Exercise 1.4

1. Prove that the following are irrational.

(i) 12 (ii) 3+5 (iii) 6 + 2 (iv) 5 (v) 3 + 25

Instruction

Using Proof by contradiction: Assume 12 is rational. Then 12 = where p, q are integers with q ≠ 0 and gcd(p,q) = .

From 12 = p/q, we get: q = p2 ⇒ 2 = . Since p and q are integers and p ≠ 0, q/p is .

But we know that 2 is .

This is a contradiction. Therefore, our assumption is wrong. Hence, 12 is irrational.

(ii) Proof by contradiction: Assume 3 + 5 is rational. Let 3 + 5 = r, where r is rational.

Then: 3 = . Squaring both sides: = r2 + +

⇒ -2 - r2 = -2rsqrt5 ⇒ = 5

Since r is rational and r ≠ 0, the left side 2+r22r is . But 5 is .

This is a contradiction. Therefore, 3 + 5 is irrational.

(iii) Proof by contradiction: Assume 6 + 2 is rational. Let 6 + 2 = r, where r is rational.

Then: 2 = r - 6. Since r is rational and 6 is rational, (r - 6) is rational. But 2 is irrational.

This is a contradiction. Therefore, 6 + 2 is irrational.

(iv) Proof by contradiction: Assume 5 is rational. Then 5 = where p, q are positive integers with gcd(p,q) = .

Squaring both sides: = . Therefore: = p2. This means p2 is divisible by .

Since 5 is , if 5 divides p2, then 5 divides .

Let p = 5k for some integer k. Substituting: 5q2 = 5k2 = . Therefore: q2 = 5k2

This means q2 is divisible by 5. Since 5 is prime, if 5 divides q2, then 5 divides .

So both p and q are divisible by 5, which contradicts our assumption that gcd(p,q) = .

Therefore, 5 is .

(v) Proof by contradiction: Assume 3+25 is rational. Let 3+25 = r, where r is rational. Then: 25 = r - 3. Therefore: 5 =

Since r is rational and 3 is rational, (r - 3) is . Since 2 is rational and non-zero, r−32 is .

But we know that 5 is . Therefore, 3+25 is .

2. Prove that p + q is an irrational, where p, q are primes.

Instruction

Assume p + q is rational, where p and q are distinct primes. Let p + q = r, where r is rational. Then: p =

Squaring both sides: p = + +

= q

Since r is rational and r ≠ 0, the left side is . But q is since q is . This is a contradiction.

Case when p = q: If p = q, then p + q = .

Assume 2p is rational. Then p is . But p is since p is prime.

Therefore, p + q is irrational for any primes p, q.

Chapter 1: Real Numbers