Class 09 | Ganita Manjari 02 | Ch. 09 Propositions and their Converses | Solutions

Class 09 | Ganita Manjari | Part II | Ch 09: Propositions and their Converses | Solutions

📘 Ganita Manjari | Part II CBSE 2026-27

Key Concepts at a Glance

★ A proposition is a statement that is either true or false (never both).

★ Many propositions have the form "if X then Y", also written "X implies Y", or "Y when X" — all three mean the same thing.

★ The converse of "if X then Y" is "if Y then X" — the two parts swap places.

★ A proposition being true tells you nothing about whether its converse is true. Any combination can happen: both true, only one true, or both false.

★ A counterexample is a single example that satisfies X but not Y — one is enough to prove "if X then Y" is false.

★ To prove a proposition true, you need an argument that works for every case — checking a few examples is never enough.

★ When a proposition and its converse are both true, X and Y "imply each other" (e.g. n is a perfect square ⟺ n has an odd number of factors).

Proposition: X⇒Y 

Converse: Y⇒X

Both true ⟺ X⇔Y  (X and Y imply each other)

This chapter opens with a familiar geometry result — if two sides of a triangle are equal, the angles opposite them are equal — and asks whether flipping it around still gives a true statement. From there it builds the vocabulary of propositions, converses and counterexamples, with examples from everyday life (wet roads and rain), number theory (multiples, perfect squares, Fermat's numbers 22n+1) and geometry (triangle areas, the Baudhāyana–Pythagoras Theorem).

A good habit for every question in this chapter: first write the proposition and its converse clearly as "if … then …" sentences, then deal with each one separately — prove it if it's true, or give one clean counterexample if it's false.

In-Text Examples & Think and Reflect

Answers to the questions the chapter poses inside its worked examples, along with all three Think and Reflect boxes, in the order they appear in the textbook.

Think and Reflect

TRWe have proved the first statement in an earlier grade. Is the second statement true? Can you prove it?

Yes, Statement 2 is true — it is the converse of Statement 1, and in this case both the proposition and its converse hold.

Given: △ABC with ∠B=∠C. To prove: AB=AC.

Construction: Following the hint, draw the altitude AD from vertex A (the vertex holding the third angle) to BC, so that AD⊥BC.

Proof: Compare △ABD and △ACD.

∠ABD=∠ACD (given, ∠B=∠C)

∠ADB=∠ADC=90∘ (AD is an altitude)

AD=AD (common side)

So △ABD≅△ACD by the AAS congruence criterion, and hence the corresponding sides are equal: AB=AC.

E2Example 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.
Justify why P is true. Determine whether Q is true, and if not, give a counterexample.

P is true. If n is a multiple of 6, then n=6k for some whole number k. So n=3×(2k), which is 3 times a whole number — so n is a multiple of 3.

Q is false. Counterexample: 9 is a multiple of 3 (9=3×3) but it is not a multiple of 6. (Any odd multiple of 3, such as 3, 15 or 21, works just as well.)

P is true (6k = 3 × 2k). Q is false — counterexample: 9 is a multiple of 3 but not of 6.

Example 3 — P: If n is a perfect square, then it has an odd number of factors. Q: If n has an odd number of factors, then it is a perfect square.
Which of them are true? The factor-partner argument in the Discussion box — does it prove P or Q?

The argument proves Q. It starts from "n has an odd number of factors" and ends with "n is a perfect square" — that is exactly the direction of Q. Factors come in factor-partner pairs; if no pair repeated a number, the factors would split neatly into pairs and their count would be even. An odd count forces some pair (f,f), so n=f×f.

P is also true. If n=f×f, then (f,f) is a factor-partner pair with a repeated number. There cannot be a second such pair (g,g), because g×g=f×f forces g=f for positive numbers. So exactly one factor is "unpaired" and all the others come in distinct pairs — giving an odd total. For example, 36 has factor pairs (1, 36), (2, 18), (3, 12), (4, 9), (6, 6): that's 4 × 2 + 1 = 9 factors.

Both P and Q are true — "n is a perfect square" and "n has an odd number of factors" imply each other. The Discussion argument proves Q only.

E4Example 4 — P: If two triangles have the same area, then they are congruent. 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.

Two triangles with equal area (6 square units each) that are clearly not congruent.

