Question

Show that 3-root5 is an irrational

Answer 1

Assume that 3 - √5 is rational and let be x.

$\displaystyle x=3-\sqrt{5} \\ \\ \\ x^2=(3-\sqrt{5})^2 \\ \\ \\ x^2=9+5-6\sqrt{5} \\ \\ \\ x^2=14-6\sqrt{5} \\ \\ \\ 6\sqrt{5} = 14-x^2 \\ \\ \\ \sqrt{5}=\frac{14-x^2}{6}$

Seems that √5 can be written as a fraction at the last step. But it's absolutely wrong.

Thus a contradiction occurs here.

Hence proved!!!