added all new homework for Neha's class

9 files changed, 1776 insertions, 1 deletions

diff --git a/gupta/Makefile b/gupta/Makefile
index 1188bc9..2adfb32 100644
--- a/gupta/Makefile
+++ b/gupta/Makefile
@@ -7,7 +7,7 @@ PDFLATEX = pdflatex
 .tex.pdf:
 	$(PDFTEX) $<
 
-all: hw1.pdf hw2.pdf hw3.pdf hw4.pdf hw5.pdf
+all: hw1.pdf hw2.pdf hw3.pdf hw4.pdf hw5.pdf hw6.pdf hw7.pdf hw8.pdf hw9.pdf hw10.pdf hw11.pdf hw12.pdf portfolio.pdf
 
 clean:
 	rm -f *.pdf *.log *.aux
diff --git a/gupta/hw10.tex b/gupta/hw10.tex
new file mode 100644
index 0000000..20a0a33
--- /dev/null
+++ b/gupta/hw10.tex
@@ -0,0 +1,240 @@
+\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{\vskip18pt plus 6pt minus 6pt\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{\bigskip\par\penalty-100\item{#1}}
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-04-07 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 10 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Judson 3.5: 2, 7, 14, 28, 13, 35, 45, 48}
+
+\problem{2.}
+Which of the following multiplication tables defined on the set $G =
+\{a,b,c,d\}$ form a group? Support your answer in each case.
+
+\answer
+
+$(a)$ is not a group because it does not have an identity element.
+
+$(b)$ is a group because it has an identity element ($a$), and every
+element has an inverse (itself), and I have verified it is associative. % prove later
+
+$(c)$ is a group because it has an identity element ($a$), and every
+element has an inverse ($(a,b,c,d)\mapsto(a,d,c,b)$). I have also
+verified that this multiplication table is associative.
+
+$(d)$ is not a group because it has an identity element ($a$), but $d$
+has no inverse.
+
+\endanswer
+
+\problem{7.}
+Let $S = \bb R\setminus \{-1\}$ and define a binary operation on $S$ by
+$a*b = a+b+ab.$ Prove that $(S,*)$ is an abelian group.
+
+\answer
+This group has identity $0$ because $a*0 = a+0+0a = a.$
+Every element has an inverse $-{a\over a+1}$ which is defined because
+$a\neq -1,$ so $a+1\neq 0.$ $$-{a\over a+1}*a = -{a\over a+1} + a -
+{a^2\over a+1} = {a^2+a\over a+1} - {a\over a+1} - {a^2\over a+1} = 0.$$
+The operation is also associative. $a*b = (a+1)(b+1),$ so
+$$(a*b)*c = (a+1)(b+1)(c+1) = a*(b*c).$$
+\endanswer
+
+\problem{14.}
+Given the groups $\bb R^*$ and $\bb Z,$ let $G = \bb R^* \times \bb Z.$
+Define a binary operation $\circ$ on $G$ by
+$(a,m)\circ (b,n) = (ab, m+n).$
+Show that $G$ is a group under this operation.
+
+\answer
+This operation is associative because multiplication and addition are
+associative, so $((a,m)\circ (b,n)) \circ (c,l) = (a,m)\circ ((b,n)\circ
+(c,l)) = (abc,m+n+l).$
+
+The element $(a,m)$ has inverse $(1/a,-m),$ and the group has identity
+$(1,0).$
+\endanswer
+
+\problem{28.}
+Prove the remainder of Proposition 3.21: if $G$ is a group and
+$a,b\in G,$ then the equation $xa = b$ has a unique solution in $G.$
+
+\answer
+By definition of the group $a$ has an inverse $a^{-1}\in G,$ so
+multiplying the right side by $a^{-1}$ gives
+$$xaa^{-1} = xe = x = ba^{-1}.$$
+\endanswer
+
+\problem{31.}
+
+Show that if $a^2 = e$ for all elements $a$ in a group $G,$ then $G$
+must be abelian.
+
+\answer
+Let $G$ be a group such that $g^2 = e$ for all elements $g\in G.$
+We will show that $G$ is abelian.
+Let $a,b\in G.$ We will show that $ab = ba.$
+$$ab = eabe = bbabaa = b(ba)(ba)a = ba.$$
+\endanswer
+
+\problem{35.}
+
+Find all the subgroups of the symmetry group of an equilateral triangle.
+
+\answer
+We have the trivial subgroup $\langle e\rangle,$ the rotation subgroup
+$\langle \rho_1 \rangle,$ the reflection subgroups $\langle
+\mu_1\rangle,$ $\langle\mu_2\rangle,$ $\langle\mu_3\rangle,$ and the
+group $G.$
+\endanswer
+
+\problem{45.}
+
+Prove that the intersection of two subgroups of a group $G$ is also a
+subgroup of $G.$
+
+\answer
+Let $H$ and $I$ be two subgroups of $G.$
+We will show that $H\cap I$ is a subgroup by showing the three
+conditions.
+
+Let $a,b\in H\cap I.$ We will show that $ab\in H\cap I.$
+From definition of $a,b,$ we know that $a,b\in H,$ so $ab\in H.$
+Similarly, $a,b\in I,$ so $ab\in I.$
+By definition of $H$ and $I$ as subgroups, $a^{-1}\in I,$ and $a^{-1}\in
+H,$ so $a^{-1}\in I\cap H.$
+Finally, because $I$ and $H$ are subgroups, $e\in I$ and $e\in H,$ so
+$e\in H\cap I.$
+
+Therefore, $ab\in H\cap I,$ and $H\cap I$ is a subgroup.
+\endanswer
+
+\problem{48.}
+
+Let $G$ be a group and $g\in G.$ Show that
+$$Z(G) = \{x\in G: gx = xg\hbox{ for all }g\in G\}$$
+is a subgroup of $G.$ This subgroup is called the center of $G.$
+
+\answer
+To show that this is a subgroup, we need to show that $x,y\in Z(G)$
+implies $xy\in Z(G),$ that $x^{-1}\in Z(G),$ and that $e\in Z(G).$
+
+Let $x,y\in Z(G)$ and $z\in G.$ We already know that $xy\in G,$ so we
+will show that $xyz = zxy.$
+Because $y\in Z(G),$ $yz = zy.$
+Similarly, because $x\in Z(G),$ $x(yz) = (zy)x,$ so $xyz = zxy.$
+We have shown that $Z(G)$ is closed under group operations.
+
+We will also show that since $xz = zx,$ we have $x^{-1}z = zx^{-1}.$
+We left multiply and right multiply by $x^{-1}$ to get
+$$x^{-1}xzx^{-1} = x^{-1}zxx^{-1} \to zx^{-1} = x^{-1}z.$$
+
+$e\in Z(G)$ because $eg = g = ge.$
+
+We have now shown that $Z(G)$ is a subgroup.
+\endanswer
+
+\section{Judson 4.5: 2a-c, 5, 23}
+
+\problem{2a-c.}
+Find the order of each of the following elements: (a) $5\in\bb Z_{12},$
+(b) $\sqrt 3\in\bb R,$ and (c) $\sqrt 3\in\bb R^*.$
+
+\answer
+\item{a.}
+Because 5 is coprime to 12, the order of this element is 12.
+
+\item{b.}
+This element has infinite order.
+
+\item{c.}
+This element has infinite order.
+\endanswer
+
+\problem{5.}
+Find the order of every element in $\bb Z_{18}.$
+\answer
+The coprime elements $\{1,5,7,11,13,17\}$ have order 18.
+
+The elements $\{2,4,8,10,14,16\}$ have order 9.
+
+The elements $\{3,9,15\}$ have order 6.
+
+The elements $\{6,12\}$ have order 3.
+
+The element $9$ has order 2.
+
+And the element $0$ has order 1.
+\endanswer
+
+\problem{23.}
+Let $a,b\in G.$ Prove the following statements.
