26 lines
736 B
TeX
26 lines
736 B
TeX
\beginsection 2.1
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Show that $\sqrt3$, $\sqrt5$, $\sqrt7$, $\sqrt{24}$, and
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$\sqrt{31}$ are not rational numbers.
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\medskip
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The solution to $x^2-3=0$ is $\sqrt3$.
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By the Rational Zeros Theorem, if $\sqrt3=p/q$ then
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$q$ divides $a_1=1$ and $p$ divides $a_0=3$.
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Since $q$ must be $\pm1$, the only possible values for
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$p/q$ are $\pm1$ and $\pm3$, none of which are
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solutions to $x^2-3=0$.
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Therefore $\sqrt3$ must be irrational.
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$\sqrt5$: $p/q=\pm1,\pm5$, none of which are
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solutions to $x^2-5=0$.
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$\sqrt7$: $p/q=\pm1,\pm7$, none of which are
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solutions to $x^2-7=0$.
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$\sqrt24$: $p/q=\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12$,
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none of which are solutions to $x^2-24=0$.
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$\sqrt31$: $p/q=\pm1,\pm31$, none of which are
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solutions to $x^2-31=0$.
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