Question

prove that root 2 + root 5 is irrational

Answer 1

√2²+√5²+2(√5)(√2) = p²/q²

2+5+2√10 = p²/q²

7+2√10 = p²/q²

2√10 = p²/q² - 7

√10 = (p²-7q²)/2q

p,q are integers then (p²-7q²)/2q is a rational number.

Then √10 is also a rational number.

But this contradicts the fact that √10 is an irrational number.

.°. Our supposition is false.

√2+√5 is an irrational number.

Hence proved.
hope you understand

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Answer 2

Question :-

Prove that √2 + √5 is irrational.

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 :)

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