exercise 1.2 Irrationality Proofs
1. Prove that √5 is irrational
Proof:
Let √5 be rational. So, we can write √5 = p/q, where p and q are co-prime integers, q ≠ 0.
Squaring both sides,
5 = p2 / q2 …(1)
⇒ 5q2 = p2
This means p2 is divisible by 5, so p must also be divisible by 5.
Let p = 5k, put the value in equation 1
5q2 = (5k)2
5q2 = 25k2
q2 = 5k2
Thus, q2 is divisible by 5, which implies q is also divisible by 5.
Hence, both p and q are divisible by 5. Here contradiction arises because our original assumption was that p and q are co-prime.
Therefore, √5 is irrational.
2. Prove that √2 + √3 is irrational
Proof:
Let √2 + √3 be rational. So, we can write √2 + √3 = r, where r is rational.
⇒ √2 = r – √3
⇒ √2 (is rational) = r (rational) – √3 (irrational)
Since r is rational, r – √3 would be irrational, so √2 would be irrational, which is a contradiction as √2 is irrational.
Hence, √2 + √3 is irrational.
3. Prove that the following are irrational
(i) √2 + √5
Let √2 + √5 be rational. So, √2 + √5 = r
⇒ √2 = r – √5
Since r is rational, r – √5 is irrational. So √2 would be irrational. Contradiction occurs as √2 is irrational.
Therefore, √2 + √5 is irrational.
(ii) √3 – √2
Let √3 – √2 be rational. So, √3 – √2 = s
⇒ √3 = s + √2
Since s is rational, s + √2 is irrational, which is false. Contradiction arises as √3 is irrational.
Therefore, √3 – √2 is irrational.
(iii) √7 – √5
Let √7 – √5 be rational. So, √7 – √5 = t
⇒ √7 = t + √5
Since t is rational, t + √5 is irrational, which is false. Contradiction arises as √7 is irrational.
Therefore, √7 – √5 is irrational.