Question

Prove that $\sqrt{2} - 3\sqrt{5}$ is an irrational number!

Answer 1

Correct question:

To prove √2 - 3√5 is an irrational number.

Proof:

$\sf Let's\ assume\that\\\sqrt{2}\ -\ 3\sqrt{5}\ is\ rational.\\\\\\\\sqrt{2}\ -\ 3\sqrt{5}\ =\ \dfrac{a}{b},\ where\ 'a'\ and\ 'b'\ are\ co\ -\ prime\ intgers\ and\ b\ is\ \ne\ 0.\\\\\\\sqrt{2}\ -\ 3\sqrt{5}\ =\ \dfrac{a}{b}\\\\\\sqrt{2}\ -\ \dfrac{a}{b}\ =\ 3\sqrt{5}\\\\\\\dfrac{\sqrt{2b}\ -\ a}{b}\ =\ 3\sqrt{5}\\\\\\\dfrac{\sqrt{2b}\ -\ a}{3b}\ =\ \sqrt{5}\\\\\\</p><p></p><p>\sf Since\ 'a'\ and\ 'b'\ are\ integers\ and\ \dfrac{\sqrt{2b}\ -\ a}{3b}\ is\ rational,\\\\\\\implies\ \sqrt{5}\ is\ rational\ as\ well.\\\\\\This\ contradicts\ the\ fact\ that\ \sqrt{5}\ is\ irrational.\\\\\\This\ contradictions\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\\therefore\ Our\ assumption\ is\ wrong.\\\\\\\bf\sqrt{2}\ -\ 3\sqrt{5}\ is\ irrational.$

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