Question

Prove that underroot 3-underroot 5 is irrational.​Please quickly I need this answer

Answer 1

Answer:

See Explanation

Step-by-step explanation:

Let us assume that $\sqrt 3 - \sqrt 5$ is a rational number.

So, it can be expressed in the form of $\frac{p}{q}$ where p and q are integers.

$\therefore \sqrt 3 - \sqrt 5=\frac{p}{q}$

$\therefore \sqrt 3 =\frac{p}{q}+ \sqrt 5$

$\therefore \sqrt 3 =\frac{p+q\sqrt 5}{q}$

Squaring both sides:

$(\sqrt 3) ^2 =\bigg(\frac{p+q\sqrt 5}{q}\bigg ) ^2$

$\therefore 3 =\frac{p^2 +5q^2 +2\sqrt 5 pq}{q^2 }$

$\therefore 3q^2 ={p^2 +5q^2 +2\sqrt 5 pq}$

$\therefore 0 = p^2 +5q^2 -3q^2 +2\sqrt 5 pq$

$\therefore - 2\sqrt 5 pq=p^2 +2q^2$

$\therefore \sqrt 5 = \bigg (\frac{p^2 +2q^2}{-2pq} \bigg)$

$\therefore \sqrt 5 = - \bigg(\frac{p^2 +2q^2}{2pq} \bigg)$

Since, p and q are integers so $- \bigg(\frac{p^2 +2q^2}{2pq} \bigg)$ is rational.

$\therefore \sqrt 5$ is also rational.

But it contradicts the fact that $\sqrt 5$ is irrational.

Hence, our assumption that $\sqrt 3 - \sqrt 5$ is rational is wrong.

$\therefore \sqrt 3 - \sqrt 5$ is irrational.