Question

Prove that (√2+√3) is an irrational

Answer 1

Heya friend !

We can prove this by contradiction method,

For this we will consider $\sqrt{2} + \sqrt{3}$ to be rational.

So,

$\sqrt{2} + \sqrt{3} = \frac{a}{b}$                        

(where, a and b are both co - prime numbers having no common factor other than 1)

$\sqrt{2} =\frac{a}{b} - \sqrt{3}$

Now, squaring both sides, we get,

2 = $\frac{a^{2} }{b^{2} }$ + 3 - $2 (\sqrt{3} ) (\frac{a}{b} )$

$2\sqrt{3} (\frac{a}{b} ) - \frac{a^{2} }{b^{2} } = 1$

$2\sqrt{3}(\frac{a}{b}) = 1 + \frac{a^{2} }{b^{2} }$

$2\sqrt{3} (\frac{a}{b}) = \frac{a^{2}+b^{2} }{b^{2} }$

So,

$\sqrt{3} = \frac{a^{2}+b^{2} }{2ab}$

[ as 'a' and 'b' are integers, So,  $\frac{a^{2}+b^{2} }{2ab }$ is rational

So,

$\sqrt{3}$ is also rational, which contradicts the fact that $\sqrt{3}$ is irrational.

So,

Our contradiction is wrong and hence,

√2+√3 is irrational

Thanks !

#BAL #answerwithquality