Learnify Academy is a tuition centre in Bahrain. Classes are for students in Bahrain only.Tuition classes in Bahrain only

Factors and perfect squares · 2 marks

Consider the following argument.
Each factor of a number has a ‘partner’ factor such that their product yields the given number, e.g., 5 is a factor of 35, and 5 × 7 = 35. Here, 7 is the partner factor of 5, and vice-versa. Let us focus on factor-partner pairs of numbers, e.g., (1, 12), (2, 6), (3, 4) are the pairs for the number 12.
If a number has an odd number of factors, there must be a factor-partner pair in which the same number repeats (e.g., the partner factor of 5 in 25). If not, the given number will have an even number of factors since each factor can be paired with its factor pair.
Thus, if a number has an odd number of factors, then it is a perfect square.

What does this argument prove? Statement P or Q?

Answer: It proves Statement Q: if a number has an odd number of factors, then it is a perfect square. The argument starts from ‘odd number of factors’ and ends at ‘perfect square’. It does not prove P.

Step-by-step solution

Given: Proposition P: If n is a perfect square, then it has an odd number of factors.; Converse Q: If n has an odd number of factors, then it is a perfect square.

Idea: An argument proves ‘if X then Y’ when it assumes X and arrives at Y. So look at what it starts from and where it ends.

  1. Where does the argument start? ‘If a number has an odd number of factors …’ — it assumes the number has an odd number of factors.½ mark
  2. Where does it end? ‘… then it is a perfect square.’½ mark
  3. So it shows: odd number of factors ⇒ perfect square. That is the statement ‘If n has an odd number of factors, then it is a perfect square’.½ mark
  4. This is Statement Q. It does not prove P, which goes the other way (perfect square ⇒ odd number of factors); P needs its own argument.½ mark
The argument proves Statement Q (odd number of factors ⇒ perfect square), not Statement P.

Answer to write in the exam

Assumes: the number has an odd number of factors.

Concludes: the number is a perfect square.

Odd number of factors ⇒ perfect square

∴ The argument proves Statement Q (not P).

Common mistakes that cost marks

  • Answering P because the argument talks about perfect squares. What matters is the direction: what is assumed and what is concluded.
  • Thinking that proving Q also proves P. A statement and its converse need separate proofs.

How this can come in the exam

MCQ (1 mark)

An argument begins ‘Suppose a triangle has three equal angles’ and ends ‘… so the triangle has three equal sides’. Which statement does it prove?

  1. If a triangle has three equal sides, then it has three equal angles.
  2. If a triangle has three equal angles, then it has three equal sides.
  3. Both of these statements.
  4. Neither statement.
Show answer

(B) If a triangle has three equal angles, then it has three equal sides.
It assumes ‘three equal angles’ and concludes ‘three equal sides’, so it proves ‘if three equal angles, then three equal sides’ only.

Short answer (2 marks)

Riya argues: ‘Let n = 4k. Then n = 2(2k), so n is even.’ Which proposition has she proved: ‘If n is even, then n is a multiple of 4′ or ‘If n is a multiple of 4, then n is even’? Is the other one true?

Show answerShe assumed n is a multiple of 4 and concluded n is even, so she proved ‘If n is a multiple of 4, then n is even’ (1 mark). The other is false: 6 is even but not a multiple of 4 (1 mark).

Try one yourself

An argument starts with ‘Let n be odd’ and ends with ‘so n2 is odd’. Write the proposition it proves, and write its converse.

Show answer

It proves ‘If n is odd, then n2 is odd’. Converse: ‘If n2 is odd, then n is odd’ (also true, but it needs its own argument).

More questions like this

All Mathematical reasoning questions · All maths questions