The decimal expansion of pq will be terminating precisely when the prime factors of q are only 2, only 5 or both 2 and 5. Can you explain why?

Step-by-step solution

Idea: A terminating decimal is exactly a fraction with a power of 10 as denominator, and 10 = 2 × 5. Here pq is in lowest terms.

1. If q has only 2s and 5s: say q = 2^m × 5ⁿ. Multiply top and bottom by the missing factors (5s if m > n, 2s if n > m) so both powers equal the larger one, k. The denominator becomes 2^k5^k = 10^k.1 mark
2. A fraction over 10^k is a decimal with k decimal places, e.g. 320 = 32² × 5 = 3 × 52² × 5² = 15100 = 0.15. So it terminates.1 mark
3. If the decimal terminates: with k decimal places it equals N10^k. Reducing to lowest terms only removes factors, so q divides 10^k = 2^k5^k and can contain no prime other than 2 and 5. Any other prime (3, 7, 11, …) in q therefore makes the decimal non-terminating.1 mark

Answer to write in the exam

Common mistakes that cost marks

• Forgetting that pq must be in lowest terms: 615 = 25 = 0.4 terminates although 15 = 3 × 5.
• Explaining only one direction (2s and 5s ⇒ terminating) and not why other primes stop it terminating.
• Multiplying only the denominator by 5s and not the numerator, which changes the value.

How this can come in the exam

Try one yourself

Write 7125 as a decimal by making the denominator a power of 10.

More questions like this

• Convert 0.35 into the form pq.
• Convert 0.6 into the form pq.
• Convert 0.45 into the form pq.
• Convert 0.16 into the form pq.
• Convert 2.357 into the form pq.

All Number systems questions · All maths questions