Question

7. Prove that 3+5root2) is an irrational number, given that root 2 is an irrational number ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: \sqrt{2} \: is \: irrational.$

$\rm :\longmapsto\:\: Let \: assume \: that \: 3 + 5 \sqrt{2} \: is \: not \: irrational.$

So,

$\rm :\implies\:3 + 5 \sqrt{2} \: is \: rational.$

$\rm :\longmapsto\:Let \: 3 + 5 \sqrt{2} \: = \dfrac{x}{y}$

where,

$\red{ \sf \: x \: and \: y \: are \: integers \: such \: that \: y \: \ne \: 0 \: and \: hcf(x,y) = 1}$

$\rm :\longmapsto\:\: 5 \sqrt{2} \: = \dfrac{x}{y} - 3$

$\rm :\longmapsto\:\: 5 \sqrt{2} \: = \dfrac{x - 3y}{y}$

$\rm :\longmapsto\:\: \sqrt{2} \: = \dfrac{x - 3y}{5y}$

As x and y are integers,

So, x - 3y and 5y are also integers.

$\rm :\longmapsto\:\: \dfrac{x - 3y}{5y} \: is \: rational$

$\bf\implies \: \sqrt{2} \: is \: rational.$

which is contradiction to the fact as it is given that

$\rm :\longmapsto\: \: \sqrt{2} \: is \: irrational.$

$\rm :\longmapsto\:Hence, \: our \: assumption \: is \: wrong.$

$\bf\implies \:3 + 5 \sqrt{2} \: is \: irrational.$

Hence, Proved

Additional Information :-

Irrational numbers

• Irrational numbers are those numbers whom decimal representation is non terminating and non repeating. Basically, square root of prime numbers are always Irrational.

Rational numbers

• Rational numbers are those numbers whom decimal representation is either terminating or non - terminating but repeating.