Question

Prove that 6+√2 is irrational?
No copied answer ❌.

Answer 1

$\huge \fcolorbox{black}{lightblue}{Solution.}$

$Let \: us \:assume \: to \: the \: contrary \: that \: 6 \times + \sqrt{5} \: is \: rational.$

So, we can find coprime integers a and b ( ≠ 0 )

such that

$6 + \sqrt{2} = \frac{a}{b}$

$6 - \frac{a}{b} = \sqrt{2}$

$Since \: <strong>a</strong> \: and \: <strong>b</strong> \: are \: integers. \: \\ and \: we \: get \frac{a}{b} \: is \: rational \: and \: so \: 6 - \frac{a}{b} \: is \: rational \: and \: so \: \sqrt{2} \: is \: rational. \\ But \: this \: contradicts \: the \: fact \: that \: \sqrt{2} \: is \: irrational.$

$So, \: we \: conclude \: that \: 6 + \sqrt{2 } \: is \: irrational.$

Answer 2

Answer:

Let us assume that 6+√2 is rational.

That is , we can find coprimes a and b (b≠0) such that

$6 + \sqrt{2} = \frac{a}{b}$

⇒$\sqrt{2} = \frac{a}{b} - 6$

⇒$\sqrt{2} = \frac{a-6b}{b}$

Since , a and b are integers ,  is rational ,and so √2 is rational.

But this contradicts the fact that √2 is irrational.

So, we conclude that 6+√2 is irrational.