MathsCBSE Class 9 · Propositions and their Converses

If-Then Statements, Converses and Counterexamples: Class 9

Q: Can a statement and its converse both be false?

Yes. For example "if a > b then a^2 > b^2" and its converse are both false.

Updated 17 October 2026 · Fact-checked

A proposition is a statement that is definitely true or definitely false. An "if P then Q" statement has the converse "if Q then P". A statement and its converse can both be true, one true, or both false. One example that satisfies P but not Q (a counterexample) is enough to show an "if P then Q" statement false. If a statement and its converse are both true, we write "P if and only if Q".

The forward statement is true, but its converse fails: x = −3 gives x² = 9 without x = 3, and one such case is enough.

Illustration drawn for QuestionBuddy.

Understand If-Then Statements, Converses and Counterexamples

Propositions. A proposition is a sentence that is either true or false, never both. " $7$ is prime" is a true proposition and " $9$ is prime" is a false one. "Is $7$ prime?" and "Take out your notebooks" are not propositions.

If-then form. Many mathematical statements are written "if $P$ , then $Q$ " (also " $P \Rightarrow Q$ "), where $P$ is the hypothesis (what is assumed) and $Q$ the conclusion. The same statement can be said in other ways: "All squares are rectangles" means "if a figure is a square, then it is a rectangle"; " $P$ only if $Q$ " means "if $P$ then $Q$ "; "whenever $P$ , $Q$ " and " $Q$ if $P$ " mean the same; and "a number is divisible by 10 whenever its last digit is 0" means "if the last digit is 0 then the number is divisible by 10". Be careful with "only if": " $n$ is even only if $n^{2}$ is even" has hypothesis " $n$ is even".

The converse. The converse of "if $P$ then $Q$ " is "if $Q$ then $P$ ": hypothesis and conclusion swap. The statement "If $x = 3$ , then $x^{2} = 9$ " is true, but its converse "if $x^{2} = 9$ , then $x = 3$ " is false, because $x = - 3$ also works. A statement and its converse are different claims: the truth of one does not tell you the truth of the other.

Four possibilities. For a statement and its converse: (1) both true, such as "if a triangle is equilateral then all its angles are $6 0^{\circ}$ " and its converse; (2) the statement true and the converse false, as above; (3) the statement false and the converse true, such as "if a number is even then it is a multiple of $4$ " (false) with the converse "if a number is a multiple of $4$ then it is even" (true); (4) both false, such as "if $a > b$ then $a^{2} > b^{2}$ " and "if $a^{2} > b^{2}$ then $a > b$ ".

Counterexamples. To show that "if $P$ then $Q$ " is false, you need one example where $P$ is true and $Q$ is false: a counterexample. For " $n^{2} + n + 41$ is prime for every whole number $n$ ": at $n = 40$ the value is $1681 = 4 1^{2}$ , which is not prime. For "if $a > b$ then $a^{2} > b^{2}$ ": take $a = 1$ , $b = - 2$ . Examples that fit the statement never prove it for all cases, but one counterexample disproves it.

If and only if. When both "if $P$ then $Q$ " and "if $Q$ then $P$ " hold, we say " $P$ if and only if $Q$ " ("iff", $P \Leftrightarrow Q$ ): the two statements are equivalent. A proof of an "iff" statement has two parts, one for each direction. Example: " $n^{2}$ is divisible by $4$ if and only if $n$ is even". If $n = 2 k$ then $n^{2} = 4 k^{2}$ ; if $n$ is odd then $n^{2}$ leaves remainder $1$ when divided by $4$ , so $n^{2}$ is not divisible by $4$ .

Using a rule correctly. If the rule is "if red badge then allowed in the lab", then a person with a red badge must be allowed. But a person who is allowed in need not have a red badge: that would be the converse. Reading a statement as if the converse were also given is a common error in proofs and in everyday reasoning.

Method. Identify $P$ and $Q$ , rewrite in if-then form, swap for the converse, test each by examples and by proof, and, if false, produce a counterexample.

