Prove that √5 is an irrational number.
Answer: Suppose √5 = pq in lowest terms. Then p2 = 5q2, so 5 divides p; writing p = 5k gives q2 = 5k2, so 5 divides q. Then 5 is a common factor of p and q: a contradiction. Hence √5 is irrational.
Step-by-step solution
Idea: Proof by contradiction, as for √2, with 5 in place of 2. The key fact is: if 5 divides p2, then 5 divides p. (Squares of numbers leaving remainders 1, 2, 3, 4 on division by 5 leave remainders 1, 4, 4, 1, never 0.)
- Assume √5 is rational: √5 = pq, with p, q integers, q ≠ 0, and p, q having no common factor other than 1.½ mark
- Squaring: 5 = p2q2, so p2 = 5q2. Thus 5 divides p2, and since 5 is prime, 5 divides p. Let p = 5k.1 mark
- Then 25k2 = 5q2, so q2 = 5k2. Thus 5 divides q2, and so 5 divides q.1 mark
- So 5 is a common factor of p and q, which contradicts our assumption. Hence √5 is irrational.½ mark
Assuming √5 = p/q in lowest terms forces 5 to divide both p and q, a contradiction. Hence √5 is irrational.
Answer to write in the exam
Suppose √5 = pq, p, q integers, q ≠ 0, HCF(p, q) = 1
5 = p2q2 ⇒ p2 = 5q2
5 | p2 ⇒ 5 | p (5 is prime); let p = 5k
25k2 = 5q2 ⇒ q2 = 5k2 ⇒ 5 | q
5 is a common factor of p and q, contradicting HCF(p, q) = 1
∴ √5 is irrational.
Common mistakes that cost marks
- Arguing with ‘even’ as in the √2 proof. Here the divisor that matters is 5.
- Writing ‘p2 = 5q2, so p = 5q‘. That does not follow; the correct step is 5 divides p.
- Using √5 ≈ 2.236 as evidence. A decimal approximation proves nothing about irrationality.
How this can come in the exam
MCQ (1 mark)
If p2 = 5q2 for integers p, q, then
- 2 divides p
- 5 divides p
- p = 5q
- q = 5p
Show answer
(B) 5 divides p
5 divides p2 and 5 is prime, so 5 divides p.
Short answer (3 marks)
Given that √5 is irrational, prove that 2 + 3√5 is irrational.
Show answer
Suppose 2 + 3√5 = r, rational (1 mark). Then √5 = r − 23 (1 mark), which is rational since rationals are closed under subtraction and division by 3. This contradicts √5 being irrational, so 2 + 3√5 is irrational (1 mark).Try one yourself
Prove that √11 is irrational.
Show answer
If √11 = pq in lowest terms, p2 = 11q2, so 11 | p; p = 11k gives q2 = 11k2, so 11 | q: contradiction. So √11 is irrational.