+
+\item{a.} The order of $a$ is the same as the order of $a^{-1}.$
+
+\answer
+Let $a$ be an element of order $n.$
+This means $a^m \neq e$ for $1\leq m < n,$ but $a^n = e.$
+By definition of inverses, $(a^{-1})^n a^n = e \to (a^{-1})^n = e.$
+Similarly, $(a^{-1})^m a^m = e.$
+Where $1\leq m < n,$ we have $a^m \neq e.$
+We assume for the sake of contradiction $(a^{-1})^m = e.$
+We would obtain $a^m = e$ giving a contradiction, so
+$(a^{-1})^m \neq e,$ and $a^{-1}$ is order $n.$
+
+\endanswer
+
+\item{b.} For all $g\in G,$ $|a| = |g^{-1}ag|.$
+
+\answer
+
+We will show that $|a| = |g^{-1}ag|.$
+Thus, we will show $(g^{-1}ag)^n = e$ if and only if $a^n = e.$
+
+$(\Rightarrow)$
+
+Let $a^n = e.$
+$$(g^{-1}ag)^n = g^{-1}a^ng = g^{-1}g = e.$$
+
+$(\Leftarrow)$
+
+Let $(g^{-1}ag)^n = g^{-1}a^ng = e.$
+This implies $g^{-1}a^n = g^{-1},$ so $a^n = e.$
+
+\endanswer
+
+\item{c.} The order of $ab$ is the same as the order of $ba.$
+
+\answer
+
+To show that $|ab| = |ba|,$ we will show that $(ab)^n = e$ if and only
+if $(ba)^n = e.$
+WLOG, we will show that $(ab)^n = e$ only if $(ba)^n = e.$
+
+Let $(ab)^n = e.$
+$$(ab)^{n+1} = (ab)^n(ab) = ab = a(ba)^nb \Longrightarrow (ba)^n = e.$$
+
+\endanswer
+
+\bye
diff --git a/gupta/hw11.tex b/gupta/hw11.tex
new file mode 100644
index 0000000..c2b2b7f
--- /dev/null
+++ b/gupta/hw11.tex
@@ -0,0 +1,134 @@
+\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{\vskip18pt plus 6pt minus 6pt\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{\bigskip\par\penalty-100\item{#1}}
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-04-14 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 11 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Judson 5.4: 1b, 2c, 2p, 3a, 3c, 24}
+
+\problem{1b.}
+
+Write the following permutation in cycle notation:
+$$\pmatrix{1&2&3&4&5\cr 2&4&1&5&3}.$$
+
+\answer
+This is $(1 2 4 5 3).$
+\endanswer
+
+\problem{2c.}
+
+Compute $(1 4 3)(2 3)(2 4).$
+
+\answer
+$$\pmatrix{1&2&3&4\cr 2&3&1&4}$$
+\endanswer
+
+\problem{2p.}
+
+Compute $[(1 2 3 5)(4 6 7)]^{-1}.$
+
+\answer
+$$\pmatrix{1&2&3&4&5&6&7\cr 5&1&2&7&3&4&6}$$
+\endanswer
+
+\problem{3a.}
+Express the following permutation as a product of transpositions and
+identify it as even or odd: $(1 4 3 5 6).$
+
+\answer
+$$(1 4 3 5 6) = (1 4)(4 3)(3 5)(5 6),$$
+and since there are 4 transpositions, this is an even permutation.
+\endanswer
+
+\problem{3c.}
+
+Express the following permutation as a product of transpositions and
+identify it as even or odd: $(1 4 2 6)(1 4 2).$
+
+\answer
+$$(1 4 2 6)(1 4 2) = (1 4)(4 2)(2 6)(1 4)(4 2),$$
+and since there are 5 transpositions, this is an odd permutation.
+
+\problem{24.}
+
+Show that a 3-cycle is an even permutation.
+
+\answer
+Let us have a 3-cycle $(a_1 a_2 a_3).$
+This can be written as $(a_1 a_2)(a_2 a_3)$ because $a_2$ ends in the
+position of $a_3,$ $a_1$ ends in the position of $a_2,$ and $a_3$ ends
+in the position of $a_1.$
+This is two transpositions, so this is an even permutation.
+\endanswer
+
+\section{Judson 6.5: 5d, 5b}
+
+\problem{5d.}
+
+List the left and right cosets of the subgroups of $A_4$ in $S_4.$
+
+\answer
+The left and right cosets are $\{A_4, (1\,2)A_4\}.$
+\endanswer
+
+\problem{5b.}
+
+List the left and right cosets of the subgroups of $\langle 3\rangle$ in
+$U(8).$
+
+\answer
+$\langle 3\rangle = \{{\rm id}, 3\},$ and since this group is abelian,
+it has the same left and right cosets.
+$5\langle 3\rangle = \{5, 7\},$ and this completes the partition of the
+group.
+\endanswer
+
+\section{Problem not from the textbook}
+
+\problem{1.}
+
+Let $H$ be a subgroup of $G$ and suppose that $g_1,g_2\in G.$
+Prove that $g_1H = g_2H$ if and only if $g_2\in g_1H.$
+
+\answer
+Let $H$ be a subgroup of $G$ and $g_1,g_2\in G.$
+We will show that $g_1H = g_2H$ if and only if $g_2\in g_1H.$
+
+$(\Rightarrow)$
+
+Let $g_1H = g_2H.$
+We will show that $g_2\in g_1H.$
+
+$e\in H,$ so $g_2e = g_2\in g_2H,$ and by equality, $g_2\in g_1H.$
+
+$(\Leftarrow)$
+
+Let $g_2 \in g_1H.$
+This means there is $h\in H$ such that $g_2 = g_1h.$
+We then know that $g_2H = g_1hH = g_1H$ because $hH = H$ by group
+closure.
+
+\endanswer
+\bye
diff --git a/gupta/hw12.tex b/gupta/hw12.tex
new file mode 100644
index 0000000..5ae4d54
--- /dev/null
+++ b/gupta/hw12.tex
@@ -0,0 +1,88 @@
+\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{\vskip18pt plus 6pt minus 6pt\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{\bigskip\par\penalty-100\item{#1}}
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-04-26 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 12 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Ross Chapter 2, 8.1d, 8.2c, 8.7b}
+
+\problem{8.1d} Prove that $\lim {n+6\over n^2-6} = 0.$
+
+\answer
+Let $\epsilon > 0.$
+We will show $N$ such that for all $n>N,$ $|{n+6\over n^2-6}-0| <
+\epsilon.$
+
+Where $n>6,$
+$$|{n+6\over n^2-6}| = {n+6\over n^2-6},$$
+and $n^2-6>n^2-36,$ so
+$${n+6\over n^2-6} < {n+6\over n^2-36} = {1\over n-6}.$$
+Choosing $n > N = 6+1/\epsilon,$ and since $n-6 > N-6,$
+$${1\over n-6} < {1\over N-6} = {1\over 6+1/\epsilon-6} = {1\over
+1/\epsilon} = \epsilon,$$
+showing that $\lim {n+6\over n^2-6} = 0.$
+\endanswer
+
+\problem{8.2c} Find the limit of $c_n = {4n+3\over 7n-5}$ and then prove
+your claim.
+
+\answer
+
+$c_n$ converges to $4/7.$
+
+Let $\epsilon > 0.$
+For all
+$$n > N = {41\over 49\epsilon} + {5\over 7},$$
+we can show that $|c_n-4/7| < \epsilon.$
+$$|c_n-4/7| = \left|{4n+3\over 7n-5}-{4\over 7}\right| = \left|{28n+21 -
+(28n-20)\over 49n-35}\right| = \left|{41\over 49n-35}\right|,$$
+and from $n > N,$ (and where $N\geq 1$) $49n-35 > 49N-35 > 0,$ so
+$$\left|{41\over 49n-35}\right| > {41\over 49N-35} =
+{41\over 49({41\over 49\epsilon}+{5\over 7})-35} =
+{41\over {41\over\epsilon} + 35-35} =
+{41\epsilon\over 41} = \epsilon.$$
+We have now shown that $\lim c_n = 4/7.$
+
+\endanswer
+
+\problem{8.7b} Show that $s_n = (-1)^n n$ does not converge.
+
+\answer
+{\bf Disproof.}
+
+Assume for the sake of contradiction that $s_n$ converges to $s.$
+Let $\epsilon = 1.$
+We must have $N$ such that for all $n>N,$ $|s_n-s| < 1$ (implying also
+$|s_{n+1}-s| < 1$).
+However,
+$$|s_n-s_{n+1}| = |s_n-s+s-s_{n+1}| \leq |s_n-s| + |s_{n+1}-s| < 1 + 1 =
+2.$$
+Yet, $|s_{n+1}-s_n| = |(-1)^n(-(n+1)-n)| = 2n+1,$
+and where $n>1,$ we obtain $2n+1>3$ and from the earlier inequality
+$2n+1<2,$ completing a contradiction.
+We have then shown that $s_n$ does not converge.
+\endanswer
+
+\bye
diff --git a/gupta/hw6.tex b/gupta/hw6.tex
new file mode 100644
index 0000000..4362d26
--- /dev/null
+++ b/gupta/hw6.tex
@@ -0,0 +1,227 @@
+\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 6 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Hammack 10: 4, 8, 20, 26}
+Prove the following statements with either induction, strong induction,
+or proof by smallest counterexample.
+
+\problem{4.} If $n\in\bb N,$ then $1\cdot 2 + 2\cdot 3 + 3\cdot 4 +
+4\cdot 5 + \cdots + n(n+1) = {n(n+1)(n+2)\over 3}.$
+
+\answer
+We will show that $\sum_{k=1}^n k(k+1) = {n(n+1)(n+2)\over 3}$ by
+induction.
