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\theoremstyle{definition}
\newtheorem*{defn}{Definition}
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\newtheorem*{recall}{Recall}
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\begin{document}

%----------------------------------------------------------------------------------------
\section*{4. Direct Proof}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}{Proofs}
\begin{itemize}
\item Specialized vocabulary\pause
\item More world-building: divisibility\pause
\item Then we prove stuff!!!
\end{itemize}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\subsection*{4.1 Theorems and their Friends}
\begin{frame}{Theorems}
\begin{defn}
A  a statement that is true and has been proved to be true can be a \dbox{theorem}, a \dbox{proposition}, a \dbox{lemma}, or a \dbox{corollary}.\pause

\textcolor{dblue}{Theorem: most significant.\\}\pause
\textcolor{dblue!80}{Proposition: slightly less significant, or perhaps quite basic.\\}\pause
\textcolor{dblue!60}{Lemma: only use is to prove a theorem or proposition.\\}\pause
\textcolor{green}{Corollary: follows directly from a theorem. (If you know a theorem, then its corollary should be extremely easy to prove.) Can also be significant.}
\end{defn}\pause

{\bf Pythagorean Theorem:} If a right triangle has hypotenuse of length $c$, and other sides of length $a$ and $b$ respectively, then $a^2+b^2=c^2$.

{\bf Fundamental Theorem of Algebra:} If $P(x)$ is a polynomial, with real coefficients, in the single variable $x$, then $P(x)$ can be factored into the product of linear and quadratic functions.

{\bf Fundamental Theorem of Calculus, Part 1:} If $f$ is continuous on $[a,b]$, then the function $g(x)$, defined by $g(x)=\int_a^x f(t)dt$, with domain $a \leq x \leq b$, is continuous on $[a,b]$, differentiable on $(a,b)$, and $g'(x)=f(x)$.

{\bf Theorem:} There are infinitely many prime numbers.
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Statements that Aren't Theorems}
A conjecture is a statement that someone thinks is true, but that has not been proven to be true or false.\pause\vfill

{\bf Goldbach Conjecture:} Every even integer greater than 2 can be written as the sum of two primes.\pause\vfill

{\bf Conjecture:} $P \neq NP$\pause\\
{\bf Conjecture:} $P = NP$\pause\vfill

{\bf Collatz Conjecture:}  
Given any positive integer $n$, the series of operations:
\begin{itemize}
\item if $n$ is even, replace it with $\frac{n}{2}$
\item if $n$ is odd, replace it with $3n+1$
\end{itemize}
will eventually result in $n=1$.
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Definitions}
\begin{defn}
A \dbox{definition} is an unambiguous explanation of the meaning of  a mathematical word of phrase.
\end{defn}\pause
A definition isn't true or false: it's simply agreed on.\\\pause\vfill

{\bf Definition:} Let $\mathbb N$ be the set of natural numbers, $\{1,2,3,\ldots\}$\pause\vfill

{\bf Definition:} Let $a$ be the largest element in the set $\{a_1,a_2,a_3\}$.\pause\vfill

{\bf Definition:} A set is a collection of objects.\\
\textcolor{mauve}{Bad definition: is the set of all sets a set?}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\subsection*{4.2 Definitions}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}{Parity}
\begin{defn}
An integer $n$ is \dbox{even} if $n=2a$ for some integer $a \in \mathbb Z$.\vfill

An integer $n$ is \dbox{odd} if $n=2a+1$ for some integer $a \in \mathbb Z$. 
\end{defn}\pause

True or False: 0 is both odd and even.\\ \pause
\textcolor{lblue}{False. There is no integer $a$ so that $0=2a+1$, so $0$ is not odd.}\pause
\begin{defn}
Two integers have the  \dbox{same parity} if they are both even or both odd.\\ If one integer is even and another integer is odd, they have \dbox{opposite parity}.
\end{defn}\pause
Show that the sum of two numbers of opposite parity is odd.\\ \pause
\textcolor{lblue}{Let $n$ be an even number and $m$ be an odd number.\\
By the definitions of even and odd, there exist integers $a$ and $b$ such that $n=2a$ and $m=2b+1$.\\
Then $n+m=(2a)+(2b+1)=2(a+b)+1$.\\
Since $a+b$ is an integer, by the definition of odd, $n+m$ is odd.}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{A Note on Definitions}
We use a special convention with definitions.\pause
\begin{defn}
An integer $n$ is \dbox{even} if \onslide<5->{\textcolor{mauve}{and only if}} $n=2a$ for some integer $a \in \mathbb Z$.\vfill

