Question

prove that √5 is irrational​

Answer 1

Let us prove that √5 is an irrational number, by using the contradiction method. So, say that √5 is a rational number can be expressed in the form of pq, where q ≠0. So, let √5 equals pq. Where p, q are co-prime integers i.e. they do not have any common factor except '1'.

Answer 2

Let's take √5 as rational

we can write √5 = a/b such that , a & b are integers , b ≠ 0 and there are no common factors of a and b .

Multiply by b on both sides we get ;

b√5 = a

to remove the root , squaring on both sides , we get

5b² = a² _____ ( 1 )

That means 5 is a factor of a² .

For any prime no. p which is a factor of a² then it will be the factor of a also .

So , 5 is a factor of a _____ (2)

Hence , we can write a = tc for some integer c .

Putting the value of a in eqn. ( 1 ) we get

5b² = (5c)²

5b² = 25c²

Dividing by 5 ; we get ,

b² = 5c²

It means 5 is a factor of b²

Thus , 5 is a factor of b _____ (3)

From ( 1 ) , (2) and (3) , we can say that 5 is a factor of both a and b

This contradicts the fact that a and b has no common factors .

Thus our assumption is wrong .

Hence , √5 is an irrational number .