\section{Proofs by Contradiction}

\definition{}
A very common proof technique is \textit{proof by contradiction}.
It works as follows: 

\vspace{2mm}

Say we want to prove a statement $P$. Assume that $P$ is false, and show that this implies a false statement.
In other words, we show that $P$ can't \textit{not} be true. \par
If it's false, we either get a known impossibility ($1 = 2$, pigs fly, et cetera), \par
or we find that (not $P$) $\implies$ (not (not $P$)), which is a contradiction in itself.


\problem{}
Show that the set of integers has no maximum using a proof by contradiction.

\begin{solution}
	Assume there is a maximal integer $x$. \par
	$x + 1$ is also an integer. \par
	$x + 1$ is larger than $x$, which contradicts our original assumption!

	\vspace{2mm}

	This is a \textit{proof by infinite descent}, a special type of proof by contradiction.\par
	Such proofs have the following structure:
	\begin{itemize}
		\item Assume there is a smallest (or largest) object with property $X$.
		\item Show that we have an even smaller object that has the same property $X$.
	\end{itemize}
\end{solution}

\vfill

\definition{}
We say a number $x \in \mathbb{R}$ is \textit{rational} if we can write $x$ as $\nicefrac{p}{q}$, \par
where $p, q$ are integers with no common factors.

\problem{}
Show that $\sqrt{2}$ is irrational. \par
\hint{Start by chasing definitions. \say{Not irrational} $=$ \say{rational.}}

\begin{solution}
	Suppose $\sqrt{2}$ is rational. Then, there exist $p, q$ so that $\sqrt{2} = \frac{p}{q}$.

	\vspace{2mm}

	Multiply by $q$ and square to find that $2q^2 = p^2$, which implies that $p^2$ is even. \par
	This then implies that $p$ is even, \par
	which implies that $p^2$ is divisible by 4, \par
	which implies that $q^2$ is divisible by 2, \par
	and thus we see that $q$ is also even.

	\vspace{2mm}

	$p$ and $q$ are both even, so they cannot be coprime. \par
	Therefore, we cannot write $\sqrt{2}$ as $\nicefrac{p}{q}$ for comprime $p, q$, \par
	and $\sqrt{2}$ is therefore irrational.
\end{solution}


\vfill
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