Question

prove that 2+5root 3 is an irrational number, it is being that root 3 is an irrational number .​

Answer 1

Step-by-step explanation:

Let 2 + 5√3 = a, where a is a rational number.

√3 = (a – 2)/5

Which is a contradiction as LHS is irrational and RHS is rational.

∴ 2 + 5√3 can not be rational.

Hence, 2 + 5√3 + is irrational.

Answer 2

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

$\rm :\longmapsto\: \sqrt{3} \: is \: irrational$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:2 + 5 \sqrt{3} \: is \: irrational.$

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

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

So,

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

So,

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

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

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

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

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

As,

$\rm :\longmapsto\:x \: and \: y \: are \: integers.$

So,

$\rm :\longmapsto\:x - 2y \: is \: also \: an \: integer.$

$\rm :\implies\:\dfrac{x - 2y}{5y} \: is \: rational.$

$\rm :\implies\: \sqrt{3} \: is \: rational$

$\rm :\longmapsto\:which \: is \: contradiction \: as \: it \: is \: given \: \sqrt{3} \: is \: irrational.$

Hence,

• Our assumption is wrong,

Thus,

$\rm :\longmapsto\:2 + 5 \sqrt{3} \: is \: irrational.$

Additional Information :-

Irrational numbers :- Irrational numbers are the real numbers whom decimal representation is neither terminating nor repeating. Basically, square root of prime numbers all are Irrational.

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