Prove that √5 is an irrational number.

Prove that √5 is an irrational number. Hence, show that -3 + 2√5 is an irrational number.

Solution:

Let us consider √5 be a rational number, then
√5 = p/q, where ‘p’ and ‘q’ are integers, q ≠ 0 and p, q have no common factors (except 1).
So,
5 = p² / q²
p² = 5q² …. (1)
As we know, ‘5’ divides 5q², so ‘5’ divides p² as well. Hence, ‘5’ is prime.
So 5 divides p
Now, let p = 5k, where ‘k’ is an integer
Square on both sides, we get
p² = 25k²
5q² = 25k² [Since, p² = 5q², from equation (1)]
q² = 5k²
As we know, ‘5’ divides 5k², so ‘5’ divides q² as well. But ‘5’ is prime.
So 5 divides q
Thus, p and q have a common factor 5. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √5 is not a rational number.
√5 is an irrational number.
Now, let us assume -3 + 2√5 be a rational number, ‘r’
So, -3 + 2√5 = r
-3 – r = 2√5
(-3 – r)/2 = √5
We know that, ‘r’ is rational, ‘(-3 – r)/2’ is rational, so ‘√5’ is also rational.
This contradicts the statement that √5 is irrational.
So, -3 + 2√5 is irrational number.
Hence proved.

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