Prove that, √3 is an irrational number.

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Prove that, √3 is an irrational number. Hence, show that 2/5×√3 is an irrational number.

Solution:

Let us consider √3 be a rational number, then
√3 = p/q, where ‘p’ and ‘q’ are integers, q ≠ 0 and p, q have no common factors (except 1).
So,
3 = p2 / q2
p2 = 3q2 …. (1)
As we know, ‘3’ divides 3q2, so ‘3’ divides p2 as well. Hence, ‘3’ is prime.
So 3 divides p
Now, let p = 3k, where ‘k’ is an integer
Square on both sides, we get
p2 = 9k2
3q2 = 9k2 [Since, p2 = 3q2, from equation (1)]
q2 = 3k2
As we know, ‘3’ divides 3k2, so ‘3’ divides q2 as well. But ‘3’ is prime.
So 3 divides q
Thus, p and q have a common factor 3. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √3 is not a rational number.
√3 is an irrational number.
Now, let us assume (2/5)√3 be a rational number, ‘r’
So, (2/5)√3 = r
5r/2 = √3
We know that, ‘r’ is rational, ‘5r/2’ is rational, so ‘√3’ is also rational.
This contradicts the statement that √3 is irrational.
So, (2/5)√3 is irrational number.
Hence proved.

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