Question

Prove the below numbers are irrational
1. 3√2 / 5
2. 3+5√2

Answer 1

$\sf Let's\ first\ prove\ that\ \dfrac{3\sqrt{2}}{5}\ is\ irrational.\\\\\\Let's\ assume\ that\ \dfrac{3\sqrt{2}}{5}\ is\ rational.\\\\\\\dfrac{3\sqrt{2}}{5}\ =\ \dfrac{a}{b},\ where\ 'a'\ and\ 'b'\ are\ co\ -\ prime\ integers\ and\ b\ \ne\ 0.\\\\\\\dfrac{3\sqrt{2}}{5}\ =\ \dfrac{a}{b}\\\\\\3\sqrt{2}\ =\ \dfrac{5a}{b}\\\\\\\sqrt{2}\ =\ \dfrac{5a}{3b}\\\\\\Since\ 'a'\ and\ 'b'\ are\ integers\ and\ \dfrac{5a}{3b}\ is\ rational,\\\\\\\implies\ \sqrt{2}\ is\ rational\ as\ well.$

$\sf This\ contradicts\ the\ fact\ that\ \sqrt{2}\ is\ irrational.\\\\\\This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\\therefore\ Our\ assumption\ is\ wrong.\\\\\\\dfrac{3\sqrt{2}}{5}\ is\ irrational.$

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$\sf Now,\ let's\ prove\ that\ 3\ +\ 5\sqrt{2}\ is\ irrational.\\\\\\Let's\ assume\ that\ 3\ +\ 5\sqrt{2}\ is\ rational.\\\\\\3\ +\ 5\sqrt{2}\ =\ \dfrac{a}{b},\ where\ 'a'\ and\ 'b'\ are\ co\ -\ prime\ integers\ and\ b\ \ne\ 0.\\\\\\3\ +\ 5\sqrt{2}\ =\ \dfrac{a}{b}\\\\\\5\sqrt{2}\ =\ \dfrac{a}{b}\ -\ 3\\\\\\5\sqrt{2}\ =\ \dfrac{a\ -\ 3b}{b}\\\\\\\sqrt{2}\ =\ \dfrac{a\ -\ 3b}{5}\\\\\\Since\ 'a'\ and\ 'b'\ are\ integers\ and\ \dfrac{a\ -\ 3b}{5}\ is\ rational,$

$\implies\ \sf \sqrt{2}\ is\ rational\ as\ well.\\\\\\This\ contradicts\ the\ fact\ that\ \sqrt{2}\ is\ irrational.\\\\\\This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\\therefore\ Our\ assumption\ is\ wrong.\\\\\\3\ +\ 5\sqrt{2}\ is\ irrational.$

Answer 2

$\huge\underline{SoLution:-}$

1). 3√2/5

Let us assume on the contrary that 3√2/5 is rational. then there exist co- prime positive integers a and b such that

3√2/5 = a/b

$\implies$ 3√2 = 5a/b

$\implies$ √2 = 5a/3b

$\implies$ √2 is irrational .

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

2). 3+5√2

Let us assume onthe contrary that 3 + 5√2is rational. then there exist co- prime positive integers a and b such that

3 + 5√2 = a/b

$\implies$ 5√2 = a/b - 3

$\implies$ √2 = a - 3b/5b

$\implies$ √5 is irrational.

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

$\huge\underline{ThAnks...}$