Question

prove 2-3$\sqrt{x}$5 is an irrational number

Answer 1

Correct question:

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

Proof:

$\sf Let's\ assume\ that\ 2\ -\ 3\sqrt{5}\ is\ rational.\\\\\\2\ -\ 3\sqrt{5}\ =\ \dfrac{a}{b},\ where\ 'a'\ and\ 'b'\ are\ co\ -\ prime\ intgers\ and\ b\ is\ \ne\ 0.\\\\\\2\ -\ 3\sqrt{5}\ =\ \dfrac{a}{b}\\\\\\2\ -\ \dfrac{a}{b}\ =\ 3\sqrt{5}\\\\\\\dfrac{2b\ -\ a}{b}\ =\ 3\sqrt{5}\\\\\\\dfrac{2b\ -\ a}{3b}\ =\ \sqrt{5}$

$\sf Since\ 'a'\ and\ 'b'\ are\ integers\ and\ \dfrac{2b\ -\ a}{3b}\ is\ rational,\\\\\\\implies\ \sqrt{5}\ is\ rational\ as\ well.\\\\\\This\ contradicts\ the\ fact\ that\ \sqrt{5}\ is\ irrational.\\\\\\This\ contradictions\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\\therefore\ Our\ assumption\ is\ wrong.\\\\\\\bf 2\ -\ 3\sqrt{5}\ is\ irrational.$

Answer 2

$\large\underline\mathrm{Correct\: Question:-}$

Prove that 2 - 3√5 is an irrational number.

$\large\underline\mathrm{GivEn:-}$

↪ 2 $-$ 3√5

$\large\underline\mathrm{To\: Prove:-}$

2 - 3√5 is an irrational number

$\large\underline\mathrm{Proof:-}$

Let us assume on the contrary that 2$-$ 3√5 is rational. Then, there exist co-prime positive integer a and b such that

2 $-$ 3√5 = $\large{\sf{\frac{a}{b}}}$

$\implies$ 2 $-$ $\large{\sf{\frac{a}{b}}}$ = 3√5

$\implies$ $\large{\sf{\frac{2b - a}{3b}}}$ = √5

$\implies$ √5 is a rational number. [ a,b are integers , so, $\large{\sf{\frac{2b - a}{3b}}}$ is a rational ]

This contradicts the fact that √5 is irrational. So, our supposition is incorrect. Hence, 2-3√5 is an irrational number.

$\huge\underline\mathrm{ThAnKs...}$