Question

Prove that√3 + √7 is an Irrational

Answer 1

Answer:

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Step-by-step explanation:

√3+√7 is irrational.

Let us assume that √3+√7 is rational.

That is , we can find coprimes a and b (b≠0) such that

Therefore,

Squaring on both sides ,we get

Rearranging the terms ,

Since, a and b are integers ,  is rational ,and so √3 also rational.

But this contradicts the fact that √3 is irrational.

This contradiction has arisen because of our incorrect assumption that √3+√7 is rational.

Hence, √3+√7 is irrational.

Answer 2

Let us assume on the contrary that √3 + √7 is rational number.

Then there exist co-prime positive integers a and b such that

$\sqrt{3} + \sqrt{7} = \frac{a}{b} \\ \frac{a}{b} - \sqrt{3} = \sqrt{7} \\ squaring \: on \: both \: sides \\( \frac{a}{b} - \sqrt{3} ) ^{2} = (\sqrt{7}) ^{2} \\ \frac{ {a}^{2} }{ {b}^{2} } - \frac{2a}{b} \sqrt{3} + 3 = 7 \\ \frac{ {a}^{2} }{ {b}^{2} } - 4 = \frac{2a}{b} \sqrt{3} \\ \frac{ {a}^{2} - 4 {b}^{2} }{ {b}^{2} } = \frac{2a}{b} \sqrt{3} \\ \frac{ {a}^{2} - 4b ^{2} }{2ab} = \sqrt{3} \\ a\: and \: b \: are \: integers \\ therefore \: \frac{ {a}^{2} - 4b ^{2} }{2ab} \: is \: rational$

Therefore √ 3 is rational number.

But this contradict the fact that √3 is irrational. So our assumption is wrong.

Hence √3 + √7 is irrational