Question

Prove that (6-3√5) is irrational.​

Answer 1

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

Proof:

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

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

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

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

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

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

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

$\tt{ \dfrac{a + 6b}{3b}}$ is a rational number because a and b are integers.

This implies that $\tt{\sqrt{5}}$ is also rational. But this is a contradiction to the fact that $\tt{\sqrt{5}}$ is irrational. The contradiction is arisen due to wrong assumption.

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

Answer 2

Answer:

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