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Prove that $\sqrt{5}-\sqrt{3}$ is not a rational number.
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Let $\sqrt{5}-\sqrt{3}$ be a rational number of form $\frac{a}{b}$, where $b \equiv 0$
Squaring on both sides
\begin{aligned} &(\sqrt{5}-\sqrt{3})^{2}=\left(\frac{a}{b}\right)^{2} \\ &(\sqrt{5})^{2}+(\sqrt{3})^{2}-2(\sqrt{5})(\sqrt{3})=\frac{a^{2}}{b^{2}} \\ &5+3+2 \sqrt{15}=\frac{a^{2}}{b^{2}} \\ &8+2 \sqrt{15}=\frac{a^{2}}{b^{2}} \\ &2 \sqrt{15}=\frac{a^{2}}{b^{2}}-8 \end{aligned}
$\sqrt{15}=\frac{\mathrm{a}^{2}-8 \mathrm{~b}^{2}}{2 \mathrm{~b}^{2}}$
since $\sqrt{15}$ is irrational,$\frac{a^{2}-8 b^{2}}{2 b^{2}}$ is rational
Since LHS $=\mathrm{RHS}$, contradiction arises,
Therefore $\sqrt{5}-\sqrt{3}$ is irrational
by Diamond (44,715 points)

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