Question

√2 + √5
prove that they are irrational
please help its urgent
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Answer 1

$\large{\mathfrak{\underline{\underline{Answer :-}}}}$

Let us assume √2 + √5 is a rational number.

A rational number can be written in the form of p/q, where p and q are integers.

⇒ √2 + √5 = p/q

Squaring both side

⇒ (√2 + √5)² = (p/q)²

______________________

Using identity :-

(a + b)² = a² + b² + 2ab

________________[Put values]

⇒ (√2)² + (√5)² + 2(√2)(√5) = (p/q)²

⇒ 2 + 5 + 2√10 = p²/q²

⇒ 7 + 2√10 = p²/q²

⇒ 2√10 = p²/q² - 7

Taking LCM

⇒ 2√10 = (p² - 7q²)/q²

√10 = (p² - 7q²)/2q²

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

Then √10 is a rational number.

But this contradiction facts that √10 is a rational number.

∴ Our supposition is wrong. So,

⟹ √2 + √5 is a rational number.

Answer 2

ANSWER:-

Given:

√2 + √5.

To prove:

Prove that they are irrational number.

Solution:

Let √2 + √3 is a rational number which can be written in the form p/q where p & q are co-prime integers & q ≠0.

$= > \sqrt{2} + \sqrt{5} = \frac{p}{q} \\ Squaring \: on \: both \: sides ,\: we \: get ;\\ \ \\ = > ( \sqrt{2} + \sqrt{5} ) {}^{2} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ = > ( \sqrt{2} ) {}^{2} + ( { \sqrt{5} )}^{2} + 2 \times \sqrt{2} \times \sqrt{5} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ = > 2 + 5 + 2 \sqrt{10} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ = > 7 + 2 \sqrt{10} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ = > 2 \sqrt{10} = \frac{ {p}^{2} }{ {q}^{2} } - 7 \\ Therefore \\ = > \sqrt{10} = \frac{ {p}^{2} }{2 {q}^{2} } - \frac{7}{2} \\ So, \\ = > L.H.S = \sqrt{10} \: is \: an \: irrational \: number . \\ and \\ = > \frac{ {p}^{2} }{ {2q}^{2} } - \frac{7}{2} \: is \: a \: rational \: number. \: \: [R.H.S]$

Therefore,

L.H.S.≠ R.H.S.

So, our contradiction has arisen because of incorrect assumption.

So,

√2 + √5 is not a rational number.Hence, it will be an irrational number.

Hope it helps ☺️