Question

prove that √5-√3 is a irrational number

Answer 1

Concept:

Irrational numbers are the set of real numbers that cannot be expressed in the form of a fraction, $\frac{p}{q}$ where p and q are integers. The denominator q is not equal to zero (q ≠ 0). Also, the decimal expansion of an irrational number is neither terminating nor repeating.

Given:

A number $\sqrt{5}-\sqrt{3}$

To prove:

The given number is irrational.

Solution:

Lets assume that $\sqrt{5}-\sqrt{3}$ is a rational number.

Any rational number can be represented as $\frac{p}{q}$  where q≠o

Therefore,

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

⇒$\sqrt{5}=\frac{p}{q}+\sqrt{3}$

Squaring on both sides

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

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

⇒$\frac{5-(\frac{p}{q} )^{2}-3}{2} *\frac{q}{p}= \sqrt{3}$

LHS is an irrational number and RHS is a rational number

Therefore

LHS≠RHS

Hence our assumption is wrong.

Therefore the given number is irrational number.