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diff --git a/gupta/hw5.tex b/gupta/hw5.tex new file mode 100644 index 0000000..bb1bd43 --- /dev/null +++ b/gupta/hw5.tex @@ -0,0 +1,217 @@ +\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 |