Show that is irrational.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rational Numbers

Question

Show that $(5 - \sqrt{3})$ is irrational.

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We Have to prove that $5 - \sqrt{3}$ is irrational.

We will prove it by Contradiction method.

Let us assume the opposite, i.e,

5 - \sqrt{3}

is rational

As we know that any rational no. can be written in the form of

\frac{p}{q}

where p and q (q ≠ 0) are co-prime (no common factor other than 1).

Hence,

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

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

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

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

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

Here,

\frac{5 q - p}{q}

is rational.

But

\sqrt{3}

is irrartional.

Since, Rational ≠ Irrational

This is contradiction.

∴ Our assumption is incorrect.

Hence,

5 - \sqrt{3}

is irrational.

Hence Proved.