+
+For the case $n = 1,$ $1\cdot 2 = {1(1+1)(1+2)\over 3} = 6/3 = 2.$
+
+In the inductive case, assume that this statement is true for some $n
+\geq 1,$ giving
+$$\sum_{k=1}^n k(k+1) = {n(n+1)(n+2)\over 3}.$$
+We will show that this statement is true for $n+1:$
+$$\sum_{k=1}^{n+1} k(k+1) = {(n+1)(n+2)(n+3)\over 3}.$$
+This is true only if
+$$\sum_{k=1}^{n+1} k(k+1) - \sum_{k=1}^n k(k+1) = (n+1)(n+2) =
+{(n+1)(n+2)(n+3)\over 3} - {n(n+1)(n+2)\over 3}.$$
+
+$${(n+1)(n+2)(n+3)\over 3} - {n(n+1)(n+2)\over 3} =
+{(n+3-n)(n+1)(n+2)\over 3} = (n+1)(n+2),$$
+thus proving the statement for $n+1.$
+
+Therefore, this identity is true for all $j\in\bb N.$
+\endanswer
+
+\problem{8.} If $n\in\bb N,$ then ${1\over 2!} + {2\over 3!} + {3\over
+4!} + \cdots + {n\over (n+1)!} = 1 - {1\over (n+1)!}$
+
+\answer
+We will show this statement for all $n\in\bb N$ by induction.
+
+In the base case $n = 1,$ ${1\over 2!} = 1/2 = 1 - {1\over 2!},$ so the
+statement is true.
+
+In the inductive case, assume that the statement is true for some $n\geq
+1.$ We will show the statement for $n+1:$
+$${1\over 2!} + {2\over 3!} + \cdots + {n+1\over (n+2)!} = 1 - {1\over
+(n+2)!}.$$
+Then,
+$$1-{1\over (n+2)!} = 1-{n+2\over(n+2)!}+{n+1\over (n+2)!} = 1-{1\over
+(n+1)!} + {n+1\over (n+2)!}.$$
+By the inductive hypothesis,
+$$1-{1\over (n+1)!} + {n+1\over (n+2)!} = {1\over 2!} + {2\over 3!} +
+\cdots + {n\over (n+1)!} + {n+1\over (n+2)!},$$
+proving our inductive step.
+
+Therefore, for all $j\in\bb N,$ this identity is true.
+\endanswer
+
+\problem{20.} Prove that $(1+2+3+\cdots+n)^2 = 1^3 + 2^3 + 3^3 + \cdots
++ n^3$ for every $n\in\bb N.$
+
+\answer
+We will show this identity for all $n\in\bb N$ by induction.
+
+In the base case, $1^2 = 1^3,$ so the statement is immediate.
+
+In the inductive case, assume $(1+2+3+\cdots+k)^2 = 1^3 + 2^3 + 3^3 +
+\cdots + k^3$ for some $k\in\bb N.$
+We will show that $(1+2+3+\cdots+k+(k+1))^2 = 1^3 + 2^3 + 3^3 + \cdots +
+(k+1)^3.$
+
+$$(1+2+\cdots+k+(k+1))^2 = (1+2+\cdots+k)^2 + 2(1+2+\cdots+k)(k+1) +
+(k+1)^2.$$
+With the result $1+2+\cdots+k = {k(k+1)\over 2},$
+we get this equal to
+$$(1+2+\cdots+k)^2 + k(k+1)(k+1) + 1(k+1)^2,$$
+and from the inductive hypothesis, this is equal to
+$$1^3 + 2^3 + \cdots + k^3 + k(k+1)(k+1) + 1(k+1)^2 = 1^3 + 2^3 + \cdots
++ k^3 + (k+1)^3.$$
+
+We have thus proven that for any $n\in\bb N,$ $(1+2+3+\cdots+n)^2 = 1^3
++ 2^3 + 3^3 + \cdots + n^3.$
+\endanswer
+
+\problem{26.} Concerning the Fibonacci sequence, prove that
+$\sum_{k=1}^n F_k^2 = F_nF_{n+1}.$
+
+\answer
+We will show this identity, for all $n\in\bb Z$ s.t. $n\geq 1,$ by
+induction.
+
+For $n=1,$ $F_1^2 = 1 = F_1F_2.$
+
+Assume for some $n\geq 1,$ $\sum_{k=1}^n F_k^2 = F_nF_{n+1}.$
+We will show the statement for $n+1:$
+$$\sum_{k=1}^{n+1} F_k^2 = F_{n+1}F_{n+2}.$$
+We get the following from the definition of the Fibonacci numbers above
+2:
+$$F_{n+1}F_{n+2} = F_{n+1}(F_n + F_{n+1}) = F_{n+1}^2 + F_{n+1}F_n,$$
+and from our inductive hypothesis,
+$$F_{n+1}^2 + F_{n+1}F_n = F_{n+1}^2 + \sum_{k=1}^n F_k^2 =
+\sum_{k=1}^{n+1} F_k^2,$$
+completing our inductive step.
+
+This shows that for all $j\in\bb Z,$ where $j\geq 1,$
+$$\sum_{k=1}^j F_k^2 = F_jF_{j+1}.$$
+\endanswer
+
+\section{Hammack 11.1: 4, 14, 15}
+
+\problem{4.}
+Let $A = \{a,b,c,d\}.$ Suppose $R$ is the relation
+$$\vbox{\halign{&\quad$#$\quad\cr
+    R&=&\{(a,a),(b,b),(c,c),(d,d),(a,b),(b,a),(a,c),(c,a),\cr
+    &&(a,d),(d,a),(b,c),(c,b),(b,d),(d,b),(c,d),(d,c)\}\cr
+}}$$
+Is R Reflexive? Symmetric? Transitive? If a property does not hold, say
+why.
+
+\answer
+This is reflexive, symmetric, and transitive.
+\endanswer
+
+\problem{14.}
+Suppose $R$ is symmetric and transitive relation on set $A,$ and there
+is an element $a\in A$ for which $aRx$ for every $x\in A.$ Prove that
+$R$ is reflexive.
+
+\answer
+For every $x\in A,$ $aRx$ is given and $xRa$ by symmetry. By
+transitivity, $xRx$ for every element, so the relationship is reflexive.
+\endanswer
+
+\problem{15.} Prove or disprove: if a relation is symmetric and
+transitive, then it is also reflexive.
+
+\answer
+The empty relation $\emptyset$ is vacuously symmetric and transitive but
+on a nonempty set containing $x,$ the empty relation doesn't contain
+$(x,x).$
+\endanswer
+
+\section{Hammack 11.2: 6, 8, 10, 12}
+
+\problem{6.} Let $A = \{1,2,3,4,5,6\},$ and consider the following
+equivalence relation on $A:$ $$R =
+\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(2,3),(3,2),(4,5),(5,4),(4,6),
+(6,4),(5,6),(6,5)\}.$$
+List the equivalence classes of $R.$
+
+\answer
+The equivalence classes are $\{2,3\},$ $\{4,5,6\},$ and $\{1\}.$
+\endanswer
+
+\problem{8.} Define a relation $R$ on $\bb Z$ as $xRy$ if and only if
+$x^2+y^2$ is even. Prove $R$ is an equivalence relation. Describe its
+equivalence classes.
+
+\answer
+$R$ is reflexive, symmetric, and transitive, so it is an equivalence
+relation.
+The relation is reflexive because $x^2 + x^2 = 2(x^2),$ and $x^2\in\bb
+Z,$ so the sum is even.
+The relation is symmetric because $x^2 + y^2 = y^2 + x^2,$ so the left
+side is even if and only if the right side is even.
+And for some $j,k\in\bb Z,$ $xRy$ and $yRz$ implies $x^2+y^2 = 2j$ and
+$y^2+z^2 = 2k,$ so $x^2 + y^2 + y^2 + z^2 - 2y^2 = x^2 + z^2 =
+2(j+k-y^2).$
+The number $j+k-y^2\in\bb Z,$ so $xRz,$ and the relation is transitive.
+
+The equivalence classes are the even numbers and the odd numbers.
+\endanswer
+
+\problem{10.} Suppose $R$ and $S$ are two equivalence relations on a set
+$A.$ Prove that $R\cap S$ is also an equivalence relation.
+
+\answer
+The equivalence relations $R$ and $S$ on $A$ are, by definition,
+reflexive, symmetric, and transitive.
+$R\cap S$ is therefore reflexive because $x\in A$ implies $(x,x)\in R$
+and $(x,x)\in S,$ (by reflection), so $(x,x)\in R\cap S.$
+
+Only if $xRy$ and $xSy,$ $x(R\cap S)y,$ and by symmetry, $yRx$ and
+$ySx,$ so $y(R\cap S)x,$ giving symmetry of the new relation.
+
+Similarly, if $x(R\cap S)y$ and $y(R\cap S)z,$ $xRy$ and $xRz,$ so by
+transitivity, $xRz,$ and similarly, $xSz,$ so $x(R\cap S)z,$ making the
+intersection of any equivalence relations also an equivalence relation.
