If a proposition is true, then is its converse always true?
Step-by-step solution
Idea: A proposition ‘if X then Y’ and its converse ‘if Y then X’ make two different claims. Knowing that X always leads to Y tells us nothing about whether Y always leads to X, so the converse has to be checked on its own.
- The converse of ‘if X then Y’ is ‘if Y then X’: the two parts swap places. These are different claims, so the truth of one does not settle the truth of the other.½ mark
- Take a true proposition: ‘If a figure is a square, then it has four sides.’ Every square has four sides, so it is true.½ mark
- Its converse is ‘If a figure has four sides, then it is a square.’ A rectangle 5 cm × 3 cm has four sides but is not a square. This one case (a counterexample) is enough to show the converse is false.½ mark
- So the converse of a true proposition is not always true. Sometimes it is: ‘If two sides of a triangle are equal, then the angles opposite them are equal’ and its converse are both true. But the converse must be proved separately each time.½ mark
Answer to write in the exam
No.
Proposition: If a figure is a square, then it has four sides. (True)
Converse: If a figure has four sides, then it is a square.
Counterexample: a rectangle 5 cm × 3 cm has four sides but is not a square.
∴ The converse of a true proposition need not be true.
Common mistakes that cost marks
- Assuming the converse is true just because the proposition is true. ‘If X then Y’ and ‘if Y then X’ are different claims.
- Thinking one counterexample is not enough. A single case where the statement fails shows that it is false.
- Writing the converse wrongly, e.g. ‘If a figure is not a square, then it does not have four sides’. That is a different statement, not the converse.
How this can come in the exam
The converse of ‘If a number ends in 0, then it is divisible by 5’ is:
- If a number is divisible by 5, then it ends in 0.
- If a number does not end in 0, then it is not divisible by 5.
- If a number is not divisible by 5, then it does not end in 0.
- A number ends in 0 only if it is divisible by 5.
Show answer
(A) If a number is divisible by 5, then it ends in 0.
Swap the ‘if’ part and the ‘then’ part. (This converse is false: 15 is divisible by 5 but ends in 5.)
Assertion (A): The proposition ‘If a number is a multiple of 10, then it is even’ is true.
Reason (R): The converse of a true proposition is always true.
- 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
(C) Assertion (A) is true but Reason (R) is false.
A is true: 10k = 2 × 5k. R is false: the converse ‘If a number is even, then it is a multiple of 10’ fails for 4.
Try one yourself
Write the converse of ‘If an animal is a cow, then it has four legs.’ Is the converse true?
Show answer
Converse: ‘If an animal has four legs, then it is a cow.’ It is false: a dog has four legs but is not a cow.
More questions like this
- Statement 1: If two sides of a triangle are equal, then the angles opposite the equal sides are equal.
Statement 2: If two angles of a triangle are equal, then the sides opposite the equal angles have equal lengths.
We have proved the first statement in an earlier grade. Is the second statement true? Can you prove it?
(Hint: Draw the altitude from the vertex containing the third angle.) - Proposition P: If it rains, then the road is wet.
Converse Q: If the road is wet, then it has rained. - 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. - 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?