Prove that is on irrational number and also prove that is irrational number.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rational Numbers

Question

Prove that $\sqrt{2}$ is on irrational number and also prove that $3 + 5 \sqrt{2}$ is irrational number.

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Consider the problem

$\sqrt{2}$ is non terminating, repeating terms so $\sqrt{2}$ is irrational.

For

3 + 5 \sqrt{2}

Let us consider that the

3 + 5 \sqrt{2}

is rational

3 + 5 \sqrt{2}

is rational. it must be in the form of

\frac{p}{q}

Then

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

5 \sqrt{2} = \frac{p}{q} - 3

\sqrt{2} = \frac{p - 3 q}{5}

And

\sqrt{2}

irrational

Therefore,

3 + 5 \sqrt{2}

is irrational.