NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 09 Propositions and their Converses Examples

Examples

In-text Question. If a proposition is true, then is its converse always true?
Answer:
Not necessarily. Knowing that a proposition is true tells us nothing about its converse; the converse has to be tested on its own.
Take the true proposition “If a number is a multiple of 6, then it is a multiple of 3”. Its converse, “If a number is a multiple of 3, then it is a multiple of 6”, is false: 9 is a multiple of 3 but not of 6.
On the other hand, Statement 1 and its converse Statement 2 (about equal sides and equal angles of a triangle) are both true, as shown in the next question.
So a converse can be true or false; it is not automatically true.

Think and Reflect 1. 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.)
Answer:
Yes, Statement 2 is true. Here is a proof.

Given: △ABC with ∠B = ∠C.
To prove: AB = AC.
Construction: The third angle is ∠A. From A, draw the altitude AD ⊥ BC. (∠B and ∠C are equal and together are less than 180°, so each is acute; hence D falls between B and C.)
Proof: Compare △ADB and △ADC.
1. ∠ABD = ∠ACD (given, ∠B = ∠C).
2. ∠ADB = ∠ADC = 90° (AD is an altitude).
3. AD = AD (common side).
4. So △ADB ≅ △ADC by the AAS criterion.
5. Hence AB = AC (CPCT).
Therefore, sides opposite equal angles are equal, and Statement 2 is true.

Teacher's Note:
1. The altitude splits the triangle into two right triangles that can be compared.
2. Explain why D lies between B and C: both base angles are acute.
3. Use AAS, not SSS: we do not yet know that AB = AC.

Example 1. Proposition P: If it rains, then the road is wet.
Converse Q: If the road is wet, then it has rained.
Which of these is true? Give a counterexample for the false one.
Answer:
P is true: whenever it rains, rain water falls on the road and makes it wet.
Q is false: a road can be wet without any rain.
Counterexample: a water tanker spills water on the road (or a pipe bursts, or someone washes a car there). The road is wet, yet it has not rained.
A situation like this, which shows that a statement need not hold, is called a counterexample.
Conclusion: P is true; its converse Q is false.

Fermat's Claim. Fermat claimed that all numbers of the form \( 2^{2^n} + 1 \) are prime (n = 0, 1, 2, 3, …); for example, \( 2^{2^3} + 1 = 257 \). Euler disproved this by showing that \( 2^{2^5} + 1 \) is composite. Verify this counterexample.
Answer:
Work out the first few values:
n = 0: 2¹ + 1 = 3 (prime)
n = 1: 2² + 1 = 5 (prime)
n = 2: 2⁴ + 1 = 17 (prime)
n = 3: 2⁸ + 1 = 257 (prime)
n = 4: 2¹⁶ + 1 = 65537 (prime)
n = 5: 2³² + 1 = 4294967297
Euler found that 4294967297 = 641 × 6700417.
So for n = 5 the number has the factor 641 and is not prime. One such value is enough to show that Fermat's claim is false: n = 5 is a counterexample.

Example 2. 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.
Proposition P is true. Justify why this is so. 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.
Answer:
(i) Why P is true. A multiple of 6 can be written as n = 6k for some integer k.
Then n = 6k = 3 × (2k), and 2k is an integer, so n is a multiple of 3.
Examples: 12 = 3 × 4, 18 = 3 × 6, 30 = 3 × 10. P is true.
(ii) The converse Q. No, a multiple of 3 need not be a multiple of 6.
Counterexample: 9 = 3 × 3 is a multiple of 3, but 9 ÷ 6 leaves remainder 3, so 9 is not a multiple of 6. (Any odd multiple of 3, such as 3, 15, 21, also works.)
Q is false.

Example 3. Here 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?
Answer:
Let us test a few numbers first.
n = 9: factors 1, 3, 9; 3 factors (odd); perfect square (3²).
n = 12: factors 1, 2, 3, 4, 6, 12; 6 factors (even); not a perfect square.
n = 16: factors 1, 2, 4, 8, 16; 5 factors (odd); perfect square (4²).
n = 18: factors 1, 2, 3, 6, 9, 18; 6 factors (even); not a perfect square.
n = 36: factors 1, 2, 3, 4, 6, 9, 12, 18, 36; 9 factors (odd); perfect square (6²).
In each case the perfect squares have an odd number of factors, and the numbers with an odd number of factors are perfect squares.
Both P and Q are true. The proofs are given in the next two parts: the Discussion argument proves Q, and a modified version proves P.

