Prove that √2 is an irrational number.
Answer:
- Let's assume that √2 is a rational number.
i.e. √2 = a/b (a and b are the two integers, such that, b is not equal to zero(0), and a and b do not have common factor other than 1) - a = √2b
Squaring both side,
a2 = 2b2
Therefore, a2 is divisible by 2 and it can be said that a is divisible by 2.
Let a = 2m, where m is an integer.
Squaring both side,
a2 = (2m)2
Now,
(2m)2 = 2b2
b2 = 2m2
This means that b2 is divisible by 2 and hence, b is divisible by 2.
This implies that a and b have a common factor and this is contradiction to the fact that a and b are a co-prime. - Therefore, √2 is an irrational number.