P is false. Counterexample: a right triangle with legs 3 and 4 (sides 3, 4, 5) has area 21​×4×3=6. A triangle with base 6 and height 2 has area 21​×6×2=6 too. The areas are equal, but the second triangle has a side of length 6, which the first doesn't have — so they are not congruent.

Q is true. Congruent triangles can be placed exactly on top of each other, so they cover exactly the same region of the plane — and hence have the same area. (Equivalently: corresponding sides and altitudes are equal, so 21​×base×height gives the same value for both.)

P is false (counterexample: triangles with base 4, height 3 and base 6, height 2 both have area 6 but aren't congruent). Q is true.

Think and Reflect

TRIt can also happen that both the proposition and converse are false! Can you give an example?

Number example. P: If a number is a multiple of 4, then it is a multiple of 6. This is false — 4 itself is a multiple of 4 but not of 6.
Converse Q: If a number is a multiple of 6, then it is a multiple of 4. This is also false — 6 is a multiple of 6 but not of 4.

Geometry example. P: If a quadrilateral is a rhombus, then it is a rectangle. False — a rhombus with angles 60° and 120° is not a rectangle.
Converse Q: If a quadrilateral is a rectangle, then it is a rhombus. False — a 4 cm × 2 cm rectangle does not have all sides equal.

Everyday example. P: If a person is a cricketer, then they are a teacher. Q: If a person is a teacher, then they are a cricketer. Both are clearly false — each needs just one person who is one but not the other.

Example: "multiple of 4 ⇒ multiple of 6" (counterexample 4) and its converse "multiple of 6 ⇒ multiple of 4" (counterexample 6) are both false.

E5Example 5 — Converse of the Baudhāyana–Pythagoras Theorem: Let a, b, c be the sidelengths of △ABC with a2+b2=c2. Construct a right triangle XYZ whose perpendicular sides YZ and XZ have lengths a and b. What can we say about the length of XY? What can we say about △ABC and △XYZ?

Length of XY. △XYZ is right-angled at Z, so by the Baudhāyana–Pythagoras Theorem: XY2=YZ2+XZ2=a2+b2. But we are given a2+b2=c2, so XY2=c2, giving XY=c.

Comparing the triangles. Now the three sides match:

BC=YZ=a, CA=ZX=b, AB=XY=c

So △ABC≅△XYZ by the SSS congruence criterion. Corresponding angles of congruent triangles are equal, so ∠C=∠Z=90∘.

Hence △ABC is right-angled (at C, the vertex opposite the longest side c), and the converse of the Baudhāyana–Pythagoras Theorem is true.

XY = c, so △ABC ≅ △XYZ (SSS), which gives ∠C = ∠Z = 90°. The converse of the Baudhāyana–Pythagoras Theorem is true.

Think and Reflect

TR1. Identify real-life examples in which a proposition is true but not the converse. Give nice counterexamples!
2. Give more examples from geometry as well as number theory in which a proposition is true but not its converse. Give appropriate counterexamples.

1. Real-life examples

(a) P: If a fruit is a mango, then it is a fruit that grows on a tree. (True.) Converse: If a fruit grows on a tree, then it is a mango. False — an apple grows on a tree too.

(b) P: If a city is in Kerala, then it is in India. (True.) Converse: If a city is in India, then it is in Kerala. False — Jaipur is in India but not in Kerala.

(c) P: If it is Sunday, then schools are closed. (True for regular schools.) Converse: If schools are closed, then it is Sunday. False — schools are also closed on public holidays such as 15 August.

2. Geometry examples

(a) P: If a quadrilateral is a square, then it is a rhombus. (True — all sides are equal.) Converse: If a quadrilateral is a rhombus, then it is a square. False — a rhombus with angles 60° and 120°.

(b) P: If two angles are vertically opposite, then they are equal. (True.) Converse: If two angles are equal, then they are vertically opposite. False — the three 60° angles of an equilateral triangle are equal but not vertically opposite.

2. Number theory examples

(a) P: 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.

(b) P: If p is a prime greater than 2, then p is odd. (True — an even number greater than 2 has 2 as a factor.) Converse: If a number greater than 2 is odd, then it is prime. False — 9 = 3 × 3.

The pattern in every example: P is true for every case, but its converse breaks down on a single well-chosen counterexample (e.g. rhombus → square fails for a 60°–120° rhombus; divisible by 3 → divisible by 9 fails for 6).