Example 3 (Discussion). Each factor of a number has a ‘partner’ factor such that their product is the number (e.g., (1, 12), (2, 6), (3, 4) for 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 of 5 in 25); otherwise the factors would pair off and their number would be even. 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:
Look at where the argument begins and where it ends.
It starts with “n has an odd number of factors” and ends with “n is a perfect square”.
The chain of reasoning is:
(i) n has an odd number of factors
⇒ (ii) some factor-partner pair repeats a number, say (f, f)
⇒ (iii) n = f × f, a perfect square.
So the argument proves Statement Q (the converse). It does not show that every perfect square has an odd number of factors, which is Statement P.

Teacher's Note:
1. To see what an argument proves, read its first assumption and its final conclusion.
2. Going from (i) to (iii) is not the same as going from (iii) to (i).

Example 3 (Is P true?). Is P true? Modify the argument to prove it.
Answer:
Yes, P is also true. Clearly (iii) ⇒ (ii): if n = f × f, then (f, f) is a factor-partner pair. The question is whether (ii) ⇒ (i).
Step 1 (an example). For 36 = 6 × 6 the pairs are (1, 36), (2, 18), (3, 12), (4, 9) and (6, 6).
The first four pairs give 2 different factors each, 8 in all; the pair (6, 6) gives just 1. Total = 8 + 1 = 9, an odd number.
Step 2 (only one repeated pair). If there were two repeated pairs (f, f) and (g, g), each would add 1 factor and the total would be even. So we must rule this out.
Suppose n = f × f = g × g with f < g. Then f × f < g × f < g × g, which gives n < n, impossible.
So a number has at most one repeated factor pair.
Step 3 (conclusion). If n = f × f, every factor other than f belongs to a pair of two different numbers, and f is counted once.
Number of factors = (an even number) + 1, which is odd. Hence P is true.
Together with the Discussion, ‘n is a perfect square’ and ‘n has an odd number of factors’ imply each other.

Teacher's Note:
1. The gap in the original argument is the possibility of two repeated pairs.
2. Step 2 closes that gap using f × f < g × g when f < g.

Example 4. Proposition P: If two triangles have the same area, then they are congruent.
Converse Q: If two triangles are congruent, then they have the same area.
Determine if these statements are true or not. Justify the true statements and give a counterexample for each false statement.
Answer:
P is false.
Counterexample: take right △PQR with ∠Q = 90°, QR = 6 cm, PQ = 2 cm, and right △LMN with ∠M = 90°, MN = 4 cm, LM = 3 cm.

Area of △PQR = \( \frac{1}{2} \) × 6 × 2 = 6 cm²
Area of △LMN = \( \frac{1}{2} \) × 4 × 3 = 6 cm²
The areas are equal. In △LMN, by the Baudhāyana–Pythagoras theorem, LN = √(3² + 4²) = √25 = 5 cm, so its longest side is 5 cm.
But △PQR has a side QR = 6 cm, longer than every side of △LMN. The sides cannot be matched, so the triangles are not congruent.
Q is true. Congruent triangles can be placed one on top of the other so that they cover each other exactly, so they enclose the same region. (Also, their corresponding bases and heights are equal, so ½ × base × height is the same.)
Conclusion: P is false; its converse Q is true.

Teacher's Note:
1. One pair of equal-area, non-congruent triangles is enough to disprove P.
2. To show two triangles are not congruent, compare their longest sides.

Think and Reflect 2. It can also happen that both the proposition and converse are false! Can you give an example?
Answer:
Example (numbers)
Proposition: If a number is a multiple of 4, then it is a multiple of 6. False: 8 is a multiple of 4 but not of 6.
Converse: If a number is a multiple of 6, then it is a multiple of 4. False: 6 is a multiple of 6 but not of 4.
Example (geometry)
Proposition: If a triangle is isosceles, then it is right-angled. False: an equilateral triangle (each angle 60°) is isosceles but has no right angle.
Converse: If a triangle is right-angled, then it is isosceles. False: the 3 cm, 4 cm, 5 cm triangle is right-angled but has no two sides equal.
In both examples the proposition and its converse are false.