+\endanswer
+
+\problem{12.} Prove or disprove: If $R$ and $S$ are two equivalence
+relations on a set $A,$ then $R\cup S$ is also an equivalence relation
+on $A.$
+
+\answer
+{\bf Disproof.}
+
+Let $R$ be a relation with equivalence classes $\{1,2\}, \{3\}$ and $S$
+be a relation with equivalence classes $\{1\}, \{2,3\}.$ $\{(1,2),
+(2,3)\} \subseteq R\cup S,$ but $(1,3)\not\in R\cup S$ because
+$1\mathbin{\not\hskip-3pt R} 3$ and $1\mathbin{\not\hskip-2pt S} 3.$
+\endanswer
+
+\bye
diff --git a/gupta/hw7.tex b/gupta/hw7.tex
new file mode 100644
index 0000000..231e3fd
--- /dev/null
+++ b/gupta/hw7.tex
@@ -0,0 +1,294 @@
+\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
+2022-03-03 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 7 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Hammack 11.3: 2, 4}
+
+\problem{2.}
+List all the partitions of the set $A = \{a,b,c\}.$ Compare your answer
+to the answer to Exercise 6 of Section 11.2
+\answer
+All partitions are $\{\{\{a,b,c\}\}, \{\{a,b\},\{c\}\},
+\{\{a\},\{b,c\}\}, \{\{b\}, \{a,c\}\}, \{\{a\},\{b\},\{c\}\}\}.$
+There is exactly one of these for each of the equivalence relations on
+exercise 6.
+\endanswer
+
+\problem{4.}
+Suppose $P$ is a partition of set $A.$ Define a relation $R$ on $A$ by
+declaring $xRy$ if and only if $x, y \in X$ for some $X\in P.$ Prove $R$
+is an equivalence relation on $A.$
+Then prove that $P$ is the set of equivalence classes of $R.$
+\answer
+We want to show that $R$ is an equivalence relation. We will show that
+it is reflexive, symmetric, and transitive.
+
+\smallskip
+(Reflexive)
+
+Let $a\in A.$ We will show that $aRa.$
+$P$ is a partition of $A,$ so $\bigcup_{p\in P} p = A,$ and so there
+must be $X\in A$ s.t. $a,a\in X.$
+We thus obtain $aRa.$
+
+\smallskip
+(Symmetric)
+
+Let $a,b\in A$ such that $aRb.$ We will show that $bRa.$
+From the definition of $R,$ there is $X\in P$ s.t. $a\in X$ and $b\in
+X.$
+Since $b,a\in X,$ $bRa.$
+
+\smallskip
+(Transitive)
+Let $a,b,c\in A$ such that $aRb$ and $bRc.$ We will show tht $aRc.$
+From the definition of $R,$ there are $X,Y\in P$ s.t. $a,b\in X$ and
+$b,c\in Y,$ but if we assume $X\neq Y,$ since $P$ is a partition, $X\cap
+Y = \emptyset,$ so either $b\not\in X$ or $b\not\in Y,$ so we must know
+that $X = Y.$
+Therefore, $c\in X,$ ($a,c\in X$) and $aRc.$
+
+\medskip
+We also want to show that the equivalence classes of $R$ is $P.$
+Let $Q$ be the equivalence classes of $R,$ defined as
+$\{\{x\in A: xRa\} : a\in A\}.$
+We will show that $P = Q.$
+
+\smallskip
+$(Q\subseteq P)$
+
+Let $a\in A.$
+By definition of $Q,$ $X = \{x\in A: xRa\} \in Q.$
+$xRa$ iff $x,a\in Y$ for some $Y\in P.$
+
+We will show that $X\subseteq Y$ and $Y\subseteq X$ to show that $X =
+Y$ and thus $X\in P.$
+
+$x\in X$ gives $xRa,$ so $x\in Y,$ so $X\subseteq Y.$
+$x\in Y$ gives $xRa,$ so $x\in X,$ and $Y\subseteq X.$
+Therefore, $X = Y,$ and we have shown $Q\subseteq P.$
+
+\smallskip
+$(Q\supseteq P)$
+
+Let $a\in A.$
+As established earlier, for some $X\in P,$ the element $a\in X,$ and
+$xRa$ iff $x\in X,$ so $X = \{x\in A: xRa\} \in Q,$ proving $P\subseteq
+Q.$
+
+We have now shown that $R$ is an equivalence relation and the
+equivalence classes of $R$ is $P.$
+
+\endanswer
+
+\section{Hammack 11.4: 4, 6}
+
+\problem{4.}
+Write the addition and multiplication tables for $\bb Z_6.$
+\answer
+
+{\tabskip5pt
+\line{\hfil
+\vbox{\halign{&\strut $#$\cr
+    \times &[0]&[1]&[2]&[3]&[4]&[5]\cr
+    [0]    &[0]&[0]&[0]&[0]&[0]&[0]\cr
+    [1]    &[0]&[1]&[2]&[3]&[4]&[5]\cr
+    [2]    &[0]&[2]&[4]&[0]&[2]&[4]\cr
+    [3]    &[0]&[3]&[0]&[3]&[0]&[3]\cr
+    [4]    &[0]&[4]&[2]&[0]&[4]&[2]\cr
+    [5]    &[0]&[5]&[4]&[3]&[2]&[1]\cr
+}}
+\hfil
+\vbox{\halign{&\strut $#$\cr
+    +  &[0]&[1]&[2]&[3]&[4]&[5]\cr
+    [0]&[0]&[1]&[2]&[3]&[4]&[5]\cr
+    [1]&[1]&[2]&[3]&[4]&[5]&[0]\cr
+    [2]&[2]&[3]&[4]&[5]&[0]&[1]\cr
+    [3]&[3]&[4]&[5]&[0]&[1]&[2]\cr
+    [4]&[4]&[5]&[0]&[1]&[2]&[3]\cr
+    [5]&[5]&[0]&[1]&[2]&[3]&[4]\cr
+}}\hfil
+}}
+\endanswer
+
+\problem{6.}
+Suppose $[a],[b]\in\bb Z_6,$ and $[a]\cdot[b]=[0].$ Is it necessarily
+true that either $[a] = [0]$ or $[b]=[0]?$
+
+\answer
+No, it isn't. $[4],[3]\in\bb Z_6,$ and $[4]\neq [0]$ and $[3]\neq [0],$
+but $[4]\cdot[3]=[12]=[0].$
+\endanswer
+
+\section{Hammack 12.1: 4, 6, 9, 12}
+
+\problem{4.}
+There are eight different functions $f: \{a,b,c\}\to\{0,1\}.$ List them
+all. Diagrams will suffice.
+\answer
+$\{(a,0), (b,0), (c,0)\}$
+
+$\{(a,0), (b,0), (c,1)\}$
+
+$\{(a,0), (b,1), (c,0)\}$
+
+$\{(a,0), (b,1), (c,1)\}$
+
+$\{(a,1), (b,0), (c,0)\}$
+
+$\{(a,1), (b,0), (c,1)\}$
+
+$\{(a,1), (b,1), (c,0)\}$
+
+$\{(a,1), (b,1), (c,1)\}$
+
+\endanswer
+
+\problem{6.}
+Suppose $f: \bb Z \to \bb Z$ is defined as $f = \{(x,4x+5) : x\in\bb
+Z\}.$ State the domain, codomain and range of $f.$ Find $f(10).$
+\answer
+The domain and codomain are $\bb Z,$ and the range is $\{4x+5 : x\in\bb
+Z\}.$
+$f(10) = 45.$
+\endanswer
+
+\problem{9.}
+Consider the set $f=\{(x^2,x), x\in\bb R\}.$ Is this a function from
+$\bb R$ to $\bb R?$ Explain.
+\answer
+$(1,-1)\in f,$ and $(1,1)\in f,$ so this is not a function because it
+has multiple outputs for one input.
+\endanswer
+
+\problem{12.}
+Is the set $\theta = \{\left((x,y),(3y,2x,x+y)\right) : x,y\in\bb R\}$ a
+function? If so, what is its domain, codomain, and range?
+\answer
+The domain is $\bb R^2,$ the codomain is $\bb R^3,$ and the range is
+$\{(3y,2x,x+y) : x,y\in\bb R\} = \{(x,y,z)\in\bb R^3 : 6z-3y-2x = 0\}.$
+\endanswer
+
+\section{Hammack 12.2: 4, 12, 14}
+
+\problem{4.}
+A function $f: \bb Z \to \bb Z \times \bb Z$ is defined as $f(n) = (2n,
+n+3).$ Verify whether this function is injective and whether it is
+surjective.
+\answer
+This function is injective but not surjective.
+
+\smallskip
+(Injective)
+
+Let $m,n\in\bb Z$ such that $f(n)=f(m)$ or $(2n,n+3)=(2m,m+3).$
+Therefore, $2n=2m \to n = m.$
+
+\smallskip
+(Surjective)
+
+$(0, 4) \not\in f(\bb Z),$ so this function is not surjective.
