Prove that is an irrational number.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rational Numbers

Question

Prove that $5 - \sqrt{3}$ is an irrational number.

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Let us assume on the contrary that $5 - \sqrt{3}$ is
rational. Then, there exist prime positive integers a
and b such that

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

\Rightarrow

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

\Rightarrow

\sqrt{3}

is rational [∵a,b are integers

∴ \frac{5 b - a}{b}

is a rational number]

This contradicts the fact that

\sqrt{3}

is irrational.
So, our assumption is incorrect. Hence,

5 - \sqrt{3}

is
an irrational number.