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Exercise Set 9.1

For Questions 1–12: frame the converse of each proposition, then decide whether each of the two statements is true. Justify the true statements and give a counterexample for each false one.

1If two lines are parallel, then the corresponding angles formed by a transversal are equal.

Lines l and m cut by transversal t; ∠1 and ∠2 are corresponding angles.

Converse: If the corresponding angles formed by a transversal on two lines are equal, then the two lines are parallel.

Proposition — true. This is the corresponding-angles property of parallel lines, established in earlier grades: when l∥m, a transversal t makes ∠1=∠2.

Converse — true. Suppose ∠1=∠2 but l and m are not parallel. Then they meet at some point R, and together with t they form a triangle. In that triangle, ∠1 is an exterior angle and ∠2 is one of its interior opposite angles. An exterior angle equals the sum of the two interior opposite angles, so ∠1=∠2+∠R. With ∠1=∠2, this forces ∠R=0∘ — impossible for a triangle. So l and m cannot meet: they are parallel.

Proposition: true. Converse (equal corresponding angles ⇒ lines parallel): also true.

2If a quadrilateral is a square, then all its angles are equal.

Converse: If all the angles of a quadrilateral are equal, then it is a square.

Proposition — true. Every angle of a square is 90°, so all four are equal.

Converse — false. Counterexample: a rectangle measuring 4 cm × 2 cm. All four of its angles are 90° (equal), but its sides are not all equal, so it is not a square.

Proposition: true. Converse: false — counterexample: a 4 cm × 2 cm rectangle (all angles 90°, but not a square).

3*Given any △ABC, let us bisect the angles at B and C. The bisectors meet at the incentre I of the triangle. Now extend the bisectors beyond I till they meet the opposite sides at E and F respectively, as shown (Fig. 9.1).
Proposition: If AB = AC, then IE = IF.

Converse: If IE = IF, then AB = AC.

Proposition — true. Here E lies on AC (on the bisector BE) and F lies on AB (on the bisector CF).

Since AB=AC, ∠B=∠C, so their halves are equal: ∠IBC=∠ICB and ∠IBF=∠ICE.

In △IBC, ∠IBC=∠ICB, so IB=IC (sides opposite equal angles — the TR result above).

Now compare △IBF and △ICE:

∠IBF=∠ICE (each is half of the equal angles B and C)

IB=IC (shown above)

∠BIF=∠CIE (vertically opposite angles, since BE and CF cross at I)

So △IBF≅△ICE (ASA), and therefore IF=IE.

Converse — false. Surprisingly, IE = IF can happen even when AB ≠ AC. It happens whenever ∠A=60∘. Counterexample: the 30°–60°–90° triangle with ∠A=60∘, ∠B=90∘, ∠C=30∘.

B = (0, 0), A = (0, 1), C = (√3, 0): ∠A = 60°, ∠B = 90°, ∠C = 30°. The highlighted segments IE and IF are equal, yet AB = 1 ≠ AC = 2.

Place the triangle on a coordinate plane: B=(0,0), A=(0,1), C=(3​,0). Then AB=1, BC=3​, AC=2, so AB=AC.

For a right triangle, the inradius is r=21+3​−2​=23​−1​, so I=(r,r).

E (bisector of ∠B, the line y=x, meets AC, the line 3​x​+y=1): E=(23−3​​,23−3​​).

F (bisector of ∠C meets AB, the y-axis): the bisector makes a 15° angle with BC, and tan15∘=2−3​, so F=(0, 3​(2−3​))=(0, 23​−3).

IE2=2(23−3​​−23​−1​)2=2(2−3​)2=14−83​

IF2=(23​−1​)2+(23​−3−23​−1​)2=4(4−23​)+(52−303​)​=14−83​

So IE=IF≈0.379, even though AB=AC.

Why ∠A = 60° works: ∠FIE=∠BIC=90∘+2∠A​=120∘, so ∠FAE+∠FIE=180∘ and A, F, I, E lie on one circle. In that circle, chords IF and IE subtend equal angles 2∠A​ at A (AI bisects ∠A), so IF = IE — no matter what B and C are.

