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+\newfam\bbold
+\def\bb#1{{\fam\bbold #1}}
+\font\bbten=msbm10
+\font\bbsev=msbm7
+\font\bbfiv=msbm5
+\textfont\bbold=\bbten
+\scriptfont\bbold=\bbsev
+\scriptscriptfont\bbold=\bbfiv
+\font\bigbf=cmbx12 at 24pt
+
+\def\answer{\smallskip{\bf Answer.}\par}
+\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
+depth 0pt{\parfillskip0pt\medskip}}
+\let\endanswer\endproof
+\def\section#1{\medskip\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in%
+\noindent{\bf #1}}
+\let\impl\to
+\def\nmid{\hskip-3pt\not\hskip2.5pt\mid}
+\def\problem#1{\par\penalty-100\item{#1}}
+
+\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