Question

prove root 2 + root 5 is irrational ​

Answer 1

$\Huge{\underline{\underline{\blue{\mathfrak{ Solution }}}}}$

$\sf Let\ \sqrt{2} + \sqrt{5} = Rational = x\\\\Thus,\ \sqrt{2} + \sqrt{5} = x\\\\\\\rightarrow\ \sqrt{2} + \sqrt{5} + 2\sqrt{10} = x^{2} \\\\\rightarrow\ \sqrt{10} = \frac{x^{2} - 7}{2} \\\\\rightarrow \sqrt{10} = Rational\\ \\\\Since,\ \sqrt{10}\ can\ never\ be\ rational.\\\\\\\\Thus,\ \sqrt{2} + \sqrt{5} = Irrational$

Answer 2

Solution :-

Let us assume that √2 + √5 is rational.

i.e, √2 + √5 = a/b where 'a' and 'b' are co primes and b ≠ 0

$\tt \sqrt{2} + \sqrt{5} = \dfrac{a}{b}$

$\tt \sqrt{2} = \dfrac{a}{b} - \sqrt{5}$

Squaring on both sides

$\tt (\sqrt{2})^{2} = (\dfrac{a}{b} - \sqrt{5})^{2}$

$\tt \sqrt{4} = (\dfrac{a}{b})^{2} - 2( \dfrac{a}{b})( \sqrt{5}) + {( \sqrt{5})}^{2}$

[Since (x - y)² = x² - 2xy + y² and above in RHS x = a/b, y = √5 ]

$\tt 2 = \dfrac{a^{2} }{b^{2} } - 2\sqrt{5}. \dfrac{a}{b} + \sqrt{25}$

$\tt 2 = \dfrac{a^{2} }{b^{2} } - 2\sqrt{5}. \dfrac{a}{b} + 52$

$\tt 2 \sqrt{5}. \dfrac{a}{b} = \dfrac{a^{2} }{b^{2} }+ 5 - 22$

$\tt 2 \sqrt{5}. \dfrac{a}{b} = \dfrac{a^{2} }{b^{2} }+ 32$

Taking LCM in RHS

$\tt 2 \sqrt{5}. \dfrac{a}{b} = \dfrac{a^{2} + 3b^{2} }{b^{2} }2$

$\tt 2 \sqrt{5} = \dfrac{a^{2} + 3b^{2} }{b^{2} } \times \dfrac{b}{a}2$

$\tt 2 \sqrt{5} = \dfrac{a^{2} + 3b^{2} }{b } \times \dfrac{1}{a}2$

$\tt 2 \sqrt{5} = \dfrac{a^{2} + 3b^{2} }{ab}2$

$\tt \sqrt{5} = \dfrac{a^{2} + 3b^{2} }{2ab}$

Since 'a' and 'b' are integers Right Hand Side i.e

$\tt \dfrac{ {a}^{2} + 3{b}^{2} }{2ab}$

is a rational number.

So, Left Hand Side of the equation is a rational number.

But, this contradicts the fact that √5 is irrational.

This contradiction has arised because of our wrong assumption that √2 + √5 is rational.

So we can conclude that √2 + √5 is irrational.

Hence proved :)