Question

prove that √3 is an irrational number.​

Answer 1

Step-by-step explanation:

Let us assume on the contrary that

3

is a rational number.

Then, there exist positive integers a and b such that

3

=

b

a

where, a and b, are co-prime i.e. their HCF is 1

Now,

3

=

b

a

⇒3=

b

2

a

2

⇒3b

2

=a

2

⇒3 divides a

2

[∵3 divides 3b

2

]

⇒3 divides a...(i)

⇒a=3c for some integer c

⇒a

2

=9c

2

⇒3b

2

=9c

2

[∵a

2

=3b

2

]

⇒b

2

=3c

2

⇒3 divides b

2

[∵3 divides 3c

2

]

⇒3 divides b...(ii)

From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

Hence,

3

is an irrational number.