# Prove that<

Let us suppose that √2 is an rational number and represent it in the form of where p and q are relatively prime and q≠0.
So,

On squaring both sides we get,

p2 = 2q2

As 2q2 is even then p2 is also even and p is also even.

Let, p2 = 4a2

4a2 = 2q2

q2 = 2a2

As 2a2 is even then q2 is also even and q is also even.

If p and q both are even, then they are further calculated which is not possible because of we assumed p and q are relatively prime number.

It means our assumption is wrong. So, by contradiction we can say that √2 is irrational number.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
All Grammar41 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

<span lang="EN-USRS Aggarwal - Mathematics