\section{Introduction}


\definition{}
A \textit{proof} is a mathematical argument that irrefutably
demonstrates the truth of a given proposition.

\vspace{2mm}

Every proof involves some sort of \textit{implication}, denoted $\implies$. \par
The statement \say{$A$ implies $B$} (written $A \implies B$), means that $B$ is true whenever $A$ is true.

\problem{}<trueimplies>
Which of the following are true? \par
\note{You don't need to provide a proof.}

\begin{itemize}
	\item $x$ is prime $\implies$ $x$ is odd.
	\item $x$ is real $\implies$ $x$ is rational.
	\item $x$ is odd $\implies$ $x$ is prime.
\end{itemize}

\vfill



\problem{}
As you saw above, $A \implies B$ does not guarantee that $B \implies A$. \par
Find two new statements $A$ and $B$ so that $A \implies B$ but $B \notimplies A$. \par
\hint{\say{new} as in \say{not from \ref{trueimplies}}}

\begin{solution}
	A fairly trite example is below. \par
	Note that \say{$X$ is a square} is a subset of the statement \say{$X$ is a rectangle.}

	\vspace{2mm}

	$X$ is a square $\implies$ $X$ is a rectangle, but $X$ is a rectangle $\notimplies$ $X$ is a square.
\end{solution}



\vfill
\pagebreak


\definition{}<iffdef>
As we just saw, implication is one-directional. \par
The statements $A \implies B$ and $B \implies A$ are independent of one another. \par

\vspace{1mm}
 
If both are true, we write $A \iff B$. This can be read as \say{$A$ if and only if $B$.} \par
In text, \say{if and only if} is often abbreviated as iff. \par

\vspace{1mm}

Bidirectional implication is the strongest relationship we can have between two statements: \par
If $A \iff B$, $A$ and $B$ are \textit{equivalent.} They are always either both true or both false.

\definition{}
The \textit{floor} of $x$ is the largest integer $a$ so that $a \leq x$. This is denoted $\lfloor x \rfloor$. \par
The \textit{ceiling} of $x$ is the largest integer $a$ so that $a \geq x$. This is denoted $\lceil x \rceil$.


\generic{Property:}
If $b_1 \leq a \leq b_2$ and $b_1 = b_2$, we must have that $b_1 = a = b_2$. \par

\vspace{1mm}

Also, if $a \leq b$ and $a \geq b$, we must have that $a = b$. \par
This is a trick we often use when showing that two quantities are equal.




\problem{}
Although $A \iff B$ looks like a single statement, we often need to prove each direction separately. \par
Show that $x \in \mathbb{Z}$ iff $\lfloor x \rfloor = \lceil x \rceil$

\begin{solution}
	\textbf{Forwards:} $x \in \mathbb{Z} ~\implies~ \lfloor x \rfloor = \lceil x \rceil$ \par
	If $x \in \mathbb{Z}$, by definition we have that $\lfloor x \rfloor = x$ and $\lceil x \rceil = x$ \par
	So, $\lfloor x \rfloor = \lceil x \rceil$

	\vspace{2mm}

	\textbf{Backwards:} $x \in \mathbb{Z} ~\rimplies~ \lfloor x \rfloor = \lceil x \rceil$ \par
	Assume that $\lfloor x \rfloor = \lceil x \rceil$, and show that $x$ is an integer. \par
	Note that $\lfloor x \rfloor \leq x \leq \lceil x \rceil$ (by definition) \par
	Since $\lfloor x \rfloor = \lceil x \rceil$, we must have that $\lfloor x \rfloor = x = \lceil x \rceil$. \par
	$\lfloor x \rfloor$ is an integer, so $x$ must be an integer.
\end{solution}


\vfill
\pagebreak

\problem{}
We don't always need to prove each direction of an iff statement separately. \par

\begin{itemize}[itemsep = 1mm]
	\item Convince yourself that we can \say{chain} iffs together: \par
	If we show that $A \iff B \iff C \iff D$, do we know that $A \iff D$?

