​ Prove that √3 + √5 is an irrational number. ​

Question

Prove that \(\sqrt3+\sqrt5\) is an irrational number.

Answer 1

If possible, let \(\sqrt3+\sqrt5\) be a rational number equal to x. Then,

\(x = \sqrt3+\sqrt5\)

x² = \((\sqrt3+\sqrt5)^2\)

= \((\sqrt3)^2+(\sqrt5)^2+2\times\sqrt3\times\sqrt5\)

= \(3+5+2\sqrt{15}\)

= \(8 + 2\sqrt{15}\)

x² - 8 = \(2\sqrt{15}\)

\(\frac{x\times x - 8}{2}\) = \(\sqrt{15}\)

Now, x is rational

x² is rational

\(\frac{x\times x-8}{2}\) is rational

\(\sqrt{15}\) is rational

But, \(\sqrt{15}\) is irrational

Thus, we arrive at a contradiction. So, our supposition that \(\sqrt3+\sqrt5\) is rational is wrong.

Hence, \(\sqrt3+\sqrt5\) is an irrational number.