Question

Prove that √5+√7 is irrational​

Answer 1

Answer:

PROVED

Explanation:

Let us assume that 5+7 is a rational number.

⇒5+7=qp, where p and q are two integers and q=0

⇒7=qp−5=qp−5q

Since, p, q and 5 are integers, so qp−5q is a rational number.

⇒7 is also a rational number.

But this contradicts the fact that 7 is an irrational number.

This contradiction has arisen due to our assumption that 5+7 is a rational number.

Hence, 5+7 is an irrational number.

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Answer 2

To Prove :-

• √5+√7 is irrational.

Proof :-

Let us assume that √5+√7 is not irrational.That means _

• √5+√7 is rational

$\sf\green{ \: Let \: \sqrt{5} + \sqrt{7} = \dfrac{x}{y}}$

• Where x and y are integers such that $\sf y \ne 0$ and x and y are co - prime.

$\sf :\implies \: \sf\: \sqrt{7} = \dfrac{x}{y} - \sqrt{5}$

On squaring both sides:-

$\sf :\implies\:\sf \: 7 = 5 + \dfrac{ {x}^{2} }{ {y}^{2} } - 2 \times \sqrt{5} \times \dfrac{x}{y}$

$\sf :\implies\:\dfrac{2 \sqrt{5}x }{y} = \dfrac{ {x}^{2} }{ {y}^{2} } - 2$

$\sf :\implies\:\dfrac{2 \sqrt{5}x }{y} = \dfrac{ {x}^{2} - {2y}^{2} }{ {y}^{2} }$

$\sf:\implies\:\sqrt{5} = \dfrac{ {x}^{2} - {2y}^{2} }{2xy}$

Here, x and y are integers.Therefore :-

$\sf :\implies\:\dfrac{ {x}^{2} - {2y}^{2} }{2xy}$ is rational number.

$\sf :\implies\: \sqrt{5} \: is \: rational.$

• Which is contradiction as √5 is irrarional.So,our assumption is wrong that $\sf \sqrt{5} +\sqrt{7}$ is not irrational.

$\sf\green{\: Hence \: \sqrt{5} + \sqrt{7} \: is \: irrational.}$

• (Proved..!!)