Prove that is not a rational number.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rationalisation

Question

Prove that $\sqrt{5} - \sqrt{3}$ is not a rational number.

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Let

\sqrt{5} - \sqrt{3}

be a rational number of form

\frac{a}{b}

,where

b

≠ 0

Squaring on both sides

(\sqrt{5} - \sqrt{3})^{2} = (\frac{a}{b})^{2}

(\sqrt{5})^{2} + (\sqrt{3})^{2} - 2 (\sqrt{5}) (\sqrt{3}) = \frac{a^{2}}{b^{2}}

5 + 3 + 2 \sqrt{1} 5 = \frac{a^{2}}{b^{2}}

8 + 2 \sqrt{1} 5 = \frac{a^{2}}{b^{2}}

2 \sqrt{1} 5 = \frac{a^{2}}{b^{2}} - 8

\sqrt{1} 5 = \frac{a^{2} - 8 b^{2}}{2 b^{2}}

since

\sqrt{1} 5

is irrational ,

\frac{a^{2} - 8 b^{2}}{2 b^{2}}

is rational

Since LHS ≠ RHS, contradiction arises,

Therefore

\sqrt{5} - \sqrt{3}

is irrational