Question

Show that √2+√3 is an irrational number...

Answer 1

Let root 2 + root 3 be x
Root 2 = X-root 3
Since we know root 2 is an irrational no
Therefore
Root2+Root3 is an irrational no

Answer 2

Let' us assume on contrary that √2 +√3 is an irrational number. Where a and b are positive integers.

√2 + √3 = $\frac{a}{b}$

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

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

$\frac{ {a}^{2} }{ {b}^{2} } \: - \sqrt{2} \times \frac{2a}{b} + 2 = 3 \\ \\ \frac{ {a}^{2} }{{b}^{2}} - 1 = \sqrt{2} \times \frac{2a}{b} \\ \\ \frac{ {a}^{2} - \: {b}^{2 } }{ {2ab} } = \sqrt{2} \\ \\ \sqrt{2} \: \: is \: a \: rational \: number \:$

√2 is a rational number since it is equal to a rational number which is in p/q form

This contradicts our assumption that √2 is an irrational number.So,our assumption is wrong

This contradicts our assumption that √2 is an irrational number.So,our assumption is wrong Hence √2 +√3 is irrational.