Proposition P: If a number is a multiple of 6, then it is a multiple of 3.
Converse Q: If a number is a multiple of 3, then it is a multiple of 6.
- (i) Proposition P is true. Justify why this is so.
- (ii) Does this now mean that all multiples of 3 are also multiples of 6? This corresponds to the converse statement Q. Determine if it is true and if not, give a counterexample.
Step-by-step solution
Idea: To justify a statement about all numbers, write a general multiple (6k) and show it has the required form. To show a statement is false, one counterexample is enough.
(i) Proposition P is true. Justify why this is so.
- Any multiple of 6 can be written as 6k, where k is a whole number.½ mark
- 6k = 3 × 2k, and 2k is also a whole number.½ mark
- So 6k is 3 times a whole number, which means it is a multiple of 3. For instance, 42 = 6 × 7 = 3 × 14. P is true.½ mark
(ii) Does this now mean that all multiples of 3 are also multiples of 6? This corresponds to the converse statement Q. Determine if it is true and if not, give a counterexample.
- No. P being true says nothing about Q. Q claims every multiple of 3 is a multiple of 6, so one multiple of 3 that is not a multiple of 6 will show Q is false.½ mark
- Take 9 = 3 × 3. It is a multiple of 3.½ mark
- 9 = 6 × 1 + 3, so 9 leaves remainder 3 when divided by 6: it is not a multiple of 6. So 9 is a counterexample and Q is false. (Any odd multiple of 3, such as 3, 15 or 21, also works.)½ mark
Check: Multiples of 6: 6, 12, 18, 24 — each is also in the list of multiples of 3 (3, 6, 9, 12, 15, 18, 21, 24). But 3, 9, 15 and 21 are multiples of 3 that are missing from the multiples of 6.
Answer to write in the exam
(i)
Let the number be 6k (k a whole number).
6k = 3 × (2k)
∴ 6k is a multiple of 3; P is true.
(ii)
Q: If a number is a multiple of 3, then it is a multiple of 6.
9 = 3 × 3 is a multiple of 3.
9 = 6 × 1 + 3, so 9 is not a multiple of 6.
∴ Q is false; counterexample: 9.
Common mistakes that cost marks
- Justifying P with a few cases only (12, 18, 24 …). Cases do not prove a statement about all numbers; use 6k = 3(2k).
- Giving 6 or 12 as a counterexample to Q. They are multiples of both 3 and 6, so they agree with Q. The counterexample must be a multiple of 3 that is not a multiple of 6.
- Deciding Q is true because many multiples of 3 (6, 12, 18 …) are multiples of 6.
How this can come in the exam
Which number is a counterexample to ‘If a number is a multiple of 4, then it is a multiple of 8’?
- 16
- 24
- 20
- 32
Show answer
(C) 20
20 = 4 × 5 is a multiple of 4, but 20 ÷ 8 = 2.5, so it is not a multiple of 8. The others are multiples of 8.
Assertion (A): Every multiple of 15 is a multiple of 5.
Reason (R): 15k = 5 × 3k for every whole number k.
- Both Assertion (A) and Reason (R) are true and R is the correct explanation of A.
- Both Assertion (A) and Reason (R) are true but R is not the correct explanation of A.
- Assertion (A) is true but Reason (R) is false.
- Assertion (A) is false but Reason (R) is true.
Show answer
(A) Both Assertion (A) and Reason (R) are true and R is the correct explanation of A.
R is true, and it shows exactly why every multiple of 15 is 5 times a whole number.
Try one yourself
Is ‘If a number is a multiple of 10, then it is a multiple of 5’ true? Is its converse true?
Show answer
The proposition is true: 10k = 5(2k). The converse ‘If a number is a multiple of 5, then it is a multiple of 10’ is false: 15 is a multiple of 5 but not of 10.
More questions like this
- In this example, n is any positive integer.
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.
You may recall that we came across these statements in the previous grade. Which of them are true? - 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? - This only proves Q. It doesn’t prove that every square number has an odd number of factors (Proposition P).
Is P true? - (i) n has an odd number of factors
which implies
(ii) the existence of a factor-partner pair in which the same number repeats, say (f, f)
which implies
(iii) n is a square number: n = f × f
We start with (iii). Clearly, (iii) implies (ii). Does (ii) imply (i)? - For example, if the number of such pairs is two — say (f, f) and (g, g) — what can we say about the number of factors?