Proposition: true (△IBF ≅ △ICE by ASA). Converse: false — counterexample: the triangle with ∠A = 60°, ∠B = 90°, ∠C = 30° has IE = IF (both 14−83​​) but AB = 1 ≠ AC = 2.

4If x = y, then a + x = a + y, where x, y and a are any three numbers. This proposition and its converse are routinely used while solving equations.

Converse: If a+x=a+y, then x=y.

Proposition — true. If x and y are the same number, then adding the same number a to each gives the same result: a+x=a+y.

Converse — true. Start from a+x=a+y and add −a to both sides (i.e. subtract a): a+x−a=a+y−a, which gives x=y.

This is exactly why, while solving equations, we can both add the same number to both sides (proposition) and cancel the same number from both sides (converse).

Proposition: true. Converse (a + x = a + y ⇒ x = y): also true — subtract a from both sides.

5If a and b are perfect squares, then ab is a perfect square.

Converse: If ab is a perfect square, then a and b are perfect squares.

Proposition — true. If a=p2 and b=q2, then ab=p2q2=(pq)2, a perfect square.

Converse — false. Counterexample: a=2, b=8. Then ab=16=42 is a perfect square, but neither 2 nor 8 is a perfect square.

Proposition: true (p²q² = (pq)²). Converse: false — counterexample: 2 × 8 = 16 is a perfect square, but 2 and 8 are not.

In Questions 6 and 7, x and y are real numbers.

6If x = y, then x2=y2.

Converse: If x2=y2, then x=y.

Proposition — true. If x and y are the same number, squaring gives the same result: x×x=y×y.

Converse — false. Counterexample: x=3, y=−3. Then x2=9=y2, but 3=−3.

(From x2=y2 we only get (x−y)(x+y)=0, i.e. x=y or x=−y. The converse would be true if x and y were both restricted to non-negative numbers.)

Proposition: true. Converse: false — counterexample: x = 3, y = −3 gives x² = y² = 9 but x ≠ y.

7If x = y, then x3=y3.

Converse: If x3=y3, then x=y.

Proposition — true. Cubing the same number gives the same result.

Converse — true. If x3=y3, then x3−y3=0, i.e.

(x−y)(x2+xy+y2)=0

So either x−y=0, or x2+xy+y2=0. Now

x2+xy+y2=(x+2y​)2+43y2​

is a sum of two squares, so it is zero only when both squares are zero: y=0 and then x=0 — in which case x=y anyway. Either way, x=y.

Unlike squaring, cubing keeps the sign of a number ((−3)3=−27, 33=27), so two different real numbers never have the same cube.

Proposition: true. Converse (x³ = y³ ⇒ x = y): also true for real numbers — the counterexample that worked for squares fails here, since (−3)³ ≠ 3³.

In Questions 8–12, n is a positive integer.

8If n is divisible by 24, then it is divisible by both 4 and 6.

Converse: If n is divisible by both 4 and 6, then it is divisible by 24.

Proposition — true. If n=24k, then n=4×(6k) and n=6×(4k), so n is divisible by both 4 and 6.

Converse — false. Counterexample: n=12. It is divisible by 4 (12=4×3) and by 6 (12=6×2), but not by 24. (4 and 6 share the factor 2, so being divisible by both only guarantees divisibility by their LCM, 12 — not by 4×6=24.)

Proposition: true. Converse: false — counterexample: 12 is divisible by 4 and 6, but not by 24.

9If n is divisible by 60, then it is divisible by both 5 and 12.

Converse: If n is divisible by both 5 and 12, then it is divisible by 60.

Proposition — true. If n=60k, then n=5×(12k) and n=12×(5k).

Converse — true. Since 12 divides n, write n=12m. Since 5 also divides n, 5 divides 12m=2×2×3×m. As 5 is prime and does not divide 2, 2 or 3, it must divide m: m=5t. Then n=12×5t=60t, so 60 divides n.

The key difference from Question 8: 5 and 12 have no common factor (other than 1), so their LCM is the full product 5×12=60.

Proposition: true. Converse: also true — since 5 and 12 share no common factor, divisibility by both means divisibility by 5 × 12 = 60.

10If n is the square of a prime number, then it has exactly 3 factors.

Converse: If n has exactly 3 factors, then it is the square of a prime number.

Proposition — true. If n=p2 with p prime, any factor of n can only have p in its prime factorisation, at most twice. So the factors are exactly 1, p and p2 — three factors. (E.g. 49: 1, 7, 49.)

