Question

Prove that 6-5√3 is irrational

Answer 1

Answer:

let us assume 6-5√3 is a rational number

6-5√3 = a/b ( a & b are coprime , a is not= 0)

6 - a/b = 5√3

6b-a/5b = √3

( as a & b are integer 6b-a/5b is integer also a rational number so rational number is equal to the ratinal number so √3 is rational )

(so our assumption is wrong that √3 is rational hence 6-5√3 is irrational)

Answer 2

To Prove : $\tt{6 - 5 \sqrt{3}}$

Proof:

Assume that $\tt{6 - 5 \sqrt{3}}$ is rational.

$\tt{\longrightarrow} \: 6 - 5 \sqrt{3} \: = \dfrac{a}{b} - - \gray{(a \: and \: b \: are \: integers)}$

$\tt{\longrightarrow} \: 6 - 5 \sqrt{3} \: = \dfrac{a}{b}$

$\tt{\longrightarrow} \: - 5 \sqrt{3} \: = \dfrac{a}{b} - 6$

$\tt{\longrightarrow} \: 5 \sqrt{3} \: = \dfrac{a}{b} + 6$

$\tt{\longrightarrow} \: 5 \sqrt{3} \: = \dfrac{a + 6b}{b}$

$\tt{\longrightarrow} \: \sqrt{3} \: = \dfrac{a + 6b}{5b}$

$\tt{ \dfrac{a + 6b}{5b}}$ is a rational number because a and b are integers.

This implies that $\tt{\sqrt{3}}$ is also rational. But this is a contradiction to the fact that $\tt{\sqrt{3}}$ is irrational. The contradiction is arisen due to wrong assumption.

Hence proved, $\tt{6 - 5 \sqrt{3}}$ is irrational.