Example 5. Here is a statement of the Baudhāyana–Pythagoras theorem. Let a, b, c be the sidelengths of a triangle. If the triangle is right-angled, then a² + b² = c². What is its converse? Is this converse true? Let △ABC be the triangle with sidelengths a, b, c such that a² + b² = c². Construct another right triangle △XYZ whose perpendicular sides YZ and XZ have sidelengths a and b. What can we say about the length of XY? (Hint: Use the Baudhāyana–Pythagoras Theorem.) What can we say about △ABC and △XYZ? (Hint: Use the SSS criterion for congruence.)
Answer:
(a) The converse. Keep the first sentence and swap the ‘if’ and ‘then’ parts: Let a, b, c be the sidelengths of a triangle. If a² + b² = c², then the triangle is right-angled (with the right angle opposite side c).


(b) Length of XY. △XYZ is right-angled at Z with YZ = a and XZ = b. By the Baudhāyana–Pythagoras theorem,
XY² = YZ² + XZ² = a² + b² = c² (given).
Lengths are positive, so XY = c.
(c) Comparing the triangles. In △ABC and △XYZ:
1. BC = YZ = a (given / construction)
2. CA = ZX = b (given / construction)
3. AB = XY = c (shown in (b))
4. So △ABC ≅ △XYZ by SSS (A ↔ X, B ↔ Y, C ↔ Z).
5. Hence ∠C = ∠Z = 90° (CPCT).
(d) Conclusion. ∠C = 90°, so △ABC is right-angled. The converse of the Baudhāyana–Pythagoras theorem is true; here both the proposition and its converse are true.

Teacher's Note:
1. The trick is to build a right triangle we know about and compare it with the given one.
2. The theorem itself is used for △XYZ; congruence then carries the right angle over to △ABC.

Think and Reflect 3. Identify real-life examples in which a proposition is true but not the converse. Give nice counterexamples!
Answer:
1. Proposition (true): If a person lives in Jaipur, then the person lives in Rajasthan.
Converse (false): If a person lives in Rajasthan, then the person lives in Jaipur.
Counterexample: someone living in Udaipur lives in Rajasthan but not in Jaipur.
2. Proposition (true): If an animal is a crow, then it is a bird.
Converse (false): If an animal is a bird, then it is a crow.
Counterexample: a sparrow is a bird but not a crow.
3. Proposition (true): If a fruit is a mango, then it has a seed.
Converse (false): If a fruit has a seed, then it is a mango.
Counterexample: an apple has seeds but is not a mango.
4. Proposition (true): If today is Diwali, then schools are closed.
Converse (false): If schools are closed, then today is Diwali.
Counterexample: schools are also closed on Sundays and during summer vacation.

Think and Reflect 4. Give more examples from geometry as well as number theory in which a proposition is true but not its converse. Give appropriate counterexamples.
Answer:
Geometry
1. Proposition: If a quadrilateral is a square, then it is a rectangle. (True: every angle of a square is 90°.)
Converse: If a quadrilateral is a rectangle, then it is a square. (False: a 6 cm × 3 cm rectangle is not a square.)
2. Proposition: If a quadrilateral is a rhombus, then its diagonals are perpendicular. (True.)
Converse: If the diagonals of a quadrilateral are perpendicular, then it is a rhombus. (False: a kite ABCD with AB = AD and CB = CD but AB ≠ BC has perpendicular diagonals and is not a rhombus.)

3. Proposition: If two angles are vertically opposite, then they are equal. (True.)
Converse: If two angles are equal, then they are vertically opposite. (False: the base angles of an isosceles triangle are equal but are not vertically opposite.)
Number theory
4. Proposition: If a number is divisible by 9, then it is divisible by 3. (True: 9k = 3 × 3k.)
Converse: If a number is divisible by 3, then it is divisible by 9. (False: 6 is divisible by 3 but not by 9.)
5. Proposition: If a number is divisible by 10, then it is even. (True: 10k = 2 × 5k.)
Converse: If a number is even, then it is divisible by 10. (False: 4 is even but not divisible by 10.)
6. Proposition: If n is a prime greater than 2, then n is odd. (True: an even number greater than 2 has the factor 2.)
Converse: If n is an odd number greater than 2, then n is prime. (False: 9 is odd, but 9 = 3 × 3.)