Question

Prove that √5is irrational number​

Answer 1

prove that

$\sqrt{5}$

is an irrational number.

.

.

.

Answer 2

To prove that √5 is irrational.

Let us assume that 5 is rational.

$\tt Let\ us\ assume\ that\ \sqrt{5}\ is\ rational.\\ \\\\\sqrt{5}\ =\ \dfrac{a}{b},\ where\ a\ and\ b\ are\ co - prime\ integers\ and\ b\ \ne\ 0.\\ \\\\\\\sqrt{5}\ =\ \dfrac{a}{b}\\\\ \\Squaring\ both\ the\ sides,\\\\\\(\sqrt{5})^2\ =\ \bigg(\dfrac{a}{b}\bigg)^2\\ \\\\5\ =\ \dfrac{a^2}{b^2}\\ \\\\a^2\ =\ 5b^2\\\\\\\implies\ 5\ divides\ a^2.\\\\\\\implies\ 5\ divides\ a.\ \ \ \bigg[If\ p\ divides\ a^2,\ then\ p\ divides\ a.\bigg]\\\\\\\\Let\ a\ =\ 5c,\ for\ some\ integer\ c.$

$\tt Substituting\ this\ value\ in\ the\ above\ obtained\ equation,\\\\\\(5c)^2\ =\ 5b^2\\\\\\25c^2\ =\ 5b^2\\\\\\b^2\ =\ 5c^2\\\\\\\implies\ 5\ divides\ b^2.\\\\\\\implies\ 5\ divides\ b.$

Thus 5 is a factor of both a and b.

This contradicts the fact that a and b are co-prime integers.

This contradiction has arisen due to our wrong assumption.

Therefore, √5 is irrational.