An integer $n$ is \dbox{odd} if \onslide<5->{\textcolor{mauve}{and only if}} $n=2a+1$ for some integer $a \in \mathbb Z$. 
\end{defn}\pause\vfill

Written:\\
\textcolor{green}{$n=2a ~\boxed{\Rightarrow}~ n$ even}\pause\vfill

Intended Meaning:\\
\textcolor{green}{$n=2a ~\boxed{\Leftrightarrow}~ n$ even}\vfill\pause\pause

This is how people write definitions. (Convention.)
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{More Definitions}
\begin{defn}Suppose $a$ and $b$ are integers. We say $a$ \dbox{divides} $b$, written
\[a|b\]
if $b=ac$ for some $c \in \mathbb Z$. In this case we also say that $a$ is a \dbox{divisor} of $b$, and $b$ is a \dbox{multiple} of $a$.\end{defn}\pause

Example: \qquad \textcolor{green}{$10=2\cdot 5$}
\begin{itemize}
\item $2|10$, $5|10$ \pause
\item $2$ is a divisor of $10$\\ \pause
\item $10$ is a multiple of 2\\ \pause
\item Is $0$ a divisor of 10? \pause  \textcolor{lblue}{No. There is no $a \in \mathbb Z$ with $0a=10$.}\\
 We write $0 \not | ~10$\pause
\item Is $-1$ a divisor of 10? \pause  \textcolor{lblue}{Yes, $-10 \in \mathbb Z$ and $(-1)(-10)=10$.}
\end{itemize}\vfill

\onslide<9->{\textcolor{mauve}{
 $\{a\in\mathbb Z:a|10\}= $  \onslide<10->{$ \{-10,-5,-2,-1,1,2,5,10\}$}}\\
 
\textcolor{mauve}{
 $\{a\in\mathbb Z:a|0\}= $ \onslide<11->{ $ \mathbb Z$}}}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{More Definitions}
\begin{defn}
A number $n \in\mathbb N$ is \dbox{prime} if it has exactly two positive divisors, $1$ and $n$.

\onslide<6->{An integer $n$ is \dbox{composite} if  $n=ab$ for two integers $a$, $b$ with $a>1$ and $b>1$.}
\end{defn}\pause
\only<2-8>{
Is 1 prime? \pause \textcolor{lblue}{ No: it only has one positive divisor.}\pause

Can a negative number be prime? \pause \textcolor{lblue}{ No: it is not in $\mathbb N$.}

\onslide<7->{What is the smallest composite number?}  \onslide<8->{\textcolor{lblue}{4}}}
\pause\pause\pause\pause

\begin{defn}
The \dbox{greatest common divisor} of integers $a$ and $b$, which are not both zero, denoted \dbox{$\gcd(a,b)$}, is the largest integer that divides both $a$ and $b$.\\
The \dbox{least common multiple} of non-zero integers $a$ and $b$, denoted \dbox{$\lcm(a,b)$}, is the smallest positive integer that is a multiple of both $a$ and $b$.
\end{defn}\pause\pause\pause

\textcolor{mauve}{$\gcd(0,-18)= \pause 18$} \hfill\pause
\textcolor{mauve}{$\lcm(-2,5)= \pause 10$}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Generally Accepted}
We can't define everything!\\\pause\vfill