+
+\endanswer
+
+\problem{12.}
+Consider the function $\theta:\{0,1\}\times\bb N \to \bb Z$ defined as
+$\theta(a,b) = a-2ab+b.$ Is $\theta$ injective? Is it surjective?
+Bijective? Explain.
+\answer
+
+\smallskip
+(Injective)
+
+Let $a,c\in\{0,1\}$ and $b,d\in\bb N$ such that $\theta(a,b) =
+\theta(c,d).$ We will show that $a=c$ and $b=d.$
+We will deal with three cases:
+
+($a=1$ and $c=1$)
+
+$\theta(1,b) = 1-2b+b = 1-2d+d = \theta(1,d).$
+The equality $1-2b+b = 1-2d+d$ becomes $b = d,$ showing that the
+function is injective.
+
+($a=0$ and $c=0$)
+
+$\theta(0,b) = b = d = \theta(0,d),$ easily showing that $b=d$ and the
+function is injective.
+
+(WLOG, $a=1$ and $c=0$)
+
+$\theta(1,b) = 1-2b+b = d = \theta(0,d).$
+With $b,d\in\bb Z,$ $b+d \geq 2,$ so the rearranged equation
+$1 = b+d$ is impossible.
+
+\smallskip
+(Surjective)
+
+Let $n\in\bb Z.$ We will show that for some $a\in\{0,1\}$ and $b\in\bb
+N,$ $\theta(a,b) = n.$
+We will deal with two cases:
+
+($n > 0$)
+
+Let $a = 0$ and $b = n.$
+$\theta(0,n) = 0-2(0)(n)+n = n.$
+
+($n \leq 0$)
+
+Let $a = 1$ and $b = -n+1.$
+$\theta(1, -n+1) = 1-2(1)(-n+1) + (-n+1) = 1+2n-2-n+1 = n.$
+
+Because $\theta$ is both injective and surjective, it is bijective.
+\endanswer
+
+\problem{14.}
+Consider the function $\theta: {\cal P}(\bb Z) \to {\cal P}(\bb Z)$
+defined as $\theta(X) = \bar X.$ Is $\theta$ injective? Is it
+surjective? Bijective? Explain.
+\answer
+
+$\theta$ is injective and surjective.
+
+\smallskip
+(Injective)
+
+Let $X$ and $Y$ be subsets of $\bb Z$ such that $\theta(X) = \theta(Y).$
+$\bar X = \bar Y$ implies $\bar{\bar X} =
+\bar{\bar Y}$ and $X = Y.$
+
+\smallskip
+(Surjective)
+
+Let $X$ be a subset of $\bb Z.$ $\bar X$ maps through $\theta$ to
+$X.$ $\theta(\bar X) = \bar{\bar X} = X.$
+
+We have shown that $\theta$ is injective, surjective, and, therefore,
+bijective.
+
+\endanswer
+
+\bye
diff --git a/gupta/hw8.tex b/gupta/hw8.tex
new file mode 100644
index 0000000..400ab7b
--- /dev/null
+++ b/gupta/hw8.tex
@@ -0,0 +1,312 @@
+\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{\vskip18pt plus 6pt minus 6pt\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{\bigskip\par\penalty-100\item{#1}}
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-03-17 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 8 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Hammack 12.4: 2, 4, 6, 8}
+
+\problem{2.} Suppose $A = \{1,2,3,4\},$ $B = \{0,1,2\},$ $C=\{1,2,3\}.$
+Let $f: A\to B$ be
+$$f = \left\{(1,0), (2,1), (3,2), (4,0) \right\},$$
+and $g: B\to C$ be $g = \left\{(0,1),(1,1),(2,3)\right\}.$ Find
+$g\circ f.$
+
+\answer
+
+$g\circ f = \left\{(0,0), (1,0), (2,2)\right\}.$
+
+\endanswer
+
+\problem{4.} Suppose $A = \{a,b,c\}.$ Let $f: A\to A$ be the function $f
+= \left\{(a,c), (b,c), (c,c)\right\},$ and let $g: A\to A$ be the
+function $g = \left\{(a,a), (b,b), (c,a)\right\}.$ Find $g\circ f$ and
+$f\circ g.$
+
+\answer
+
+$g\circ f = \left\{(a,a), (b,a), (c,a)\right\},$ and $f\circ g =
+\left\{(a,c), (b,c), (c,c)\right\}.$
+
+\endanswer
+
+\problem{6.} Consider the functions $f,g: \bb R\to \bb R$ defined as
+$f(x) = {1\over x^2+1}$ and $g(x) = 3x+2.$ Find the formulas for $g\circ
+f$ and $f\circ g.$
+
+\answer
+
+$f\circ g = {1\over (3x+2)^2+1}$ and $g\circ f = {3\over x^2+1}+2.$
+
+\endanswer
+
+\problem{8.} Consider the functions $f,g: \bb Z\times \bb Z \to \bb
+Z\times \bb Z$ defined as $f(m,n) = (3m-4n, 2m+n)$ and $g(m,n) = (5m+n,
+m).$ Find the formulas for $g\circ f$ and $f\circ g.$
+
+\answer
+
+$g\circ f = (3m-4(5m+n), 2m+n)$ and $f\circ g = (5(2m+n)+(3m-4n), 2m+n).$
+
+\section{Hammack 12.5: 6, 8}
+
+\problem{6.}
+The function $f: \bb Z\times\bb Z \to \bb Z\times\bb Z$ defined by the
+formula $f(m,n) = (5m+4n,4m+3n)$ is bijective. Find its inverse.
+
+\answer
+$f(m,n) = (r,s) = (5m+4n,4m+3n).$ Thus, $4r-5s = n$ and $4s-3r = m,$
+so the inverse is $g(r,s) = (4r-5s, 4s-3r).$
+\endanswer
+
+\problem{8.} Is the function $\theta: {\cal P}(\bb Z)\to {\cal P}(\bb
+Z)$ defined as $\theta(X) = \overline X$ bijective? If so, what is its
+inverse?
+
+\answer
+Yes, $\theta$ is bijective. This was proven in the last homework.
+
+The inverse $\theta^{-1}$ is $\theta$ since $\theta(\theta(X)) = X.$
+\endanswer
+
+\section{Hammack 12.6: 6, 7, 8, 12, 14}
+
+\problem{6.} Given a function $f: A\to B$ and a subset $Y\subseteq B,$
+is $f(f^{-1}(Y)) = Y$ always true? Prove or give a counterexample.
+
+\answer
+Let $f: \bb R\to\bb R$ be $f(x) = 1.$ $f(f^{-1}(\{2\})) = f(\emptyset) =
+\emptyset \neq \{2\}.$
+\endanswer
+
+\problem{7.} Given a function $f: A\to B$ and subsets $W,X\subseteq A,$
+prove $f(W\cap X)\subseteq f(W)\cap f(X).$
+
+\answer
+Let there be a function $f: A\to B$ and subsets $W, X\subseteq A.$
+Let $b\in f(W\cap X).$ We will show that $b\in f(W)\cap f(X)$ to show
+that $f(W\cap X)\subseteq f(W)\cap f(X).$
+From the definition of image, there is $a\in W\cap X$ s.t. $b = f(a).$
+Since $a\in W,$ $b \in f(W),$ and from $a\in X,$ $b\in f(X).$
+Therefore, $b\in f(W)\cap f(X).$
+\endanswer
+
+\problem{8.} Given a function $f: A\to B$ and subsets $W,X\subseteq A,$
+then $f(W\cap X) = f(W)\cap f(X)$ is false in general. Produce a
+counterexample.
+
+\answer
+Let $W = \{1\}$ and $X = \{2\},$ and $f(x) = 0.$
+We have $W\cap X = \emptyset,$ so $f(W\cap X) = \emptyset.$
+However, $f(W) = f(W) = \{0\} = f(W)\cap f(X).$
+This disproves $f(W\cap X) = f(W)\cap f(X)$ in general.
+\endanswer
+
+\problem{12.} Consider $f: A\to B.$ Prove that $f$ is injective if and
+only if $X = f^{-1}(f(X))$ for all $X\subseteq A.$ Prove that $f$ is
+surjective if and only if $f(f^{-1}(Y)) = Y$ for all $Y\subseteq B.$
+
+\answer
+
+Let $f: A\to B.$ We will show that $f$ is injective if and only if $X =
+f^{-1}(f(X))$ for all $X\subseteq A.$
+
+
+\smallskip
+$(\Rightarrow)$
+
+We will show this by contraposition.
