Question

Prove that √5 is irrational


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Answer 1

Let us assume that √5 is a rational number.
we know that the rational numbers are in the form of p/q form where p,q are intezers.
so, √5 = p/q
     p = √5q
we know that 'p' is a rational number. so √5 q must be rational since it equals to p
but it doesnt occurs with √5 since its not an intezer
therefore, p =/= √5q
this contradicts the fact that √5 is an irrational number
hence our assumption is wrong and √5 is an irrational number.

Answer 2

If possible, let us assume that ,$\sqrt{5}$ is irrational.

$\sqrt{5}$=a÷b
$\sqrt{5}$×b=a
b$\sqrt{5}$=a
squaring on both sides
(b$\sqrt{x})^2$ =$a^{2}$
$a^{2}$=5$b^{2}$      this is first equation
$a^{2}$ is a multiple of 5,$a^{2}$ is multiple of 5,a is also multiple of 5.
$a^{2}$=(5$l^{2}$)
$a^{2}$= $25 l^{2}$    this is second equation
equate equation :- 1&2
25l=5$b^{2}$
25l÷5=$b^{2}$
5l=$b^{2}$
$b^{2}$ is a multiple of 5 and b is also multiple of 5
As , a and b are multiples of 2 which are contradictory. Our assumption is wrong.
$\sqrt{5}$ is irrational