We accept the following facts:
\begin{itemize}
\item When we add, subtract, or multiply two integers, the result is an integer. \pause
\item It's possible to divide any two non-zero numbers and end up with a quotient and remainder.
\pause
\item We can do algebra. For example, $\frac{4x}{2}=2x$, and we can solve equations.\pause
\item Every natural number greater than 1 has a unique factorization into powers of primes.
\end{itemize}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\subsection*{4.3 Direct Proof}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}{Direct Proof of Conditional Statements}
How to prove $P \Rightarrow Q$\pause\vfill

\begin{mdframed}
{\bf Proposition:} If $P$, then $Q$.\\

\textit{Proof}: Suppose $P$.\\
\vdots\\
Therefore $Q$.\qed
\end{mdframed}

Every step should be completely justified.
\vfill\pause

\centering
\begin{tabular}{|c|c||c|}
\hline
$\mathbf{P}$&$\mathbf{Q}$&$\mathbf{P \Rightarrow Q}$\\\hline
T&T&T\\ \hline
T&F&F\\ \hline
F&T&T\\ \hline
F&F&T \\ \hline
\end{tabular}
\end{frame}
%----------------------------------------------------------------------------------------
\begin{frame}{Direct Proof}
Prove the following:\\
{\centering \textit{If $4|x$, then $\dfrac{x}{2}$ is even.}}\pause
\vfill
\begin{mdframed}
{\bf Proposition:} If $4|x$, then $\dfrac{x}{2}$ is even.\\

\textit{Proof}: Suppose $4|x$.\\
\only<-7>{\onslide<4->{\alert<4>{Then by the definition of ``divides," there exists some $b \in \mathbb Z$ such that $4b=x$.}}\\
\onslide<5->{\alert<5>{Then $\dfrac{x}{2}=\dfrac{4b}{2}=2b$}}\\