If-Then Statements, Converses and Counterexamples: formulas to remember

Implication: $P \Rightarrow Q$; "If $P$ then $Q$ "; $P$ is the hypothesis.
Converse: $Q \Rightarrow P$; Swap hypothesis and conclusion.
Counterexample to $P \Rightarrow Q$: $P true and Q false$; One such case disproves the statement.
If and only if: $P \Leftrightarrow Q$; Both $P \Rightarrow Q$ and $Q \Rightarrow P$ .
Only if: $P only if Q \equiv P \Rightarrow Q$; Not the converse.
Four cases: $(T, T), (T, F), (F, T), (F, F)$; Truth of statement and converse.

How to solve If-Then Statements, Converses and Counterexamples questions

Separate hypothesis and conclusion, then treat each direction as its own claim.

1Rewrite the sentence in "if ..., then ..." form and underline $P$ and $Q$ .
2Write the converse by swapping $P$ and $Q$ .
3Test each statement: look for a counterexample first; if none is found, try to prove it.
4To disprove, give one explicit example where the hypothesis holds and the conclusion fails.
5If both directions are true, write the result as " $P$ if and only if $Q$ " and prove both parts.
6State clearly which direction you are proving at each stage.

If-Then Statements, Converses and Counterexamples in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Converse is a different claim

Suppose a shop announces: "If you spend more than 1000 rupees, you get a free bag."

Write the converse, and say whether a customer who received a free bag must have spent more than 1000 rupees.

Try it first, then show the solutionSolution

Statement: if spend $> 1000$ , then free bag.
Converse: if free bag, then spent $> 1000$ .
The shop might also give free bags during a festival offer or to members, so a customer with a free bag need not have spent more than 1000 rupees.

Answer: The converse is "if you got a free bag, you spent more than 1000 rupees", which does not follow from the announcement.

Memory hook: The announcement is a one-way street.

Tricks to remember If-Then Statements, Converses and Counterexamples

Short memory aids. Say each one aloud once, then use it on a question the same day.

Swap and test
To form the converse swap the two parts, then test it afresh; do not assume it is true.
Use it when: Every converse question.
One example kills
One counterexample proves a general statement false; a hundred examples never prove it true.
Use it when: Disproving claims.
Both ways for iff
An "if and only if" proof has two parts.
Use it when: Equivalence proofs.

If-Then Statements, Converses and Counterexamples: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Converse and truth · 3 marks · Short answerBoard level

Write the converse of each statement and decide whether the statement and its converse are true. (a) If a number is divisible by 6, it is divisible by 3. (b) If a quadrilateral is a square, it is a rectangle. (c) If $x = 3$ , then $x^{2} = 9$ .

Try it first, then show the solutionSolution

(a) True. Converse: if a number is divisible by 3, it is divisible by 6: false (counterexample $9$ ). (1)
(b) True. Converse: if a quadrilateral is a rectangle, it is a square: false (a $2 \times 3$ rectangle). (1)
(c) True. Converse: if $x^{2} = 9$ , then $x = 3$ : false (counterexample $x = - 3$ ). (1)

Answer: In each case the statement is true and its converse is false.

Example 2: If-then forms · 3 marks · Short answerBoard level

Rewrite each statement in if-then form and write its converse: (i) All squares are rectangles. (ii) $n$ is even only if $n^{2}$ is even. (iii) A number is divisible by 10 whenever its last digit is 0.

Try it first, then show the solutionSolution

(i) If a figure is a square, then it is a rectangle. Converse: if a figure is a rectangle, then it is a square. (1)
(ii) If $n$ is even, then $n^{2}$ is even. Converse: if $n^{2}$ is even, then $n$ is even. (1)
(iii) If the last digit is 0, then the number is divisible by 10. Converse: if a number is divisible by 10, then its last digit is 0. (1)

Answer: Rewritten and converses as listed; the converse in (i) is false, in (ii) and (iii) it is true.

Example 3: An iff statement · 5 marks · Long answerHarder board level

Prove that $n^{2}$ is divisible by 4 if and only if $n$ is an even whole number.

