Question

Prove that \(\sqrt{5}-\sqrt{3}\) is not a rational number.

Answer 1

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