Prove that √2 is an irrational number.

by

Prove that √2 is an irrational number. Hence show that 3 — √2 is an irrational.

Solution:

Let us consider √2 be a rational number, then
√2 = p/q, where ‘p’ and ‘q’ are integers, q ≠ 0 and p, q have no common factors (except 1).
So,
2 = p2 / q2
p2 = 2q2 …. (1)
As we know, ‘2’ divides 2q2, so ‘2’ divides p2 as well. Hence, ‘2’ is prime.
So 2 divides p
Now, let p = 2k, where ‘k’ is an integer
Square on both sides, we get
p2 = 4k2
2q2 = 4k2 [Since, p2 = 2q2, from equation (1)]
q2 = 2k2
As we know, ‘2’ divides 2k2, so ‘2’ divides q2 as well. But ‘2’ is prime.
So 2 divides q
Thus, p and q have a common factor 2. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √2 is not a rational number.
√2 is an irrational number.
Now, let us assume 3 – √2 be a rational number, ‘r’
So, 3 – √2 = r
3 – r = √2
We know that, ‘r’ is rational, ‘3- r’ is rational, so ‘√2’ is also rational.
This contradicts the statement that √2 is irrational.
So, 3- √2 is irrational number.
Hence proved.

More Solutions:

Leave a Comment