the equivalence class of $-2/3$, $overline(-2/3)$ is ${-2/3, -3/2}$
- the equivalence class of $0$, $overline(0)$, is just ${0}$

=== d

For any $a$, $a$ has a unique prime factorization so $a R a$. So $R$ is
reflexive. If $a R b$, then $b R a$ since $a$ and $b$ have unique prime
factorizations and have the same amount of twos in their unique prime factors.
So $R$ is symmetric. If $a R b$, and $b R c$, then the unique prime
factorization of $a$ has the same amount of 2s as the prime factorization of
$b$ which has the same amount of 2s as the prime factorization of $c$. So $a$
and $c$ have the same amount of 2s in their prime factorization and $a R c$. So $R$ is transitive. Therefore $R$ is an equivalence relation.

- In $overline(7)$, we have no 2s in the prime factorization, ${3,5,9} subset overline(7)$
- In $overline(10)$, there is one 2, so ${4,6,14} subset overline(10)$
- In $overline(72)$, there are three 2s, so ${8, 24, 40} subset overline(72)$

=== i

For any $x in RR$, $x T x$ iff. $sin(x) = sin(x)$ which is true, so $T$ is
reflexive. For $x, y in RR$, $x T y$ iff. $sin(x) = sin(y)$ iff. $sin(y) =
sin(x)$ iff. $y T x$. So $T$ is symmetric. For $x,y,z in RR$, $x T y$ and $y T
z$ iff. $sin(x) = sin(y)$ and $sin(y) = sin(z)$ iff. $sin(x) = sin(z)$ iff. $x
T z$ so $T$ is transitive. Therefore $T$ is an equivalence relation on $RR$.

- $overline(0)$ is given by all $y in RR$ where $sin(y) = 0$, so ${pi n : n in ZZ}$
- $overline(pi/2)$ is given by all $y in RR$ where $sin(y) = sin(pi/2) = 1$, so ${pi/2 + 2pi n : n in ZZ}$
- $overline(pi/4)$ is given by all $y in RR$ where $sin(y) = sin(pi/4)$, so ${pi/4 + 2pi n, (3pi)/4 + 2pi n : n in ZZ}$

== 7

If $p/q R s/t$, then $p t = q s$ and hence $q s = p t$. Therefore $s/t R p/q$,
so $R$ is symmetric. $p/q R p/q$ iff. $p t = p t$ so $R$ is reflexive.

Now we seek to show $R$ is transitive. $p/q R s/t$ and $s/t R a/b$ iff. $p t =
q s$ and $s b = a t$. Considering the special case where $s = 0$, we see that
$p t = 0$ and $a t = 0$. Note that $t$ is nonzero because $s/t in QQ$, so $p =
0$ and $a = 0$. Therefore $p b = a q$ and $p/q R s/t$.

Otherwise we consider when $s != 0$, note that $p t = q s$ and $s b = a t$ iff.
$p t s b = q s a t$, and recall $t$ is nonzero. So we divide them out and $p b
= a q$ so again $p/q R a/b$. So $R$ is transitive and it is indeed an
equivalence relation.

The equivalence class of $2/3$ is all $QQ$ who reduce to $2/3$, such as $4/6,
6/9$, and so on.

= 3.3

== 2

=== a

No, $cal(P)$ is not pairwise disjoint as ${2,3}sect{3,4} = {3}$.

=== c

Yes, each element of $cal(P)$ is pairwise disjoint and their total union is
$A$.

=== e

No, $cal(P)$ is not a subset of the power set of $A$ (its elements are not
subsets of $A$). Instead $cal(P)$ is simply equal to $RR$.

== 3

=== a

Proposition: $cal(P) = {{-x,x} : x in NN union {0}}$ is a partition of $ZZ$.

#proof[
  Suppose $X in cal(P)$. Then $X = {-x,x}$ for some $x in NN union {0}$, so $X
!= emptyset$.

  Then suppose $X,Y in P$. Then $X = {-x,x}$ for some $x in NN union {0}$ and
  $Y = {-y,y}$ for some $y in NN union {0}$.

  We show that either $X$ and $Y$ are the same set, or they are disjoint. If $x
  = y$, then $X = Y$. If $x != y$, then $-x != -y$, $-x != y$, and $x != -y$.
  So $X sect Y = emptyset$ (this implies any two sets in $cal(P)$ are pairwise
  disjoint).

  Now we show that $union.big _(X in cal(P)) X = ZZ$. First note that
  $union.big _(X in P) X subset.eq ZZ$ since every $x in X$ (for all $X in cal(P)$) is an integer.

  Now let $x in ZZ$. If $x in NN union {0}$, then
  $
    {-x,x} in cal(P)
  $
  so $x in union.big _(X in cal(P)) X$. If $x in.not NN union {0}$, then $x <
  0$ and $-x in NN union {0}$. So ${-x, -(-x)} = {x,-x} in cal(P)$. So $x in
  union.big _(X in P) X$. Therefore $ZZ subset.eq union.big _(X in P) X$. So $X
  = cal(P)$.
]

== 7

=== a

These are the equivalence classes of $NN$ under relation "has the same number
of digits."

=== c

$R$ is the relation "both positive, or both zero, or both negative." E.g. $a R
b$ iff. $a > 0$ and $b > 0$, or $a < 0$ and $b < 0$, or $a = 0$ and $b = 0$.

== 9

=== d

$R = {(5,1), (1,5), (2,4),(4,2),(3,3), (5,5), (1,1),(2,2), (3,3),(4,4),(5,5)}$