+Let there be $X\in A$ such that $f^{-1}(f(X)) \neq X.$
+We will show that $f$ is not injective by showing that for some $x, y\in
+A,$ where $x\neq y,$ $f(x) = f(y).$
+
+For all $x\in X,$ we have $y = f(x) \in f(X)$ and $f(x) = y,$ so $x\in
+f^{-1}(f(X)),$
+Therefore, we know there is $z\not\in X$ such that $z\in f^{-1}(f(X)).$
+Therefore, $f(z) \in f(X),$ and for some $w\in X,$ $f(z) = f(w),$ and
+$w\neq z,$ so $f$ is not injective.
+
+\smallskip
+$(\Leftarrow)$
+
+Assume that for all $X\subseteq A,$ $f^{-1}(f(X)) = X.$
+We will show that $f$ is injective.
+Let $x,y\in A$ such that $f(x) = f(y).$
+We will show that $x = y.$
+
+From $f(x) = f(y),$ we know $f(\{x\}) = f(\{y\}),$ so $f^{-1}(f(\{x\}))
+= \{x\} = \{y\} = f^{-1}(f(\{y\})),$
+so $x = y.$
+
+
+\medskip
+We will now show that $f$ is surjective if and only if $f(f^{-1}(Y)) =
+Y$ for all $Y\subseteq B.$
+
+\smallskip
+$(\Rightarrow)$
+
+Let $f$ be surjective.
+Let $Y\subseteq B.$
+We will show that $f(f^{-1}(Y)) = Y.$
+
+Since $f$ is surjective, for all $y\in Y,$ there is $x\in A$ such that
+$f(x) = y,$ so $f(f^{-1}(Y)) \supseteq Y.$
+Also, for all $z\in f(f^{-1}(Y)),$ we know $z = f(x)$ where $f(x) = y$
+for some $y\in Y.$
+From these, $z = y \in Y,$ so $f(f^{-1}(Y)) \subseteq Y$ and therefore
+$f(f^{-1}(Y)) = Y.$
+
+\smallskip
+$(\Leftarrow)$
+
+Assume that for all $Y\subseteq B,$ $f(f^{-1}(Y)) = Y.$
+We will show that $f$ is surjective.
+To show that $f$ is surjective, we take $b\in B$ and show that we have
+$a\in A$ such that $f(a) = b.$
+
+$f(f^{-1}(\{b\})) = \{b\},$ and $f^{-1}(\{b\})$ is not empty because if
+it were, $f(f^{-1}(\{b\})) = f(\emptyset) = \emptyset \neq \{b\}.$
+Therefore, we pick $a\in f^{-1}(\{b\})$ and note that $f(a) = b,$ so $f$
+is surjective.
+
+\endanswer
+
+\problem{14.} Let $f: A\to B$ be a function, and $Y\subseteq B.$ Prove
+or disprove: $f^{-1}(f(f^{-1}(Y))) = f^{-1}(Y).$
+
+\answer
+Let $f: A\to B$ be a function, and $Y\subseteq B.$
+We will show that $f^{-1}(f(f^{-1}(Y))) = f^{-1}(Y).$
+
+$(\subseteq)$
+
+Let $x\in f^{-1}(f(f^{-1}(Y))).$
+We will show that $x\in f^{-1}(Y).$
+
+From definition of preimage, $f(x) \in f(f^{-1}(Y)),$ and from
+definition of image, $x\in f^{-1}(Y).$
+
+$(\supseteq)$
+
+Let $x\in f^{-1}(Y).$
+We will show that $x\in f^{-1}(f(f^{-1}(Y))).$
+
+From definition of preimage, for some $y\in Y,$ $f(x) = y,$ and
+$y\in f(f^{-1}(Y)).$
+And thus we have $x\in f^{-1}(f(f^{-1}(Y))).$
+We have shown that $f^{-1}(f(f^{-1}(Y))) = f^{-1}(Y).$
+
+\endanswer
+
+\section{Hammack 13.1: 2, 6, 14}
+
+\problem{2.} Show that sets $\bb R$ and $(\sqrt 2,\infty)$ have equal
+cardinality by describing a bijection from one to the other.
+
+\answer
+Let $f: \bb R \to (\sqrt 2,\infty)$ be $f(x) = e^x+\sqrt 2.$ We will
+show that $f$ is surjective, injective, and therefore bijective.
+
+\smallskip
+(Injective)
+
+Let $x,y\in\bb R$ such that $f(x) = f(y).$ We will show that $x = y.$
+$e^x+\sqrt 2 = e^y+\sqrt 2,$ so $e^x = e^y,$ and $x = y.$
+We have shown that $f$ is injective.
+
+\smallskip
+(Surjective)
+
+Let $y\in (\sqrt 2,\infty).$
+We will show there exists $x\in\bb R$ such that $f(x) = y.$
+We set $x$ to $\ln(y-\sqrt2),$ and obtain $f(x) =
+e^{\ln(y-\sqrt2)}+\sqrt2 = y-\sqrt2+\sqrt2 = y.$
+We have shown that $f$ is surjective.
+
+We now know that $f$ is bijective.
+\endanswer
+
+\problem{6.} Show that sets $\bb N$ and $S = \{{\sqrt 2\over n}: n\in\bb
+N\}$ have equal cardinality by describing a bijection from one to the
+other.
+
+\answer
+Let $f: \bb N \to S$ be defined $f(n) = {\sqrt 2\over n}.$
+
+\smallskip
+(Injective)
+
+Let $m,n\in\bb N$ such that $f(m) = f(n).$
+We will show that $m = n.$
+We obtain ${\sqrt2\over m} = {\sqrt 2\over n},$ and thus $m = n.$
+We have shown that $f$ is injective.
+
+\smallskip
+(Surjective)
+
+Let $s\in S.$
+We will show that there exists $n\in\bb N$ so that $f(n) = s.$
+By definition of $S,$ we know that there is $m\in\bb N$ such that $s =
+{\sqrt 2\over m}.$
+Letting $n = m,$ we have $f(n) = {\sqrt 2\over n} = s.$
+We have shown that $f$ is surjective.
+
+We now know that $f$ is bijective.
+\endanswer
+
+\problem{14.} Show that sets $\bb N\times\bb N$ and $S=\{(n,m)\in\bb
+N\times\bb N : n\leq m\}$ have equal cardinality by describing a
+bijection from one to the other.
+
+\answer
+Let us have $f:\bb N\times\bb N\to S$ defined as $f(n,m) = (n, n+m-1).$
+We will show that $f$ is bijective by showing that $f$ is injective and
+surjective.
+This will show that $|\bb N\times\bb N| = |S|.$
+
+\smallskip
+(Injective)
+
+Let $(n,m), (k,l)\in\bb N\times\bb N$ such that $f(n,m) = f(k,l).$
+$(n, n+m-1) = (k, k+l-1) \to n = k, n+m-1 = k+l-1.$
+Substituting $n = k$ into the latter constraint, $m = l.$
+Thus, $(n,m) = (k,l),$ and $f$ is injective.
+
+\smallskip
+(Surjective)
+
+Let $(n,m)\in S.$
+$(n, m-n+1)\in \bb N\times \bb N$ because $n\in\bb N$ from definition of
+$S$ and $n\leq m\to 1\leq m-n+1$ (and integer closure).
+
+$$f(n, m-n+1) = (n, n+(m-n+1)-1) = (n,m),$$
+so $f$ is surjective.
+
+We now know that $f$ is bijective.
+
+\endanswer
+
+\bye
diff --git a/gupta/hw9.tex b/gupta/hw9.tex
new file mode 100644
index 0000000..161e2a1
--- /dev/null
+++ b/gupta/hw9.tex
@@ -0,0 +1,192 @@
+\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{\vskip18pt plus 6pt minus 6pt\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{\bigskip\par\penalty-100\item{#1}}
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-03-17 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 8 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Hammack 13.2: 4, 6, 8, 13}
+
+\problem{4.}
+Prove that the set of all irrational numbers is uncountable.
+(Suggestion: consider proof by contradiction using Theorems 13.4 and
+13.6)
+
+\answer
+Theorem 13.4 tells us that the rationals are a countably infinite set.
+Assume for the sake of contradiction that the irrational numbers are
+countable.
+By definition of the irrational numbers, the union of the rationals and
+irrationals are the real numbers.
+By theorem 13.6, the real numbers must then be a countable set.
+However, we know that $\bb R$ is uncountably infinite, giving a
+contradiction.
+Therefore, the irrational numbers are uncountable.
+\endanswer
+
+\problem{6.}
+Prove or disprove: There exists a bijective function $f: \bb Q\to\bb
+R.$
+
+\answer
+{\bf Disproof.}
+
+By the textbook, $\bb Q$ is countably infinite, so we have a bijection
+$g: \bb N\to \bb Q.$
+For the sake of contradiction, assume there exists a bijective function
+$f: \bb Q\to \bb R.$
+
+We now have a bijection $g\circ f:\bb N\to \bb R,$ because the
+composition of bijections is bijective.
