prove that is an irrational number.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rational Numbers

Question

prove that $3 + \sqrt{5}$ is an irrational number.

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Let us assume that

3 + \sqrt{5}

is a rational number.

Now,

3 + \sqrt{5} = \frac{a}{b}

[Here a and b are co-prime numbers]

\sqrt{5} = [(\frac{a}{b}) - 3]

\sqrt{5} = [(\frac{a - 3 b}{b})]

Here,

[(\frac{a - 3 b}{b})]

is a rational number.

But we know that

\sqrt{5}

is an irrational number.

So,

[(\frac{a - 3 b}{b})]

is also a irrational number.

So, our assumption is wrong.

3 + \sqrt{5}

is an irrational number.

Hence, proved.