\documentclass{amsart}
\usepackage{amsmath,amssymb,amsthm}
\usepackage[margin=0.5in]{geometry}
\newtheorem{proposition}{Proposition}
\theoremstyle{definition}
\newtheorem{definition}{Definition}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\RR}{\mathbb{R}}
%\title{Portfolio Review}
\begin{document}
%\maketitle
\begin{section}{Miscellaneous proofs}
%%% A slightly irrational proof of the irrationality of $\sqrt{3}$.
\begin{proposition}
$\sqrt{3}$ is irrational.
\end{proposition}
\begin{proof}
Suppose $\sqrt{3}$ is rational.
Then there exists 2 integers $x$ and $y$ such that $\sqrt{3}=\frac{x}{y}$.
Suppose also that there are no common denominators between $x$ and $y$.
Then $3=\frac{x^2}{y^2}$, $y^2=3x^2$ and $x^2=3y^2$.
Thus both $x$ and $y$ are divisible by 3.
Since we already stipulated that there are no common denominators between $x$ and $y$, this results in a contradiction.
Thus $\sqrt{3}$ is irrational.
\end{proof}
\vfill
%%% Problem with variable names
\begin{proposition}
Let $a \in \ZZ$.
Prove that if $a$ is odd, then $a + 1$ is even.
\end{proposition}
\begin{proof}
By definition, if $a$ is {\em odd}, then there is an integer $x$ such that $a = 2x+1$.
Thus $a + 1 = (2x+1)+1 = 2x + 2 = 2(x + 1)$.
By definition, if $a$ is {\em even}, then there is an integer $y$ such that $a = 2y$.
Therefore, a + 1 is even.
\end{proof}
\vfill
%%% Without loss of generality?
\begin{proposition}
If 7 does not divide $ab$, then 7 divides neither $a$ nor $b$.
\end{proposition}
Contrapositive Proof.
\begin{proof}
Let $a,b \in \ZZ$. Suppose that 7 divides $a$ or 7 divides $b$.
Case 1. Without loss of generality, suppose 7 divides $a$ but 7 does not divide $b$.
By definition $a=7x$ for some $x \in \ZZ$ and $b\ne7y$ for some $y \in \ZZ$.
Hence $ab=7(xb)$.
Therefore 7 divides $ab$.
Case 2. Suppose 7 divides $a$ and 7 divides $b$.
By definition $a=7m$ and $b=7n$ for some $m,n \in \ZZ$. Then $ab=7(7mn)$.
Therefore 7 divides $ab$.
\end{proof}
\vfill
%%% Just a little confusing
\begin{proposition}
Let $a, b \in Z$. Prove that if 7 does not divide ab, then 7 divides neither a nor b.
\end{proposition}
\begin{proof}
(Contrapositive) Suppose 7 divides a or 7 divides b and $a, b \in Z$. By definition, $7m = a$ and $7n = b$ for some $m,n \in Z$. Thus, if 7 divides a, then $ab = 7(mb)$. Next, if 7 divides b, then $ab = 7(na)$. In either situation, ab is a multiple of seven. Therefore, it follows that 7 divides ab. Since the contrapositive is true, it follows that the original statement is true.
\end{proof}
\vfill
%%% Could you do this without cases?
\begin{proposition}
Let $x \in \RR$. If $x^2 + 5x < 0$, then $x < 0$.
\end{proposition}
\textbf{Contrapositive:}
Let $x \in \RR$. If $x \geq 0$, then $x^2 + 5x \geq 0$.
\begin{proof}
Let $x$ be a real number greater than or equal to 0.
Case 1: $x > 0$.
By definition of positive numbers, since $x > 0$, $5x > 0$.
By definition of squares, $x^2$ is greater than 0.
Since two positive numbers added together equal a third positive number, $x^2 + 5x$ is greater than 0.
Case 2: $x = 0$.
Since a number times 0 is 0, $5x = 5(0) = 0$.
Similarly, $0^2 = 0$.
Thus, $x^2 + 5x = 0 + 0 = 0$.
Therefore, for all real numbers $x \geq 0$, $x^2 + 5x \geq 0$.
Therefore, by contrapositive, if $x^2 + 5x < 0$, then $x < 0$.
\end{proof}
\vfill
%%% Some confusing use of symbols here, and maybe it could use more explanation?
\begin{proposition}
Let $x \in \mathbb{Z}$. If x is odd, then $8\mid(x^2 - 1)$.
\end{proposition}
\begin{proof}
Let $x \in \mathbb{Z}$ and odd.
By definition, $\exists a \in \mathbb{Z}$ such that $x = 2a+1$.
Hence, $x^2 = 4a(a+1) +1$ and $8\mid(x^2 - 1) = 8\mid4a(a+1)$.
We know that $a(a+1)$ is always even (because any number multiplied by an even produces an even), so $\exists b$ such that $a(a+1) = 2b$.
Thus $8\mid4a(a+1) = 8\mid8b$. Since 8 divides 8, we know that $8\mid(x^2 - 1)$.
\end{proof}
\vfill
%%% This one is pretty good
\begin{proposition}
Let $a, b, c \in \ZZ$.
If $a|b$ and $a|(b + c)$, then $a|c$
\end{proposition}
\begin{proof}
Suppose $a|b$ and $a|(b + c)$ for some integers $a, b, c$.
By definition, there exists $x, y \in \ZZ$ such that $ax = b$ and $ay = b + c$.
Then $ay - ax = a(y - x) = b + c - b = c$.
Therefore, since $(y - x)$ is an integer, $a|c$.
\end{proof}
\vfill
%%% Mostly correct, but what would your English teacher say?
\begin{proposition}
Suppose $x \in\ZZ$.