+However, this function cannot be surjective because we can construct a
+real number $r$ not in the image of $g\circ f$ by setting the $i$th
+digit of $r$ to be different from the $i$th digit of $g(f(i))$ (e.g. by
+choosing 0 unless the $i$th digit of $g(f(i))$ is 0).
+We have reached a contradiction, so $f$ must not exist.
+
+\endanswer
+
+\problem{8.}
+Prove or disprove: The set $\bb Z\times\bb Q$ is countably infinite.
+
+\answer
+{\bf Proof.}
+
+$\bb Q$ is countably infinite, and $\bb Z$ is countably infinite because
+$\bb Z$ is the union of $\{1-n : n\in\bb N\}\cup \bb N,$ ($\bb Z$ is
+countably infinite since the union of countably infinite sets is
+countably infinite) so by a theorem proved in the textbook, $\bb
+Z\times\bb Q$ is countably infinite.
+
+\endanswer
+
+\problem{13.}
+Prove or disprove: If $A = \{X\subseteq\bb N : X\hbox{ is
+finite}\},$ then $|A| = \aleph_0.$
+
+\answer
+{\bf Proof.}
+
+To all finite sets of natural numbers, we can bijectively assign a
+corresponding whole number ($\bb N\cup\{0\}$) defined by its binary
+representation (if $i\in X,$ the $i$th least significant bit is one,
+otherwise zero).
+This is necessarily surjective because all finite numbers have a finite
+number of ones in binary and thus a corresponding set in $A.$
+It is also injective because any sets which differ by an element must
+also differ in their binary representation by a one or a zero.
+
+Therefore $|A| = 1+|\bb N| = \aleph_0.$
+
+\endanswer
+
+\section{Hammack 13.3: 8, 10}
+
+\problem{8.}
+Prove or disprove: If $A\subseteq B$ and $A$ is countably infinite and
+$B$ is uncountable, then $B-A$ is uncountable.
+
+\answer % assume contrapositive. b-a is countable.
+We will prove this by assuming the contrapositive and showing that $B$
+is countable.
+Let $B-A$ and $A\subseteq B$ be countable sets.
+By theorem 13.6 and $(B-A)\cup A = B,$ (since $A\subseteq B$) $B$ is
+countable.
+\endanswer
+
+\problem{10.}
+Prove that if $A$ and $B$ are finite sets with $|A|=|B|,$ then any
+surjection $f:A\to B$ is also an injection. Show this is not necessarily
+true if $A$ and $B$ are not finite.
+
+\answer
+
+Let $A$ and $B$ be finite sets with $|A|=|B|=n.$
+We enumerate the elements of $A$ as $a_1,a_2,\ldots,a_n.$
+Let us have some surjection $f:A\to B.$
+We then let $f(a_i) = b_i.$
+Since this is a surjection, all $n$ elements of $B$ must be listed in
+$b_1,b_2,\ldots,b_n.$
+If $b_i = b_j,$ we must have $i=j$ because otherwise we have
+double-counted an element in $B,$ so $B$ would contain less than $n$
+elements.
+Then, we have $f(a_i) = f(a_j)$ implies $i = j$ so $a_i = a_j.$
+
+\endanswer
+
+\section{Hammack 13.4: 4, 5}
+
+\problem{4.}
+Let ${\cal F}$ be the set of all functions $\bb R\to\{0,1\}.$ Show that
+$|\bb R|<|{\cal F}|.$
+
+\answer
+We will show that $|\bb R| < |{\cal F}|$ by showing a bijection between
+${\cal F}$ and ${\cal P}(\bb R).$
+Let $f,g\in{\cal F}.$
+Let us define corresponding sets $R,S.$ $r\in R$ if and only if
+$f(r)=1,$ and $S$ similarly defined with respect to $g.$
+This is injective because if $R=S,$ $f(r)=g(r)$ for all reals, so $f=g.$
+This is surjective because any $R\subseteq\bb R$ has a corresponding $f$
+where $f(r) = 1$ if and only if $r\in R.$
+\endanswer
+
+\problem{5.}
+Consider the subset $B = \{(x,y): x^2+y^2\leq 1\}\subseteq\bb R^2.$ Show
+that $|B| = |\bb R^2|.$
+
+\answer
+
+We will show this by the Cantor-Bernstein-Schr\"oeder theorem.
+We start with the trivial injection $f:B\to \bb R^2,$ defined $f(x,y) =
+(x,y).$
+
+Then we use the injection $g:\bb R^2\to B,$ defined $g(x,y) =
+{(x,y)\over 1+(x^2+y^2)}.$
+We will show this is injective.
+Let $g(x,y) = g(w,z).$
+${(x,y)\over 1+(x^2+y^2)} = {(w,z)\over 1+(w^2+z^2)}.$
+
+${(x,y)\over 1+(x^2+y^2)} = {(w,z)\over 1+(w^2+z^2)}.$
+
+\endanswer
+
+\section{Problem not from the textbook}
+
+\problem{1.} Write out the multiplication table for the group $(\bb
+Z_{12}^*,
+\cdot).$ (Recall that $\bb Z_{12}^* = \{[k]:\gcd(k,12)=1\}.$)
+
+\answer
+\halign{\strut#\quad\vrule&&\quad#\quad\cr
+    $*$ &1 &5 &7 &11\cr\noalign{\hrule}
+    1  &1 &5 &7 &11\cr
+    5  &5 &1 &11&7 \cr
+    7  &7 &11&1 &5 \cr
+    11 &11&7 &5 &1 \cr
+}
+
+\endanswer
+
+\bye
diff --git a/gupta/portfolio.tex b/gupta/portfolio.tex
new file mode 100644
index 0000000..25b41b3
--- /dev/null
+++ b/gupta/portfolio.tex
@@ -0,0 +1,288 @@
+\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
+
+\def\clap#1{\hbox to 0pt{\hss #1 \hss}}
+\headline{\vtop{\hbox to \hsize{\strut\rlap{Math 2106 - Dr. Gupta}\hfil
+\clap{\bf Portfolio}\hfil\llap{Spring 2022}}\hrule height .5pt}}
+
+\def\section#1{\medskip\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in%
+\noindent{\bf #1}}
+\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
+depth 0pt{\parfillskip0pt\medskip}}
+\def\problem#1{\medskip\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in\item{#1}}
+\def\nmid{\hskip-3pt\not\hskip2.5pt\mid}
+
+\problem{1.} Direct proof
+
+Suppose $x,y\in\bb Z.$ If $x$ and $y$ are odd, then $xy$ is odd.
+(Homework 3---Hammack 4.4)
+
+{\bf Proof.}
+
+Let $x,y\in\bb Z$ such that $x$ and $y$ are odd.
+We will show that $xy$ is odd.
+From the definition of odd we have integers $j,k\in\bb Z$ such that
+$x=2j+1$ and $y=2k+1.$
+$$xy = (2j+1)(2k+1) = 4jk+2j+2k+1 = 2(2jk+j+k) + 1.$$
+
+By integer closure, $2jk+j+k\in\bb Z,$ so by the definition of odd, we
+have shown that $xy$ is odd.
+We have shown that the product of odd integers is odd.
+\endproof
+
+\problem{2.} Contrapositive proof
+
+Prove that if $n^2$ is even, then $n$ is even. (Quiz 3)
+
+{\bf Proof.}
+
+We will show this by contraposition.
+Let $n$ be an odd number (i.e. not even).
+We want to show that $n^2$ is an odd number.
+Since $n$ is odd, there must exist $j\in\bb Z$ such that $n=2j+1.$
+$$n^2 = (2j+1)^2 = 4j^2+4j+1 = 2(2j^2+2j) + 1.$$
+By integer closure, $2j^2+2j\in\bb Z,$ so we have shown that $n^2$ is
+odd (i.e. not even).
+We have shown that if $n^2$ is even, then $n$ is even.
+\endproof
+
+\problem{3.} Proof by contradiction
+
+$\sqrt 6$ is irrational. (Homework 4---Hammack 6.4)
+
+{\bf Proof.}
+
+Assume for the sake of contradiction that $\sqrt 6$ is rational.
+We then have that $\sqrt 6 = {p\over q}$ for some $p,q\in\bb Z$ s.t.
+$\gcd(p,q) = 1$ (there is no $m>1$ such that $m\mid p$ and $m\mid q$).
+
+Squaring both sides, we find
+$$6 = {p^2\over q^2} \to p^2 = 6q^2 = 2(3q^2),$$ from which $3q^2\in\bb
+Z$ tells us $2\mid p^2.$
+
+{\bf Lemma.}
+We will show that $2\mid p^2$ implies $2\mid p$ by contrapositive.
+Let $2\nmid p,$ so $p = 2k + 1$ for some $k\in\bb Z.$
+We then compute $p^2 = 4k^2 + 4k + 1 = 2(2k^2+2k) + 1,$ and
+$2k^2+2k\in\bb Z,$ so $2\nmid p^2.$
+\endproof
+
+By the lemma, $2\mid p$ and $4\mid p^2,$ so there exists $k\in\bb Z$
+s.t. $p^2 = 4k.$
+We then get $4k = 6q^2$ and $2k = 3q^2.$ % TODO missing proof that 2k=nm.