Try it first, then show the solutionSolution

Part 1 (if $n$ is even then $4 ∣ n^{2}$ ): let $n = 2 k$ . (½)
Then $n^{2} = 4 k^{2}$ , which is divisible by 4. (1)
Part 2 (if $4 ∣ n^{2}$ then $n$ is even): prove the contrapositive idea by showing that an odd $n$ fails. (1)
Let $n = 2 k + 1$ . Then $n^{2} = 4 k^{2} + 4 k + 1 = 4 (k^{2} + k) + 1$ , which leaves remainder $1$ on division by 4, so it is not divisible by 4. (1½)
So if $4 ∣ n^{2}$ , $n$ cannot be odd, hence $n$ is even. Both directions hold, so the statement is an "if and only if". (1)

Answer: $n^{2}$ is divisible by 4 if and only if $n$ is even.

Example 4: Counterexamples · 3 marks · Short answerHarder board level

Give a counterexample to each: (i) Every prime number is odd. (ii) $n^{2} + n + 41$ is prime for every whole number $n$ . (iii) If $a > b$ then $a^{2} > b^{2}$ .

Try it first, then show the solutionSolution

(i) $2$ is prime and even. (1)
(ii) At $n = 40$ : $1600 + 40 + 41 = 1681 = 41 \times 41$ , which is not prime. (1)
(iii) Take $a = 1$ , $b = - 2$ : $a > b$ but $a^{2} = 1 < 4 = b^{2}$ . (1)

Answer: (i) $2$ ; (ii) $n = 40$ ; (iii) $a = 1$ , $b = - 2$ .

Example 5: A rule at the lab door · 4 marks · Case-basedHOTS / competency

Case: a school rule says "If a student wears a red badge, then the student is allowed into the lab." (i) Ravi wears a red badge: what follows? (ii) Tara is allowed into the lab: can we conclude she wears a red badge? (iii) Write the converse and say whether it is guaranteed. (iv) Meena is not allowed into the lab: what can be said about her badge?

Try it first, then show the solutionSolution

(i) The rule applies, so Ravi is allowed into the lab. (1)
(ii) No. The rule says nothing about students without badges; Tara may be allowed for another reason (for example, as a teacher's helper). (1)
(iii) Converse: if a student is allowed into the lab, the student wears a red badge. It is not guaranteed by the rule. (1)
(iv) If Meena wore a red badge, the rule would let her in; since she is not allowed in, she is not wearing a red badge. (1)

Answer: (i) allowed; (ii) no; (iii) converse not guaranteed; (iv) no red badge.

Where students lose marks in If-Then Statements, Converses and Counterexamples

The slips examiners see most often, and what to write instead.

Mistake: Assuming the converse is true
Treating the statement as two-way.
Wrong working
"If $x = 3$ then $x^{2} = 9$ " so "if $x^{2} = 9$ then $x = 3$ ".
Right working
The converse is false: $x = - 3$ .
Do this: Always test the converse separately.
Usually costs: The conclusion.
Mistake: Using examples as proof
Checking a few cases.
Wrong working
$n^{2} + n + 41$ is prime for $n = 1, 2, 3$ so it is always prime.
Right working
It fails at $n = 40$ ( $1681 = 4 1^{2}$ ).
Do this: Examples can disprove but never prove.
Usually costs: The whole question.
Mistake: Wrong hypothesis for "only if"
Reversing the parts.
Wrong working
" $n$ is even only if $n^{2}$ is even" means "if $n^{2}$ is even then $n$ is even".
Right working
It means "if $n$ is even then $n^{2}$ is even".
Do this: "P only if Q" is "if P then Q".
Usually costs: 1 mark.

Exam tips

Underline hypothesis and conclusion in the statement.
Give a concrete counterexample with numbers, not just a description.
For "if and only if", write "Part 1" and "Part 2".
State clearly whether each statement and its converse is true or false.

