Question

prove that 7 over root 3 is irrational number ​

Answer 1

$\sf Let's\ assume\ that\ \dfrac{7}{\sqrt{3}}\ is\ rational.\\\\\\\dfrac{7}{\sqrt{3}}\ =\ \dfrac{a}{b},\ where\ 'a'\ and\ 'b'\ are\ co\ -\ prime\ integers\ and\ b\ is\ \ne\ 0.\\\\\\\dfrac{7}{\sqrt{3}}\ =\ \dfrac{a}{b}\\\\\\\dfrac{7b}{a}\ =\ \sqrt{3}$

$\sf Since\ 'a'\ and\ 'b'\ are\ integers\ and\ \dfrac{7b}{a}\ is\ rational,\\\\\\\implies\ \sqrt{3}\ is\ also\ rational.\\\\\\This\ contradicts\ the\ fact\ that\ \sqrt{3}\ is\ irrational.\\\\\\This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\\therefore\ Our\ assumption\ is\ wrong.\\\\\\\dfrac{7}{\sqrt{3}}\ is\ irrational.$

Answer 2

Answer:

Here is the answer......

Step-by-step explanation:

Let, 7 root3 be a rational number,

7《3 = a/b where b not equal to zero, (a,b=1)

《3 = a / 7b

We know that a / 7b is a rational number,

So the fact that root3 is irrational number.

i.e. irrational not equal to rational number.