Then $x$ is even if and only if $3x+5$ is odd.
\end{proposition}
\begin{proof}
Proving this directly we suppose $x$ is even.
By definition there is an integer $a$ such that $x=2a$.
Thus $3x+5=3(2a)+5=6a+5=2(3a+2)+1=2(c)+1$.
Where $c=3a+2$ for some integer $c$.
Hence $3x+5$ is odd by definition.
Conversely, if $3x+5$ is odd, then x is even.
Proving this using a contrapositive proof.
Suppose $x$ is not even, so then $x$ is odd.
By definition there is an integer $b$ such that $x=2b+1$.
Thus $3x+5=3(2b+1)+5=6b+3+5=6b+8=2(3b+4)=2(d)$.
Where $d=3b+4$ for some integer $d$.
Hence $3x+5$ is even by definition.
In either case $7\nmid ab$.
\end{proof}
\vfill
\end{section}
\newpage
\begin{section}{Problem 9 of worksheet 3}
\begin{proposition}
For any $a, b \in \ZZ$, it follows that $(a + b)^3 \equiv a^3 + b^3 (mod 3)$.
\end{proposition}
%%% I'd say this is an instance of unnecessary variables making the proof harder to read
\begin{proof}
Suppose $a, b \in \ZZ$. Thus, there is an integer $x$ such that $x = a^2b + ab^2$.
Hence $3x = 3a^2b + 3ab^2 = (a + b)^3 - a^3 - b^3$.
It follows that $3|(a + b)^3 - a^3 - b^3$.
Therefore, $(a + b)^3 \equiv a^3 + b^3 (mod 3)$.
\end{proof}
\vfill
%%% This one is close, but has some problems with symbols and is missing the conclusion
\begin{proof}
Suppose $a,b \in \ZZ$.
Then $(a+b)^3 \equiv a^3 + 3 a^2 b + 3ab^2 + b^3$.
Hence $(a+b)^3 - (a^3 + b^3) = 3a^2b + 3ab^2 = 3(a^2b + ab^2)$.
We see that $a^2b + ab^2$ is an integer.
Thus by definition $(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$.
\end{proof}
\vfill
%%% No math problems here, but the writing could be improved
\begin{proof}
Let $a, b,\in \ZZ$.
Then $(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$.
Hence $(a+b)^{3}-(a^{3}+b^{3}) = a^{3}+3a^{2}b+3ab^{2}+b^{3}-a^{3}-b^{3}$.
This is equivalent to $3a^{2}b+3ab^{2}=3(a^{2}b+ab^{2})$.
We see that $a^{2}b+ab^{2}$ is an integer, thus by the definition of divisibility which states that for any integer $x,y$, we say that $x$ divides $y$ if there is an integer $c$ such that $y=xc$, $3\mid[(a+b)^{3}-(a^{3}+b^{3})]$.
Therefore, by the definition of congruence, $(a+b^{3})\equiv a^{3}+b^{3}(mod 3)$.
\end{proof}
\vfill
%%% Some fairly minor problems
\begin{proof}
Suppose $a$ and $b$ are integers.
As $a$ and $b$ are integers, $(a+b)^3$ and $a^3+b^3$ are themselves integers.
As the two of them are integers, their difference is an integer.
In addition, $(a+b)^3-(a^3+b^3) = 3(a^2b+ab^2)$.
This fits Definition 2 as $(a^2b+ab^2)$ is an integer, so $n = 3(a^2b+ab^2)$ for some integer $x$.
%%% What's x and where is it used?
It follows then that $3m = 3(a^2b+ab^2)$ for some integer $m$.
Thus $3m = 3(a^2b+2ab^2)=a^3+3a^2b+3ab^2+b^3-(a^3+b^3)$.
Hence $3m=(a+b)^3-(a^3+b^3)$.
Therefore, $3|(a+b)^3-(a^3+b^3)$, and finally $ (a+b)^3 \equiv a^3 + b^3 (\mod 3)$.
\end{proof}
\vfill
%%% Probably doesn't need all of this algebra and has the division argument that I dislike so much.
%%% But nothing is really wrong.
\begin{proof}
Suppose $a,b\in \mathbb{Z}$. Now, $(a+b)^3-(a^3+b^3)=a^3+3a^2b+3ba^2+b^3-(a^3+b^3)=a^3+b^3+3(a^2b+ab^2)-(a^3+b^3)=a^3+b^3-a^3-b^3+3(a^2b+ab^2)=3(a^2b+ab^2)$ where $(a^2b+ab^2)$ is an integer under integer operation rules. By Definition 1, $3\mid3(a^2b+ab^2)$. Therefore, $3\mid(a+b)^3-(a^3+b^3)$ also. This fits the definition of congruence (Definition 2).
Thus, for any $a,b\in \mathbb{Z}$, it follows that $(a+b)^3 \equiv a^3+b^3\medspace (mod \medspace3)$.
\end{proof}
\vfill
%%% This one looks pretty good
\begin{proof}
Let $a,b \in \ZZ$. Now examine the equation $(a + b)^3$. This equation can be expanded $(a + b)^3 = a^3 + 3ab^2 + 3a^2b + b^3$. Now, if we subtract $a^3 + b^3$ from both sides we are left with $(a + b)^3 - (a^3 + b^3) = a^3 + 3ab^2 + 3a^2b + b^3 - (a^3 + b^3) = 3(ab^2 + a^2b)$. We have been shown that $(a + b)^3 - (a^3 + b^3) = 3(ab^2 + a^2b)$ so, by definition, $3 \mid (a + b)^3 - (a^3 + b^3)$. Therefore, by definition, $(a + b)^3 \equiv (a^3 + b^3)$ (mod 3).
\end{proof}
\vfill
\end{section}
\end{document}