Practice questions on If-Then Statements, Converses and Counterexamples

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The converse of "If a triangle is equilateral, then it is isosceles" is
(a) If a triangle is not equilateral, then it is not isosceles
(b) If a triangle is isosceles, then it is equilateral
(c) If a triangle is not isosceles, then it is not equilateral
(d) A triangle is equilateral and isosceles
Hint
Swap the hypothesis and the conclusion.
Answer
(b). The converse swaps the parts. It is false: a triangle with sides 5, 5, 8 is isosceles but not equilateral.
Q2 · 3 marks · Short answerBoard level
For each statement, write the converse and say if the converse is true or false (with a reason): (i) If two angles are vertically opposite, they are equal. (ii) If $x$ is a multiple of 4, then $x$ is even. (iii) If all three angles of a triangle are $6 0^{\circ}$ , the triangle is equilateral.
Hint
Look for a pair of equal angles that are not vertically opposite.
Answer
(i) Converse: equal angles are vertically opposite; false (two angles of an isosceles triangle). (ii) Converse: even implies multiple of 4; false ( $x = 6$ ). (iii) Converse: an equilateral triangle has all angles $6 0^{\circ}$ ; true.
Q3 · 3 marks · Short answerHarder board level
Statement S: "A number is divisible by 5 if its last digit is 5". Write the converse, decide the truth of S and the converse, and give a statement in "if and only if" form about divisibility by 10.
Hint
Think of 10, 20, ... as numbers divisible by 5.
Answer
S is true. Converse: "if a number is divisible by 5, its last digit is 5" is false ( $10$ ). For 10: "A whole number is divisible by 10 if and only if its last digit is 0" is true in both directions.
Q4 · 5 marks · Long answerHarder board level
Prove that a natural number $n$ is odd if and only if $n^{2}$ is odd.
Hint
Prove both directions; for the second, show that an even $n$ gives an even square.
Answer
(a) If $n$ is odd then $n^{2}$ is odd: $n = 2 k + 1$ , $n^{2} = 4 k^{2} + 4 k + 1 = 2 (2 k^{2} + 2 k) + 1$ , which is odd. (b) If $n^{2}$ is odd then $n$ is odd: suppose $n$ were even, $n = 2 m$ ; then $n^{2} = 4 m^{2}$ is even, a contradiction. So $n$ is odd. Both directions hold, so $n$ is odd if and only if $n^{2}$ is odd.
Q5 · 4 marks · Case-basedHOTS / competency
Case: a student claims "If the diagonals of a quadrilateral are perpendicular, then it is a rhombus." (i) Write this in if-then form for the true statement "the diagonals of a rhombus are perpendicular". (ii) Is the claim the converse of that statement? (iii) Show that the claim is false by using the kite with vertices $(0, 4)$ , $(3, 0)$ , $(0, - 1)$ , $(- 3, 0)$ . (iv) Name one more property that, with perpendicular diagonals, would make the converse true.
Hint
Compute the four side lengths of the kite.
Answer
(i) If a quadrilateral is a rhombus, then its diagonals are perpendicular. (ii) Yes, the claim swaps hypothesis and conclusion. (iii) The diagonals lie along the axes and are perpendicular. The sides are $5$ , $10$ , $10$ , $5$ (not all equal), so it is not a rhombus: a counterexample. (iv) If the diagonals are also bisecting each other (or the figure is a parallelogram), then perpendicular diagonals make it a rhombus.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Converse of "if P then Q"?Show answer
"If Q then P".
2.What disproves an if-then statement?Show answer
One counterexample.
3.What is iff?Show answer
Both the statement and its converse are true.
4.Is a proposition always true?Show answer
No, it is either true or false.
5.Does "P only if Q" mean "if Q then P"?Show answer
No, it means "if P then Q".

If-Then Statements, Converses and Counterexamples: frequently asked questions

Can a statement and its converse both be false?

Yes. For example "if a > b then a^2 > b^2" and its converse are both false.

Is one example enough to prove a statement?

No. Examples only illustrate. A proof must cover all cases; one counterexample is enough to disprove.

Why does the converse of a theorem need its own proof?

Because truth does not automatically flow backwards. For example, vertically opposite angles are equal, but equal angles need not be vertically opposite.