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\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
2022-02-17 at 11:59 pm}\hrule height .5pt}}

\centerline{\bigbf Homework 5 - Holden Rohrer}
\bigskip

\noindent{\bf Collaborators:} None

\section{Hammack 7: 6, 9, 12}

\problem{6.} Suppose $x,y\in\bb R.$ Then $x^3+x^2y = y^2+xy$ if and only if
$y = x^2$ or $y = -x.$

$x^2(x+y) = y(x+y).$
\answer
$(\Rightarrow)$

Let $x^3 + x^2y = y^2+xy.$
We then have $x^2(x+y) = y(x+y).$
If $y = -x,$ $x+y = 0 \impl x^2(x+y) = y(x+y).$
Otherwise, we can divide by $x+y$ because it is nonzero, giving
$y = x^2.$
Therefore, $y=-x$ or $y=x^2.$

$(\Leftarrow)$

Let $y = -x$ or $y = x^2.$
We will first consider the case $y = -x,$ then the case $y = x^2.$

With $y = -x,$ $x^3 + x^2y = x^3 - x^3 = 0 = x^2 - x^2 = y^2 + xy.$

If $y = x^2,$ $x^3 + x^2y = x^3 + x^4 = x^3 + x^4 = xy + y^2.$
\endanswer

\problem{9.} Suppose $a\in\bb Z.$ Prove that $14\mid a$ if and only if
$7\mid a$ and $2\mid a.$

\answer
$(\Rightarrow)$

Let $14\mid a.$
This gives $a = 14k$ for some integer $k.$
$a = 2(7k),$ so with $7k\in\bb Z,$ we get $2\mid a.$
Similarly, $a = 7(2k),$ so with $2k\in\bb Z,$ we get $7\mid a.$

$(\Leftarrow)$

Let $7\mid a$ and $2\mid a.$
These give $a = 7j$ for some $j$ and $a = 2k$ for some integers $j$ and
$k.$
A product of odd number $7$ and odd number $j$ cannot be even (and $a$
is even because $2\mid a$), so $j$ must be even.
Thus, there exists $l\in\bb Z$ such that $j = 2l.$
This gives $a = 7(2l) = 14l \to 14\mid a.$
\endanswer

\problem{12.} There exist a positive real number $x$ for which $x^2 < \sqrt
x.$

\answer
Observe that $x = 1/4$ gives $x^2 = 1/16$ and $\sqrt x = 1/2,$ so $1/16
< 1/2.$
\endanswer

\section{Hammack 8: 12, 22, 28}

\problem{12.} If $A,$ $B,$ and $C$ are sets, then $A-(B\cap C) =
(A-B)\cup(A-C).$

\answer
$(\subseteq)$

Let $x\in A - (B\cap C).$
This gives us $x\in A$ and $x\not\in B\cap C.$
We get $x\not\in B$ or $x\not\in C.$
WLOG, let $x\not\in B.$
$x\in A$ and $x\not\in B,$ so $x\in A-B,$ so $x\in (A-B)\cup(A-C).$

$(\supseteq)$

Let $x\in (A-B)\cup (A-C).$
This gives $x\in A-B$ or $x\in A-C.$
WLOG, let $x\in A-B.$
Therefore, $x\in A$ and $x\not\in B.$
This implies $x\not\in B\cap C,$ so $x\in A-(B\cap C).$

Since we have $A-(B\cap C) \subseteq (A-B)\cup(A-C)$ and $(A-B)\cup(A-C)
\subseteq A-(B\cap C),$
we obtain $$A-(B\cap C) = (A-B)\cup(A-C).$$
\endanswer

\problem{22.} Let $A$ and $B$ be sets. Prove that $A\subseteq B$ if and
only if $A\cap B = A.$

\answer
$(\Rightarrow)$

Let $A\subseteq B.$
Let $x\in A.$
By subset, $x\in B.$
And if and only if $x\in A$ and $x\in B,$ $x\in A\cap B,$ so $A = A\cap
B.$

$(\Leftarrow)$

Let $A\cap B = A.$
This implies $A\subseteq A\cap B,$ or $x\in A\impl x\in A\cap B\impl
x\in B.$
$x\in A\impl x\in B$ is the definition of $A\subseteq B.$
\endanswer

\problem{28.} Prove $\{12a+25b:a,b\in\bb Z\} = \bb Z.$

\answer
Let $A = \{12a+25b:a,b\in\bb Z\}.$ If $x\in A,$ $x\in\bb Z$ because the
integers are closed under addition and mulitplication.

If $x\in\bb Z,$ let $b = x$ and $a = -2x,$ giving us $12(-2x) + 25x =
x\in A.$

Therefore, since $A\subseteq\bb Z$ and $\bb Z\subseteq A,$ these two
sets are equal.
\endanswer

\section{Hammack 9: 6, 28, 30, 34}
Prove or disprove each of the following statements.

\problem{6.} If $A,$ $B,$ $C,$ and $D$ are sets, then $(A\times
B)\cup(C\times D) = (A\cup C)\times(B\cup D).$

\answer
{\bf Disproof.}

Let $A = B = \{1\}$ and $C = D = \{2\}.$
The set $A\times B = \{(1,1)\}.$
The set $C\times D = \{(2,2)\}.$
And the set $A\cup C = B\cup D = \{1,2\}.$

$$(A\times B)\cup(C\times D) = \{(1,1),(2,2)\} \neq
\{(1,1),(1,2),(2,1),(2,2)\} = (A\cup C)\times(B\cup D).$$
\endanswer

\problem{28.} Suppose $a,b\in\bb Z.$ If $a\mid b$ and $b\mid a,$ then $a =
b.$

\answer
We will show this by contrapositive.
Let $a\neq b.$ We will show that $a\nmid b$ or $b\nmid a.$

WLOG, $a > b.$
Immediately, $a\nmid b.$
\endanswer

\problem{30.} There exist integers $a$ and $b$ for which $42a + 7b = 1.$

\answer
{\bf Disproof.}

Let $a,b\in\bb Z.$
For the sake of contradiction, assume $42a + 7b = 1.$
Dividing by 7, $6a + b = 1/7.$
By closure of the integers, $6a+b\in\bb Z,$ and $1/7\not\in\bb Z,$
giving a contradiction.
\endanswer

\problem{34.} If $X\subseteq A\cup B,$ then $X\subseteq A$ or $X\subseteq
B.$

\answer
{\bf Disproof.}

Let $A = \{1\},$ $B = \{2\},$ and $X = A\cup B.$
Immediately, $X\subseteq A\cup B.$
And then, $2\in X,$ but $2\not\in A,$ so $X\not\subseteq A.$
Also, $1\in X,$ but $1\not\in B,$ so $X\not\in B.$
\endanswer

\section{Problem not from the textbok}

\problem{1.} Let $A,$ $B,$ and $C$ be arbitrary sets. Prove that if
$A-C\not\subseteq A-B,$ then $B\not\subseteq C.$

\answer
We will prove this by contrapositive.
Let $B\subseteq C.$ We will show that $A-C\subseteq A-B.$

If $x\in B,$ $x\in C,$ so by contrapositive, if $x\not\in C,$ $x\not\in
B.$

Let $y\in A-C.$
By definition of setminus, $y\in A$ and $y\not\in C.$
As established, this implies $y\not\in B.$
Therefore, with $y\in A$ and $y\not\in B,$ $y\in A-B,$ so
$A-C\subseteq A-B.$
\endanswer

\bye