Converse — true. If n has exactly 3 factors, that is an odd number of factors, so by Example 3, n is a perfect square: n=m2. Since 1 has only one factor, m>1, and then 1, m and m2 are three different factors of n — so they are all its factors.

If m were not prime, it would have a factor d with 1<d<m, and d would be a fourth factor of n — a contradiction. So m is prime, and n is the square of a prime.

Proposition: true. Converse: also true — a number with exactly 3 factors must be p² for a prime p.

11If n is a product of two unequal prime numbers, then it has exactly 4 divisors.

Converse: If n has exactly 4 divisors, then it is a product of two unequal prime numbers.

Proposition — true. If n=pq with p, q distinct primes, any divisor of n can only contain p and q in its prime factorisation, each at most once. So the divisors are exactly 1, p, q and pq — four divisors. (E.g. 15: 1, 3, 5, 15.)

Converse — false. Counterexample: n=8. Its divisors are 1, 2, 4, 8 — exactly four — but 8=23 is not a product of two unequal primes. (In general, the cube of any prime, such as 27, also has exactly 4 divisors.)

Proposition: true. Converse: false — counterexample: 8 has exactly 4 divisors (1, 2, 4, 8) but 8 = 2³.

12If n and n + 3 have no factors in common, then n is not a multiple of 3.

("No factors in common" means no common factor other than 1.)

Converse: If n is not a multiple of 3, then n and n + 3 have no factors in common.

Proposition — true. Suppose n were a multiple of 3, say n=3k. Then n+3=3(k+1) is also a multiple of 3, so 3 would be a common factor — contradicting what we're given. So n is not a multiple of 3.

Converse — true. Let d be any common factor of n and n + 3. Then d also divides their difference, (n+3)−n=3. So d=1 or d=3. If n is not a multiple of 3, then d=3 is impossible, leaving only d=1. So n and n + 3 have no common factor.

Proposition: true. Converse: also true — any common factor of n and n + 3 must divide 3, and it can't be 3 when n isn't a multiple of 3.

13There are no known "neat" expressions that generate only primes! Find counterexamples to the following claims.
(i) All numbers of the form 4n2+1 are prime.
(ii) All numbers of the form n2+n+11 are prime.
(iii) All numbers of the form 4n+3 are prime.

n	4n2+1	n2+n+11	4n+3
1	5	13	7
2	17	17	11
3	37	23	15 = 3 × 5
4	65 = 5 × 13	31	19
…		…
10		121 = 11 × 11

(i) n=4: 4(42)+1=65=5×13, not prime.

(ii) n=10: 102+10+11=121=11×11, not prime. (Values for n = 0 to 9 are all prime: 11, 13, 17, 23, 31, 41, 53, 67, 83, 101 — which is exactly why checking a few cases is never a proof! A quick way to spot this one: at n=11, every term is a multiple of 11, giving 143=11×13.)

(iii) n=3: 4(3)+3=15=3×5, not prime.

(i) n = 4 gives 65 = 5 × 13. (ii) n = 10 gives 121 = 11². (iii) n = 3 gives 15 = 3 × 5.

14Find counterexamples to the following statements.
(i) If n is a prime number, then 2n−1 is a prime number.
(ii) If n is an even number, then 2n+1 is a prime number.

(i) Checking primes in order: 22−1=3, 23−1=7, 25−1=31, 27−1=127 — all prime. But for n=11 (prime):

211−1=2048−1=2047=23×89, which is not prime.

(ii) 22+1=5 and 24+1=17 are prime, but for n=6 (even):

26+1=64+1=65=5×13, which is not prime.

(i) n = 11: 2¹¹ − 1 = 2047 = 23 × 89. (ii) n = 6: 2⁶ + 1 = 65 = 5 × 13.

15Consider the statement: "If a number is divisible by 8, then it is divisible by both 2 and 4".
(i) Justify the statement.
(ii) Recall the divisibility shortcuts that we have studied for different numbers. To check whether a given number is divisible by 8, is it enough to check whether it is divisible by 2 and 4? Why or why not?

(i) If n is divisible by 8, then n=8k. So n=2×(4k) and n=4×(2k) — n is divisible by both 2 and 4.

