## 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**

**Answer:**

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**

**Answer:**

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**

**Answer:**

Let us assume that 6 + √2 is rational. Therefore it can be expressed as p/q, where p, q are integers and q ≠ 0.

= 6 + √2 = p/q

= √2 = p - 6/q

Since p and q are integers, p - 6/q is rational, and hence √2 is rational. But it contradicts the fact that √2 is irrational. Hence, 6 + √2 is irrational.

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