	\item Does this still work if $A \iff B \implies C \iff D$?

	\item Show that $x^2 - 6x - 6 = 3 \iff x = 3$ by building a chain of iffs. \par
	\hint{You remember how to factor quadratics, right?}
\end{itemize}

\begin{solution}
	Does this still work if $A \iff B \implies C \iff D$? \par
	Of course not. $D \notimplies A$ since $C \notimplies B$.
	We can only conclude that $A \implies D$.

	\linehack{}

	$x^2 - 6x - 6 = 3$ \par
	$\iff x^2 - 6x - 9 = 0$ \par
	$\iff (x-3)^2 = 0$ \par
	$\iff x-3 = 0$ \par
	$\iff x = 0$

	Note that this is a well-defined argument. Every step is an iff statement we can rigorously justify.
	We're not hand-wavily \say{rearranging} one equation into another,
	we're building a chain of implications that eventually bring us to our result.
	This is the logic behind most algebraic proofs.
\end{solution}


\vfill

\problem{}
Another trick you may find useful is the \say{implication cycle.} \par
Convince yourself that if $A \implies B \implies C \implies D \implies A$, \par
we can conclude that $A$, $B$, $C$, and $D$ are equivalent. \par
\note{$A \iff B$ means that $A$ and $B$ are equivalent. See \ref{iffdef}.}

\vfill

\problem{Bonus}
Use an implication cycle to show that the following definitions of a \textit{squarefree integer} are equivalent.
%\hint{Show that $A \implies B \implies C \implies D \implies A$}
\begin{enumerate}
	\item $n^2$ does not divide $q$ for any $n \in \mathbb{Z}^+$, $n \neq 1$
	\item $p^2$ does not divide $q$ for any prime $p$ 
	\item $q$ is a product of distinct primes
	\item $q ~|~ n^k \implies q ~|~ n$ for all $n, k \in \mathbb{Z}^+$
\end{enumerate}

\begin{solution}
	Assume $q$ has a square factor, so that $q = an^2$ for some $a, n \in \mathbb{Z}^+$ \par
	By D, we know that $q ~|~ (an)^2 \implies q ~|~ an$ \par
	But $q ~|~ an \implies an^2 ~|~ an$ \par
	$\implies n = 1$

	\vspace{2mm}

	So, $q$ cannot have a square factor that isn`t 1.
\end{solution}


\vfill
\pagebreak

Often enough, proving a statement is simply a matter of \say{definition chasing,}
where we expand the symbols used in the statement we're proving, and then do a bit of
rearranging to arrive at the result we want.


\definition{}
Let $n, x \in \mathbb{Z}$. \par
We say \say{$n$ divides $x$} if $x = kn$ for some $k \in \mathbb{Z}$

\definition{}
Let $a, b \in \mathbb{Z}$ and $n \in \mathbb{Z}^+$. \par
We say \say{$a$ is congruent to $b$ modulo $n$} (and write $a \equiv_{n} b$) if $n$ divides $a - b$. \par

\definition{}
Let $a, b \in \mathbb{Z}$. We define $a ~\%~ b$ as the remainder of $a \div b$.

\problem{}
Let $a, b, n$ be positive integers. \par
Show that $a + b ~\equiv_n (a ~\%~ n) + (b ~\%~ n)$

\begin{solution}
	$a + b ~\equiv_n (a ~\%~ n) + (b ~\%~ n)$ \par

	\vspace{2mm}
	...can be rewritten as... \par
	$n$ divides $a + b - (a ~\%~ n) - (b ~\%~ n)$ \par
	
	\vspace{2mm}
	...which can be rearranged to... \par
	$n$ divides $(a - (a ~\%~ n)) + (b - (b ~\%~ n))$

	\vspace{2mm}
	...which is clearly true, if you think about the meaning of \say{$n$ divides $x$} and \say{$a ~\%~ b$}.
\end{solution}

\vfill
\pagebreak