Moved Proof techniques

Misc/Proof Techniques/parts/1 contradiction.tex

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+\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
+\pagebreak