(ii) No. Checking 2 and 4 is the converse of the statement — "if divisible by 2 and 4, then divisible by 8" — and that converse is false. Counterexample: 12 is divisible by 2 and by 4, but not by 8. (Also, every number divisible by 4 is already divisible by 2, so checking 2 adds nothing new; the LCM of 2 and 4 is only 4.)

The correct shortcut for 8 is to check whether the number formed by the last three digits is divisible by 8.

(i) 8k = 2(4k) = 4(2k). (ii) Not enough — 12 is divisible by 2 and 4 but not by 8. Use the last-three-digits test instead.

16Recall that a shortcut to check whether a given number is divisible by 3 is to add the digits of the number and check if the sum is a multiple of 3. Express the relationship between "a number is divisible by 3" and "sum of the digits is a multiple of 3" using "If-then" sentences.

Proposition: If a number is divisible by 3, then the sum of its digits is a multiple of 3.

Converse: If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.

Both are true, so the two statements imply each other — which is exactly why the digit-sum test works in both directions. To see why, take a 3-digit number with digits a, b, c:

100a+10b+c=(99a+9b)+(a+b+c)

The part 99a+9b=9(11a+b) is always a multiple of 3. So the number and its digit sum differ by a multiple of 3 — if either one is a multiple of 3, so is the other. The same idea works for any number of digits, since 10,100,1000,… are each 1 more than a multiple of 9.

If a number is divisible by 3, then its digit sum is a multiple of 3; and if its digit sum is a multiple of 3, then the number is divisible by 3. Both are true.

17We have identified different types of quadrilaterals — squares, rectangles, parallelograms, rhombi, kites and trapezia. One can identify more types (e.g., we could create a category of quadrilaterals that have equal-length opposite sides). Suppose we have identified a category of quadrilaterals called Q, and we have to construct a quadrilateral of this type. For this, we are to use two thin sticks, put them together as diagonals so that the quadrilateral obtained by joining their endpoints is of type Q (see Fig. 9.2).
(i) Suppose Q satisfies the following property. If a quadrilateral is of type Q, then it has equal-length diagonals. (a) Should the two sticks be of equal length? Why or why not? (b) Will it matter how the two sticks are put together?
(ii) Instead of the property mentioned above, suppose Q satisfies the following property. If a quadrilateral has equal diagonals, then it is of type Q. What will be your answers to (a) and (b) now?

(i) Property: "type Q ⇒ equal diagonals"

(a) Yes, the sticks must be of equal length. Every type-Q quadrilateral has equal diagonals. If we used unequal sticks, the diagonals would be unequal, so the result could not be of type Q. Equal diagonals are necessary.

(b) It may matter. The property only tells us something every type-Q quadrilateral has — it doesn't say equal diagonals are enough. Q might also require more. For example, if Q were "rectangles", the sticks would also have to cross at their midpoints; if Q were "squares", they would have to cross at their midpoints and at right angles. With only this property given, we can't be sure that every way of placing two equal sticks produces a type-Q quadrilateral.

(ii) Property: "equal diagonals ⇒ type Q"

(a) Using equal sticks is enough — any quadrilateral with equal diagonals is guaranteed to be of type Q. Strictly speaking, equal sticks aren't required: this property doesn't say every type-Q quadrilateral has equal diagonals, so some type-Q quadrilaterals might have unequal ones. But choosing equal sticks is a sure-fire way to succeed.

(b) No, it won't matter how the equal sticks are put together (as long as they cross each other, so that joining the endpoints gives a quadrilateral). Equal diagonals alone are sufficient to make it type Q.

This question shows the practical difference between a property and its converse: "Q ⇒ equal diagonals" makes equal diagonals a necessary condition, while "equal diagonals ⇒ Q" makes it a sufficient one.

(i) (a) Yes — equal diagonals are necessary. (b) It may matter; the property alone doesn't guarantee success. (ii) (a) Equal sticks are sufficient (not strictly required). (b) No — any crossing placement of equal sticks gives type Q.

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Extra Practice Questions

Seven extra questions in the style of the textbook's own exercise, for independent practice once you've gone through the solved questions above. For each one, write the converse, decide whether each statement is true, and justify or give a counterexample — then tap to check your answer.

1If a triangle is equilateral, then each of its angles is 60°.

Converse: If each angle of a triangle is 60°, then the triangle is equilateral.

