## Chapter 1 Real Numbers Exercise 1.3

Question 1: Prove that √5 is irrational.
Answer: Let us assume that √5 is rational. Therefore, √5 can be expressed in the form p/q, where p, q are co-prime integers and q ≠ 0.
= √5 = p/q
= p = √5q

Squaring on both sides,
= (p)² = (√5q)²
= p² = 5q²    ...(1)

Therefore, p² is divisible by 5 and by Theorem 1.3, p is also divisible by 5.

Then, there exists some integer, c, such that,
= p = 5c

Squaring on both sides,
= (p)² = (5c)²
= p² = 25c²    ...(2)

From (1) and (2),
5q² = 25c²
q² = 5c²

Hence, 5 divides q². Also 5 divides q.

Since 5 is common factor for both p and q, it contradicts our assumption that p and q are co-prime integers. Hence, our assumption that √5 is rational is wrong. Therefore, √5 is irrational.

Question 2: Prove that 3 + 2√5 is irrational.
Answer: Let us assume that 3 + 2√5 is rational. Therefore, 3 + 2√5 can be expressed as p/q, where p, q are integers and q ≠ 0.
= 3 + 2√5 = p/q
= √5 = p - 3q/2q

Since, p and q are integers, p - 3q/2q is a rational, and hence √5 is also rational. But it contradicts the fact the fact that √5 is irrational. Hence 3 + 2√5 is irrational.

Question 3: Prove that the following are irrationals:
i) 1/√2
Let us assume that 1/√2 is rational. Therefore it can be expressed as p/q, where p, q are integers and q ≠ 0.

= 1/√2 = p/q
= √2 = q/p

Since p and q are integers, q/p is rational, and hence √2 is rational. But it contradicts the fact that √2 is irrational. Hence, 1/√2 is irrational.

ii) 7√5
Let us assume that 7√5 is rational. Therefore it can be expressed as p/q, where p, q are integers and q ≠ 0.

= 7√5 = p/q
= √5 = p/7q

Since p and q are integers, p/7q is rational, and hence √5 is rational. But it contradicts the fact that √5 is irrational. Hence, 7√5 is irrational.

iii) 6 + √2