Question

Given that √2 is an irrational numbers.
Prove that (5+ 3√2) is an irrational numbers? ​

Answer 1

$\large\bf{\underline{\underline{Given\: that}}}$

$\underline{\boxed{\sf\purple{√2\: is \:irrational}}}$

$\large\bf{\underline{\underline{To\: prove:}}}$

$\underline{\boxed{\sf\purple{5 + 3√2 \:is \:irrational}}}$

$\large\bf{\underline{\underline{Assumption}}}$

$\textbf\color{teal}{Let us assume 5 + 3√2 is rational.}$

$\large\bf{\underline{\underline{Proof}}}$

As 5 + 3√2 is rational. (Assumed) They must be in the form of p/q where q≠0, and p & q are co prime.

$\tt\color{red}{Then,}$

$5+3 \sqrt{2}= \frac{p}{q}$

$3 \sqrt{2}= \frac{p}{q}$

$3 \sqrt{2} = \frac{p - 5q}{q}$

$\sqrt{2}= \frac{p-5q}{3q}$

$\bold{We\: know\: that,}$

$\sqrt{2} \ is \ irrational\ (given)$

$\frac{p-5q}{3q} \ is \ rational$

$\bold{And,}$

$\underline{\boxed{\sf\green{Rational\: ≠ \:Irrational.}}}$

Therefore we contradict the statement that, 5+3√2 is rational.

$\textbf\color{blue}{Hence proved that 5 + 3√2 is irrational.}$

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Thankyou :)

Refer the attachment for better understanding :)