Proposition — true. All sides are equal, so all angles are equal (angles opposite equal sides), and they add to 180°, so each is 3180∘​=60∘.

Converse — true. All angles are equal, so all sides are equal (sides opposite equal angles — the TR result of this chapter).

Both the proposition and its converse are true.

2If n is an odd number, then n2 is odd.

Converse: If n2 is odd, then n is odd.

Proposition — true. n=2k+1 gives n2=4k2+4k+1=2(2k2+2k)+1, which is odd.

Converse — true. If n were even, n=2k, then n2=4k2 would be even. So an odd n2 forces n to be odd.

Both true — "n is odd" and "n² is odd" imply each other.

3If a quadrilateral is a rectangle, then its diagonals are equal.

Converse: If a quadrilateral has equal diagonals, then it is a rectangle.

Proposition — true. In rectangle ABCD, △ABC ≅ △BAD (AB common, BC = AD, ∠B = ∠A = 90°, SAS), so AC = BD.

Converse — false. Counterexample: an isosceles trapezium (non-parallel sides equal) has equal diagonals but is not a rectangle.

Proposition true; converse false — an isosceles trapezium has equal diagonals.

4For a real number x: if x>3, then x2>9.

Converse: If x2>9, then x>3.

Proposition — true. If x>3, then x2=x×x>3×3=9 (multiplying positive numbers larger than 3).

Converse — false. Counterexample: x=−4 gives x2=16>9, but −4<3.

Proposition true; converse false — x = −4.

5If a number ends in 0, then it is divisible by 5.

Converse: If a number is divisible by 5, then it ends in 0.

Proposition — true. A number ending in 0 is a multiple of 10, and 10k=5×2k.

Converse — false. Counterexample: 15 is divisible by 5 but ends in 5.

Proposition true; converse false — 15.

6If n is a prime number, then n is odd.

Converse: If n is odd, then n is prime.

Proposition — false. Counterexample: 2 is prime but even.

Converse — false. Counterexample: 9 is odd but 9=3×3 is not prime.

Both the proposition and its converse are false (counterexamples 2 and 9).

7If the sum of two whole numbers a and b is even, then a and b are both even.

Converse: If a and b are both even, then a + b is even.

Proposition — false. Counterexample: 3+5=8 is even, but 3 and 5 are both odd.

Converse — true. a=2p, b=2q gives a+b=2(p+q), which is even.

Proposition false (3 + 5 = 8); converse true — here it's the other way round from most examples.

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Class 9 Maths

Propositions And Their Converses Class 9  Maths Ganita Manjari Part 2 Chapter 9 NCERT Solutions

Exercise Set 9.1

→ Frame the converse for each of the propositions in Questions 1-12. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
1. If two lines are parallel, then the corresponding angles formed by a transversal are egual.
Solution:
Converse: If the corresponding angles formed by a transversal are equal, then the two lines are parallel.
Checking the proposition:
The statement is true.
Justification: Corresponding angles are equal when two parallel lines are cut by a transversal.
Checking the Converse:
The statement is true.
Justification: If a pair of corresponding angles are equal, then the two lines are parallel.

Philosophy

2. If a quadrilateral is a square, then all its angles are equal.
Solution:
Converse: If all the angles of a quadrilateral are equal, then it is a square.
Checking the proposition:
The statement is true.
Justification: A square has four right angles, so all its angles are equal.
Checking the Converse:
The statement is false.
Counterexample: A rectangle that is not a square has four equal angles of 90°.

3. Given any △ABC, Iet us bisect the angles at B and C. The bisectors meet at the incentre I of the triangle.

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Now extend the bisectors beyond I till they meet the opposite sides at E and F respectively, as shown.
Proposition: If AB = AC, then IE = IF.
Solution:
Converse: If IE = IF, then AB = AC.
Checking the proposition:
The statement is true.
Justification:
If AB = AC, △ABC is an isosceles triangle, then ∠B = ∠C. Hence, by symmetry, the two angle bisectors BE and CF are equal in the corresponding parts around the incentre I.
Therefore, IE = IF.
Checking the proposition:
The statement is false.
Counterexample:
If IE = IF, the figure is not necessarily symmetrical.
Therefore, we cannot that AB = AC.

