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Irrational numbers · 5 marks

Try to prove the irrationality of √3 using the approach of proof by contradiction. Will the same approach work for √5, √7, or √10?

  1. 1. Try to prove the irrationality of √3 using the approach of proof by contradiction.
  2. 2. Will the same approach work for √5, √7, or √10?
Answer: If √3 = pq in lowest terms, then p2 = 3q2, so 3 divides p, then 3 divides q: a contradiction. Yes, the same approach works for √5, √7 (use 5, 7 in place of 3) and √10 (use the factor 2 of 10), because none of 5, 7, 10 is a perfect square.

Step-by-step solution

Idea: Copy the proof for √2, replacing ‘even’ (divisible by 2) by ‘divisible by 3’. The needed fact: if 3 divides p2, then 3 divides p. This is true for any prime in place of 3.

1. Try to prove the irrationality of √3 using the approach of proof by contradiction.

  1. Assume √3 = pq with p, q integers, q ≠ 0, and no common factor other than 1. Squaring: p2 = 3q2, so 3 divides p2.½ mark
  2. Fact: if 3 divides p2, then 3 divides p. (Any integer is 3m, 3m + 1 or 3m + 2. Their squares are 9m2, 9m2 + 6m + 1 and 9m2 + 12m + 4; only the first is divisible by 3.) So p = 3k.1 mark
  3. Then 3q2 = 9k2, so q2 = 3k2, and by the same fact 3 divides q.½ mark
  4. So 3 is a common factor of p and q, contradicting lowest terms. Hence √3 is irrational.1 mark
√3 is irrational.

2. Will the same approach work for √5, √7, or √10?

  1. √5 and √7: yes. Replace 3 by 5 or 7: p2 = 5q2 (or 7q2), and since 5 and 7 are prime, 5 | p2 ⇒ 5 | p and then 5 | q. Same contradiction.1 mark
  2. √10: yes, using the prime factor 2. p2 = 10q2 is even, so p = 2k; then 4k2 = 10q2 gives 2k2 = 5q2, so 5q2 is even; as 5 is odd, q2 is even, so q is even. Both even: contradiction. (The approach fails only for perfect squares like √4 = 2 or √9 = 3, which are rational.)1 mark
Yes: √5, √7 and √10 are all irrational by the same method.
√3 is irrational (assuming p² = 3q² in lowest terms forces 3 to divide both p and q). The same approach shows √5, √7 and √10 are irrational.

Answer to write in the exam

1.

Suppose √3 = pq, HCF(p, q) = 1, q ≠ 0

p2 = 3q2 ⇒ 3 | p2 ⇒ 3 | p (3 is prime); let p = 3k

9k2 = 3q2 ⇒ q2 = 3k2 ⇒ 3 | q

3 divides both p and q, contradicting HCF(p, q) = 1

∴ √3 is irrational.

2.

√5: p2 = 5q2 ⇒ 5 | p ⇒ 5 | q, contradiction

√7: p2 = 7q2 ⇒ 7 | p ⇒ 7 | q, contradiction

√10: p2 = 10q2 ⇒ p = 2k ⇒ 2k2 = 5q2 ⇒ q even, contradiction

∴ Yes; √5, √7 and √10 are irrational.

Common mistakes that cost marks

  • Using ‘even’ in the √3 proof. The argument must use divisibility by 3, not by 2.
  • Claiming ‘3 divides p2, so 3 divides p‘ with no reason. For 3 it can be checked with remainders; it works because 3 is prime (it fails for 4: 4 | 36 but 4 ∤ 6).
  • Thinking the method proves √4 or √9 irrational. They equal 2 and 3; the step ‘n | p2 ⇒ n | p‘ needs a prime n.

How this can come in the exam

MCQ (1 mark)

Which of the following is rational?

  1. √7
  2. √10
  3. √49
  4. √15
Show answer

(C) √49
√49 = 7 = 71.

Short answer (3 marks)

Prove that √7 is irrational.

Show answerSuppose √7 = pq in lowest terms; then p2 = 7q2 (1 mark). 7 is prime and divides p2, so 7 | p; write p = 7k, giving q2 = 7k2, so 7 | q (1 mark). 7 is a common factor of p and q: contradiction, so √7 is irrational (1 mark).

Try one yourself

Will the same proof show that √12 is irrational? (Hint: 12 = 4 × 3.)

Show answer

Yes: √12 = 2√3, and if 2√3 = r were rational then √3 = r2 would be rational, which is false. So √12 is irrational.

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