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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Step 1: Prove that

3 + \sqrt{5}

is an irrational number

Let’s assume

3 + \sqrt{5}

is a rational number and

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

Now,

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

\Rightarrow \sqrt{5} = \frac{a}{b} - 3

\Rightarrow \sqrt{5} = \frac{a - 3 b}{b}

Since,

\frac{a - 3 b}{b}

is a rational number.

So,

\sqrt{5}

is also a rational number, but it is a contradiction.

Therefore, our assumption is wrong.

Hence,

3 + \sqrt{5}

is an irrational number.