\onslide<3->{\alert<3>{Therefore, $\dfrac{x}{2}=2\only<3-5>{a}\only<6->{\alert<6>{b}}$ for some $\only<-5>{a}\only<6->{\alert<6>{b}} \in \mathbb Z$.}}\\
Therefore,\quad $\dfrac{x}{2}$ is even \onslide<3->{\alert<3>{by the definition of ``even".}}}
\only<8->{
Then there exists some $a \in \mathbb Z$ such that $4a=x$. Then $\dfrac{x}{2} = 2a$, so $\dfrac{x}{2}$
is even.}\qed
\end{mdframed}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}{Direct Proof}
Prove the following:\\
{\centering \textit{If $x$ is odd, then $x^2+1$ is even.}}\pause
\vfill
\begin{mdframed}
{\bf Proposition:} If $x$ is odd, then $x^2+1$ is even.\\
\textit{Proof}: Suppose $x$ is odd.\\
\onslide<3->{\textcolor{dblue}{
Then $x=2a+1$ for some integer $a$.\\
Then $x^2+1=(2a+1)^2+1=4a^2+4a+1+1=2(2a^2+2a+1)$.\\
If we define $b=2a^2+2a+1$, then $b \in \mathbb Z$ and $x=2b$.}}\\
Therefore, $x^2+1$ is even.\qed
\end{mdframed}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}{Direct Proof}
Prove the following:\\
{\centering \textit{If $x$ is even, then for any $n \in \mathbb N$, $x^n$ is even as well.}}
\vfill
\begin{mdframed}
{\bf Proposition: If $x$ is even and $n \in \mathbb N$, then $x^n$ is even.}\\
\textit{Proof}: Suppose $x$ is even and $n \in \mathbb N$.\\
\onslide<2->{\textcolor{dblue}{
Then $x = 2a$ for some integer $a$ and $n \geq 1$. \\
Since $n \geq 1$, $n-1 \geq 0$, so $x^{n-1} \in \mathbb Z$.\\
Then $x^n=x\left(x^{n-1}\right)=2a\left(x^{n-1}\right)$.\\
Let $b=a\left(x^{n-1}\right)$. Then $b \in \mathbb Z$ and $x^n=2b.$}}\\
Therefore, $x^n$ is even.\qed
\end{mdframed}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}{Direct Proof}
Prove the following:\\
{\centering \textit{If $a|b$, then $a|(b^2+5a)$.}}
\vfill
\begin{mdframed}
{\bf Proposition: If $a|b$, then $a|(b^2+5a)$.}\\
\textit{Proof}: Suppose $a|b$.\\
\onslide<2->{\textcolor{dblue}{
Then there exists some integer $x$ such that $b=ax$. \\Then $b^2+5a=(ax)^2+5a=a(ax^2+5)$.\\
If we let $y=ax^2+5$, then $y \in \mathbb Z$ and $b=ay$.}}\\
Therefore, $a|(b^2+5a)$.\qed
\end{mdframed}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}{Direct Proof}
Prove the following:\\
{\centering \textit{For any two positive real numbers $z$ and $y$, if $x \leq y$ then $\sqrt{x}\leq\sqrt{y}$.}}\\
Hint: Factor $\sqrt{x^2}-\sqrt{y^2}$. You may assume that dividing both sides of an inequality by a positive number does not change the direction of the inequality.
\vfill
\begin{mdframed}
{\bf Proposition:}\\
\textit{Proof}: Suppose $x,y \in \mathbb R$ and $0<x \leq y$.\\
\onslide<2->{\textcolor{dblue}{
Since $x \leq y$, $x-y \leq 0$.\\
Since $x$ and $y$ are nonnegative, $\sqrt{x}^2=x$ and $\sqrt{y}^2=y$.\\
Therefore, $x-y \leq 0$ tells us $\sqrt{x}^2-\sqrt{y}^2 \leq 0$. \\
Factoring, we see $(\sqrt x + \sqrt y)(\sqrt{x}-\sqrt{y}) \leq 0$. Since $\sqrt{x}+\sqrt{y}>0$, dividing both sides of this inequality by it yields
$\sqrt{x}-\sqrt{y} \leq 0$.
}}\\
Therefore, $\sqrt{x}\leq\sqrt{y}$.\qed
\end{mdframed}
\onslide<3->{\textcolor{green}{Corollary: for all nonnegative real $x$ and $y$, if $x \leq y$, then $\sqrt{x} \leq \sqrt{y}$.}}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}{Direct Proof}
Prove the following:\\
{\centering \textit{For any two nonnegative integers $z$ and $y$, $\sqrt{xy}\leq \dfrac{x+y}{2}$. (That is, the geometric mean is no greater than the arithmetic mean.)}}\\
Hint: use the result from the previous slide, $x \leq y \Rightarrow \sqrt{x} \leq \sqrt{y}$ for all nonnegative real $x$, $y$.
\vfill
\begin{mdframed}
{\bf Proposition:} For any two nonnegative integers $z$ and $y$, $\sqrt{xy}\leq \dfrac{x+y}{2}$.\\
\textit{Proof}: Suppose $x$ and $y$ are nonnegative integers.\\
\onslide<2->{\textcolor{dblue}{
Note $0 \leq (x-y)^2$.\\
Then $0 \leq x^2-2xy+y^2$, so if we add $4xy$ to both sides,
$4xy \leq x^2+2xy+y^2=(x+y)^2$.\\
Then (using the result from the last slide)
$2\sqrt{xy} \leq x+y$.
}}\\
Therefore, $\sqrt{xy}\leq \dfrac{x+y}{2}$.\qed
\end{mdframed}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\subsection*{4.4 Using Cases}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}[t]{Cases}
Prove that $1+(-1)^n(2n-1)$ is divisible by 4 for every $n \in \mathbb Z$.

\only<2>{\begin{center}\begin{tabular}{|c|c|}
\hline
$n$&$1+(-1)^n(2n-1)$\\
\hline
0&0\\\hline
1&0\\ \hline
2&4\\ \hline
3&-4\\ \hline
4&8\\ \hline
5&-8\\ \hline
6&12\\ \hline
7&-12\\ \hline
\end{tabular}\end{center}}

\onslide<3->{\vfill
\textbf{Proposition:} For every $n \in \mathbb Z$, $1+(-1)^n(2n-1)$ is a divisible by 4.\\
\onslide<4->{\textit{Proof:}\\
\onslide<5->{ Let $n$ be an integer. Then $n$ is either even or odd.
\\ \textbf{Suppose $n$ is even.}
\onslide<6->{Then $n=2a$ for some integer $a$, and $(-1)^n=1$. So,
\[1+(-1)^n(2n-1)=1+2(2a)-1=4a\]
and thus the expression is divisible by $4$.\\}

