Prove that √5 is irrational. And also Explain.
Answers
Answered by
2
v can prove this by contradiction method :
let us assume tht √5 is rational
so v can get a/b such tht - √5=a/b
⇒√5b=a ..................... (i)
squaring on both sides v get
5b²=a² ............(ii)
so 5 divides a² which v can say as 5 divides a
now, let a=5c for some integer c.
now again by squaring on both sides v get
a²=25c² ........................(iii)
now by (ii)
substituting a²
v get
5b²=25c²
i.e , b²=5c²
⇒5 divides c² which v can say tht 5 divides c.
so v can say tht a and b have 5 as a common factor.
but this contradicts d fact tht a and b r co prime( they do not have any other factor other than 1)
this contradiction arises due to our wrong consumption tht √5 is rational
hence v can conclude tht √5 is irrational.
hope it helped u!!!!!
let us assume tht √5 is rational
so v can get a/b such tht - √5=a/b
⇒√5b=a ..................... (i)
squaring on both sides v get
5b²=a² ............(ii)
so 5 divides a² which v can say as 5 divides a
now, let a=5c for some integer c.
now again by squaring on both sides v get
a²=25c² ........................(iii)
now by (ii)
substituting a²
v get
5b²=25c²
i.e , b²=5c²
⇒5 divides c² which v can say tht 5 divides c.
so v can say tht a and b have 5 as a common factor.
but this contradicts d fact tht a and b r co prime( they do not have any other factor other than 1)
this contradiction arises due to our wrong consumption tht √5 is rational
hence v can conclude tht √5 is irrational.
hope it helped u!!!!!
npriya:
ur wlcm
Similar questions