63 lines
2.0 KiB
TeX
63 lines
2.0 KiB
TeX
\section{Proofs by Contradiction}
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\definition{}
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A very common proof technique is \textit{proof by contradiction}.
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It works as follows:
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\vspace{2mm}
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Say we want to prove a statement $P$. Assume that $P$ is false, and show that this implies a false statement.
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In other words, we show that $P$ can't \textit{not} be true. \par
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If it's false, we either get a known impossibility ($1 = 2$, pigs fly, et cetera), \par
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or we find that (not $P$) $\implies$ (not (not $P$)), which is a contradiction in itself.
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\problem{}
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Show that the set of integers has no maximum using a proof by contradiction.
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\begin{solution}
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Assume there is a maximal integer $x$. \par
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$x + 1$ is also an integer. \par
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$x + 1$ is larger than $x$, which contradicts our original assumption!
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\vspace{2mm}
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This is a \textit{proof by infinite descent}, a special type of proof by contradiction.\par
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Such proofs have the following structure:
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\begin{itemize}
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\item Assume there is a smallest (or largest) object with property $X$.
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\item Show that we have an even smaller object that has the same property $X$.
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\end{itemize}
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\end{solution}
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\vfill
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\definition{}
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We say a number $x \in \mathbb{R}$ is \textit{rational} if we can write $x$ as $\nicefrac{p}{q}$, \par
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where $p, q$ are integers with no common factors.
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\problem{}
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Show that $\sqrt{2}$ is irrational. \par
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\hint{Start by chasing definitions. \say{Not irrational} $=$ \say{rational.}}
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\begin{solution}
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Suppose $\sqrt{2}$ is rational. Then, there exist $p, q$ so that $\sqrt{2} = \frac{p}{q}$.
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\vspace{2mm}
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Multiply by $q$ and square to find that $2q^2 = p^2$, which implies that $p^2$ is even. \par
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This then implies that $p$ is even, \par
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which implies that $p^2$ is divisible by 4, \par
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which implies that $q^2$ is divisible by 2, \par
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and thus we see that $q$ is also even.
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\vspace{2mm}
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$p$ and $q$ are both even, so they cannot be coprime. \par
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Therefore, we cannot write $\sqrt{2}$ as $\nicefrac{p}{q}$ for comprime $p, q$, \par
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and $\sqrt{2}$ is therefore irrational.
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\end{solution}
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\vfill
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\pagebreak |