Question

Prove that √2+√3 is irrational

Answer 1

To Prove :

√2 + √3 is irrational number.

SolutioN :

°•° Let assume √2 + √3 is rational number.

So,

$\tt : \implies \sqrt{2} + \sqrt{3} = \dfrac{p}{q}$

Where as,

• p , q are Co prime and HCF ( p , q ) = 1.

Now, Squaring both sides.

$\tt : \implies{ \Bigg(\sqrt{2} + \sqrt{3}\Bigg) }^{2} = {\Bigg(\dfrac{p}{q} \Bigg) }^{2}$

$\tt : \implies 2 + 3 + 2 \sqrt{6} = \dfrac{{p}^{2} }{{q }^{2} }$

$\tt : \implies 5 + 2 \sqrt{6} = \dfrac{{p}^{2} }{{q }^{2} }$

$\tt : \implies 2 \sqrt{6} = \dfrac{{p}^{2} }{{q }^{2} } - 5$

$\tt : \implies2 \sqrt{6} = \dfrac{{p}^{2} - 5{q}^{2} }{{q }^{2} }$

$\tt : \implies\sqrt{6} = \dfrac{{p}^{2} - 5{q}^{2} }{2{q }^{2} }$

Here, We are notice √6 is irrational number but p² - 5q² / 2q² is rational number.

# Irrational number ≠ Rational number.

Hence, Our assumption was wrong √2 +√3 is irrational number.

Hence ProveD.

Answer 2

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