\textbf{Suppose $n$ is odd.} 
\onslide<7->{Then $n=2a+1$ for some integer $a$, and $(-1)^n=-1$. So,
\[1+(-1)^n(2n-1)=1-2(2a+1)+1=-4a=4(-a)\]
and thus the expression is divisible by $4$.}
\qed
\vfill}}}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\begin{frame}[t]{Cases}
\textbf{Proposition:} If $a \in \mathbb Z$ is a multiple of 4, then $a=1+(-1)^n(2n-1)$ for some $n \in \mathbb N$.\\

\only<2>{\begin{center}\begin{tabular}{|c|c|}
\hline
$n$&$1+(-1)^n(2n-1)$\\
\hline
0&0\\\hline
1&0\\ \hline
2&4\\ \hline
3&-4\\ \hline
4&8\\ \hline
5&-8\\ \hline
6&12\\ \hline
7&-12\\ \hline
\end{tabular}\end{center}}

\onslide<3->{\vfill\textit{Proof:}
Suppose $a \in \mathbb Z$ is a multiple of 4. Then $a=4b$ for some integer $b$.\\
\textbf{Suppose $b \geq 0$.} In this case, let $n=2b$. Then 
\[1+(-1)^n(2n-1)=1+(2(2b))-1)=4b=a.\]
\textbf{Suppose $b < 0$.} In this case, let $n=1-2b$.  Note $1-2b>0$, since $b<0$, so $n \in \mathbb N$. Also, $n$ is odd, so:
\[1+(-1)^n(2n-1)=1+(-1)(2(1-2b))-1)=1-(2-4b-1)=4b=a.\]

\qed\vfill}
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\subsection*{4.5 Treating  Similar Cases}
\begin{frame}[t]{``Without Loss of Generality"}
\textbf{Proposition:} Given any three (not necessarily distinct) positive numbers, it is possible to choose two so that their sum is strictly greater than the third.\\ \pause \vfill
\textcolor{green}{Example: given $1, 1, 7.8$:\quad\pause \textcolor{mauve}{$1<1+7.8$}}\\ \pause
\textcolor{green}{Example: given $5, 3, 8$:\quad \pause\textcolor{mauve}{$3<5+8$, and also $5<3+8$}}\\ \pause\vfill
\textit{Proof: }
Let $a$, $b$, and $c$ be positive numbers, and WLOG (``without loss of generality") let $a \leq b$.\\

Then $b+c \geq a+c >a$ (since $c>0$).
\qed\vfill
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\subsection*{4.5 Treating  Similar Cases}
\begin{frame}[t]{Triangle Inequality}
\textbf{Proposition:} $\forall x,y \in \mathbb Z$, $|x+y|\leq|x|+|y|$\\ \pause \vfill
\textcolor{green}{Example:  $|1+1|\leq 1+1$} \pause\\
\textcolor{green}{Example: $|2-1|\leq2+1$}\pause\vfill
\textit{Proof: }\\
\onslide<4->{\textbf{Suppose $x,y \geq 0$.
\vfill
\textbf{Suppose $x,y < 0$.
\vfill
\textbf{Suppose $x \geq 0,y < 0$.}}}}
\qed\vfill
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\subsection*{}
\begin{frame}{More Examples}
\textbf{Proposition: } $\forall n \in \mathbb N$, $n^2-3n+9$ is odd.\vfill

\textbf{Proposition: } Suppose $\gcd(a,b)>1$ and $b$ is prime. Then $b=\gcd(a,b)$.
\vfill
\textbf{Proposition: } For every $a \in \mathbb Z$, if $a^2 | a$, then $a^2=|a|$.
\vfill
\textbf{Proposition: } Every odd integer is the difference of two squares. \\
(For example: $9=5^2-4^2$, $13=7^2-6^2$.)
\vfill
\end{frame}
%----------------------------------------------------------------------------------------
%----------------------------------------------------------------------------------------
\end{document} 