Question

6+√2 prove that irrational

Answer 1

as root 2 is irrational therefore any number added to a irrational number is irrational
hence 6+root 2 is irrational

Answer 2

Answer:

$\Large{\underline{\underline{\bf{Given:-}}}}$

we know that √2 is irrational

$\Large{\underline{\underline{\bf{To Prove:-}}}}$

6+√2

$\Large{\underline{\underline{\bf{Solution:-}}}}$

$let \: 6 + \sqrt{2} is \: a \: rational \: no. \\ \\ 6 + \sqrt{2} = \frac{x}{y} (where \: x \: and \: y \: are \: integers) \\ \\ \sqrt{2} = \frac{x}{y} - 6 \\ \\ \sqrt{2} = \frac{x - 6y}{y}$

★√2 is a irrational no. that we know already which is equating with a rational no. but here they both are rational no.so, our assumption is wrong

Hence, 6+√2 is irrational no.

$\Large{\underline{\underline{\bf{Hence proved!}}}}$

$\huge\mathfrak\pink{Thanks}$