Question

Prove √2+3√5 is an irrational number

Answer 1

SOLUTION:-

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Given:

√2 + 3√5

To find:

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Irrational number.

Explanation:

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Let us assume that √2+ 3√5 be a rational number.

A rational number is of the form of p/q

• p= numerator
• q= denominator

Where p & q are Integers and q ≠0.

Therefore,

$\sqrt{2} + 3 \sqrt{5} = \frac{p}{q}$

Squaring both sides, we get;

$( \sqrt{2} + 3 \sqrt{5} ) {}^{2} = ( \frac{p}{q} ) {}^{2} \\ \\ ( \sqrt{2} ) {}^{2} + (3 \sqrt{5} ) {}^{2} + 2 \times \sqrt{2} \times 3 \sqrt{5} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ 2 + 9\times 5 + 6 \sqrt{10} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ 2 + 45 + 6 \sqrt{10} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ 47 + 6 \sqrt{10} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ 6 \sqrt{10} = \frac{ {p}^{2} }{ {q}^{2} } - 47 \\ \\ 6 \sqrt{10} = \frac{ {p}^{2} - 47 {q}^{2} }{ {q}^{2} } \\ \\ \sqrt{10} = \frac{ {p}^{2} - 47 {q}^{2} }{6 {q}^{2} }$

p² - 47q² & 6q² is an integers.

We know that √10 is an irrational number.

$\frac{ {p}^{2} - 47 {q}^{2} }{6 {q}^{2} } \: is \: a \: rational \: number$

&

Irrational number ≠ rational number

Thus,

Our contradiction is wrong.

So,

√2 + 3√5 is an irrational number.