+So $q^2$ must be even, and as we've shown $2\mid q^2$ gives $2\mid q.$
+Thus, $\gcd(p,q) = 2 \neq 1,$ giving a contradiction, meaning $\sqrt 6$
+is irrational.
+\endproof
+
+\problem{4.} Equivalence proof
+
+Let $A$ and $B$ be sets. Prove that $A\subseteq B$ if and only if $A\cap
+B = A.$ (Homework 5---Hammack 8.22)
+
+{\bf Proof.}
+
+$(\Rightarrow)$
+
+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 $A\subseteq A\cap B.$
+Let $x\in A.$
+By subset, we know that $x\in A\cap B,$ and from that, $x\in B.$
+This is the definition of $A\subseteq B.$
+\endproof
+
+\problem{5.} A proof involving sets
+
+Let $A,$ $B,$ and $C$ be arbitrary sets. Prove that if $A-C\not\subseteq
+A-B,$ then $B\not\subseteq C.$ (Homework 5 --- Problem 1)
+
+{\bf Proof.}
+
+We will show this by contrapositive.
+Let $A$, $B,$ and $C$ be arbitrary sets such that $B\subseteq C.$
+We will show that $A-C\subseteq A-B.$
+
+$(\subseteq)$
+
+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 $$A-(B\cap C) = (A-B)\cup(A-C).$$
+\endproof
+
+\problem{6.} An induction (or strong induction) proof
+
+Concerning the Fibonacci sequence, prove that $\sum_{k=1}^n F_k^2 =
+F_nF_{n+1}.$ (Homework 6---Hammack 10.26)
+
+{\bf Proof.}
+
+First, we define the Fibonacci numbers as $F_1 = F_2 = 1,$ and for
+$k>2,$ $F_k = F_{k-1} + F_{k-2}.$
+
+We will show this identity for all $n\in\bb Z,$ where $n\geq 1,$ by
+induction.
+For $n=1,$ $\sum_{k=1}^n F_k^2 = F_1^2 = 1 = 1\cdot 1 = F_1F_2.$
+
+We now assume for some $j\geq 1,$
+$$\sum_{k=1}^j F_k^2 = F_jF_{j+1},$$
+and we will show that $\sum_{k=1}^{j+1} F_k^2 = F_{j+1}F_{j+2}$ to
+complete the inductive step.
+Adding $F_{j+1}^2$ to the inductive hypothesis and simplifying gives
+$$\sum_{k=1}^{j+1} F_k^2 = F_jF_{j+1} + F_{j+1}F_{j+1} =
+F_{j+1}(F_j+F_{j+1}) = F_{j+1}F_{j+2},$$
+by definition of the Fibonacci numbers.
+
+We have shown by induction that for all $n\in\bb Z,$ s.t. $n\geq 1,$
+$$\sum_{k=1}^n F_k^2 = F_nF_{n+1}.$$
+\endproof
+
+\problem{7.} Proof a relation is an equivalence relation
+
+Suppose $R$ and $S$ are two equivalence relations on a set $A.$ Prove
+that $R\cap S$ is also an equivalence relation. (Homework 6---Hammack 11.2.10)
+
+{\bf Proof.}
+
+The equivalence relations $R$ and $S$ on $A$ are, by definition,
+reflexive, symmetric, and transitive.
+We will now show that $R\cap S$ is reflexive, symmetric, and transitive
+(so it is an equivalence relation).
+
+\smallskip
+(Reflexive)
+
+$R$ and $S$ are reflexive, so for all $a\in A,$ we know $(a,a)\in R$ and
+$(a,a)\in S.$
+Therefore, $(a,a)\in R\cap S,$ so $R\cap S$ is reflexive.
+
+\smallskip
+(Symmetric)
+
+Let $(x,y)\in R\cap S.$
+By symmetry, $(x,y)\in R$ implies $(y,x)\in R.$
+Similarly, this implies $(y,x)\in S.$
+From these, $(y,x)\in R\cap S,$ so $R\cap S$ is symmetric.
+
+\smallskip
+(Transitive)
+
+Let $(x,y),(y,z)\in R\cap S.$ We will show $(x,z)\in R\cap S.$
+
+$(x,y),(y,z)\in R,$ and by transitivity, $(x,z)\in R.$
+Similarly, $(x,z)\in S,$ so $(x,z)\in R\cap S.$
+We have shown that $R\cap S$ is transitive.
+
+\smallskip
+We have now shown that $R\cap S$ is an equivalence relation.
+\endproof
+
+\problem{8.} Injectivity or surjectivity of a function
+
+A function $f: \bb Z\to \bb Z\times\bb Z$ is defined as $f(n) =
+(2n,n+3).$ Verify whether this function is injective and whether it is
+surjective (Homework 7---Hammack 12.2.4).
+
+{\bf Proof.}
+This function is injective but not surjective.
+
+\smallskip
+(Injective)
+
+Let $m,n\in\bb Z$ such that $f(n) = f(m)$ or $(2n,n+3) = (2m,m+3).$
+Then, $2n=2m,$ and rearranging, $n=m.$
+This shows $f$ is injective.
+
+\smallskip
+(Surjective)
+
+We will show there is no $n$ such that $f(n) = (0,0),$ thereby showing
+$f$ is not surjective.
+$(2n, n+3) = (0,0)$ implies $2n = 0$ and $n = 0,$ but $n+3 = 0+3 = 3
+\neq 0,$ thereby dissatisfying the second part of the ordered pair.
+$f$ is not surjective.
+
+We have shown $f$ is injective but not surjective. \endproof
+
+\problem{9.} A proof that something is a group
+
+Let $S = \bb R\setminus \{-1\}$ and define a binary operation on $S$ by
+$a*b = a+b+ab.$ Prove that $(S,*)$ is an abelian group. (Homework
+10---Judson 3.5.7)
+
+{\bf Proof.}
+
+First, we show that the group operation is closed on $S.$
+Clearly, if $a,b\in S,$ then $a,b\in\bb R$ and $a*b\in\bb R$ by closure
+of the reals.
+However, we must show that $a+b+ab \neq -1.$
+For the sake of contradiction, we assume $a+b+ab=-1.$
+Rearranging, $0 = a+b+ab+1 = (a+1)(b+1),$ implying $a+1=0$ or $b+1=0.$
+Without loss of generality, we treat $a+1=0$ and obtain $a=-1,$ meaning
+$a\not\in S,$ giving a contradiction, and showing that $*$ is closed on
+$S.$
+
+Now, we show that $(S,*)$ is abelian.
+Again, let $a,b\in S.$
+We compute $a*b = a+b+ab = b+a+ba = b*a,$ by commutativity of addition
+and multiplication on the reals.
+
+Next, we must show that $0$ is the identity element of $(S,*).$
+With $a\in S,$
+$$0*a = 0+a+0a = a,$$
+and since $*$ is commutative, $a*0 = a,$ so $0$ is the identity.
+
+Last, we must show that for all $a\in S,$ we have $a^{-1}$ such that
+$a*a^{-1} = 0$ (this also implies $a^{-1}*a = 0$ by commutativity).
+Let
+$$a^{-1} = -{a\over 1+a} = {1\over 1+a}-1,$$
+which is guaranteed to exist because $a\neq -1,$ so $1+a\neq 0.$
+Also $a^{-1}\in S$ because ${1\over 1+a}\neq 0,$ so ${1\over 1+a}-1 =
+a^{-1}\neq -1,$ and $a^{-1}\in\bb R$ by real closure, so $a^{-1}\in S.$
+
+Computing the product,
+$$a*a^{-1} = a-{a\over 1+a} - {a^2\over 1+a} = {a(1+a)\over 1+a} -
+{a+a^2\over 1+a} = 0,$$
+so $a^{-1}\in S$ exists for all $a\in S.$
+
+\problem{10.} A proof of your own choosing
+
+Let $a,b\in G.$ Prove that the order of $ab$ is the same as the order of
+$ba$ (Homework 10---Judson 4.5.23c)
+
+{\bf Proof.}
+
+To show that $|ab|=|ba|,$ we will show that $(ab)^n=e$ if and only if
+$(ba)^n=e.$
+WLOG, we only need show that $(ba)^n=e$ if $(ab)^n=e.$
+
+Let $(a,b)^n = e.$
+$(ab)^{n+1}$ can be written as $(ab)^n ab = eab$ and as $a(ba)^nb.$
+Setting these equal,
+$$ab = a(ba)^nb \Rightarrow a^{-1}abb^{-1} = a^{-1}a(ba)^nbb^{-1}
+\Rightarrow e = (ba)^n.$$
+
+\bye