Question

Prove that 3√2 - 3 is an irrational number, given √2 is an irrational number.

Answer 1

Answer:

⭐Answer⭐

$given \: that\: \sqrt{2} \: is \: \\ irrational. \\ \\ let \: be \: assume \: that \: 3 \sqrt{ 2} - 3 \\ is \: a \: rational \: number \\ \\ therefore \\ \\ 3 \sqrt{2} - 3 = \frac{a}{b} \\ \\ \3sqrt{2} = \frac{a}{b} - 3 \\ \\ 3 \sqrt{2} = \frac{a - 3b}{b} \\ \\ \sqrt{2} = \frac{a - 3b}{3b} \\ \\ \frac{a - 3b}{3b} \: is \: a \: rational \: number \\ therefore \: \sqrt{2} \: is \: also \: a \: \\ rational \: number \: but \: it \: is \\ given \: that \: \sqrt{2} \: is \: irrational \\ \\ therefore \: our \: assumption \: is \: wrong \\ and \: 3 \sqrt{2} - 3 \: is \: an \: \\ irrational \: number.$

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Answer 2

Let 3√2 - 3 be a rational number.

Rational numbers are expressed in the form a/b,

where a and b are co- prime and b≠0

$\implies 3\sqrt{2}-3=\dfrac{a}{b}$

$\implies 3\sqrt{2} = \dfrac{a}{b} +3$

$\implies 3\sqrt{2} = \dfrac{a+3b}{b}$

$\implies \sqrt{2} = \dfrac{a+3b}{3b}$

The RHS is a rational number

=> LHS is also a rational number

=> √2 is also a rational number

But, this contradicts to the fact that it is an irrational number.

Hence, our assumption is wrong.

$\boxed{\boxed{\bold{Therefore, \ 3\sqrt{2} -3 \ is \ an \ irrational \ number}}}}}}$