4. If x = y, then a + x = a + y, where x, y and a are any three numbers.
This proposition and its converse are routinely used while solving equations.
Solution:
Proposition: If x = y, then a + x = a + y.
Converse: If a + x = a + y, then x = y.
Checking the proposition:
The proposition is true.
Justification: Since x = y, adding the same number (a) to both sides gives a + x = a + y.
Checking the converse:
The converse is true.
Justification: If a + x = a + y, subtracting (a) from both sides gives x = y.

Mathematics study materials

5. If a and b are perfect squares, then ab is a perfect square.
Solution:
Proposition: If a and b are perfect squares, then ab is a perfect square.
Converse: If ab is a perfect square, then a and b are perfect squares.
Checking the proposition:
The proposition is true.
Justification: Let a = m² and b = n²,
Then, ab = m²n² = (mn)², which is a perfect square.
Checking the converse:
The converse is false.
Counterexample: 
Take a = 2 and b =8
Then, ab = 2 × 8 = 16, which is a perfect square.
However, neither 2 nor 8 is a perfect square.
Hence, the converse is false.

In Questions 6 and 7, x and y are real numbers.

6. If x = y, then x² = y².
Solution:
Proposition: If x = y, then x² = y².
Converse: If x² = y², then x = y.
Checking the proposition:
The proposition is true.
Justification: If x = y, then squaring both sides gives x² = y².
Checking the converse:
The converse is false.
Counterexample: Take x = 2, y = –2.
Then, x² = y² = 4, but x ≠ y.
Hence, the converse is false.

Ganita Manjari PDF

7. If x = y, then x³ = y³.
Solution:
Proposition: If x = y, then x³ = y³.
Converse: If x³ = y³, then x = y.
Checking the proposition:
The proposition is true.
Justification: If x = y, then cubing both sides gives x³ = y³.
Checking the converse:
The converse is true.
Justification: If x³ = y³, then taking the cube root of both sides gives x = y.

→ In Questions 8 – 12, n is a positive integer.

8. If n is divisible by 24, then it is divisible by both 4 and 6.
Solution:
Converse: If n is divisible by both 4 and 6, then n is divisible by 24.
Checking the proposition:
The proposition is true.
Justification: Since 24 = 4 × 6, every multiple of 24 is divisible by both 4 and 6.
Checking the converse:
The converse is false.
Counterexample: Take n = 12. It is divisible by both 4 and 6, but 12 is not divisible by 24.

Free NCERT solutions

9. If n is divisible by 60, then it is divisible by both 5 and 12.
Solution:
Converse: If n is divisible by both 5 and 12, then n is divisible by 60.
Checking the proposition:
The proposition is true.
Justification: Since 60 = 5 × 12, every multiple of 60 is divisible by both 5 and 12.
Checking the converse:
The converse is true.
Justification: If n is divisible by both 5 and 12, then it is divisible by their LCM 60.
Hence, n is divisible by 60.

10. If n is the square of a prime number, then it has exactly 3 factors.
Solution:
Converse: If n has exactly 3 factors, then n is the square of a prime number.
Checking the proposition:
The proposition is true.
Justification:
Let n = p², where p is prime.
Its factors are 1, p and p².
Thus, n has exactly 3 factors.
Checking the converse:
The converse is true.
Justification: A number with exactly 3 factors must be of the form p², where p is a prime number.

Mathematics

11. If n is a product of two unequal prime numbers, then it has exactly 4 divisors.
Solution:
Converse: If n has exactly 4 divisors, then n is a product of two unequal prime numbers.
Checking the proposition:
The proposition is true.
Justification:
Let n = pq
where p and q are unequal primes.
Its divisors are 1, p, q, pq.
Thus, n has exactly 4 divisors.
Checking the converse:
The converse is false.
Counterexample: Take n = 8 = 2³.
Its divisors are 1, 2, 4, 8.
So it has exactly 4 divisors. But 8 is not the product of two unequal prime numbers.

Philosophy

12. If n and n + 3 have no factors in common, then n is not a multiple of 3.
Solution:
Converse: If n is not a multiple of 3, then n and n + 3 have no factors in common.
Checking the proposition:
The proposition is true.
Justification: If n were a multiple of 3, then n + 3 would also be a multiple of 3. Thus, they would have 3 as a common factor. Hence, if they have no common factor, n cannot be a multiple of 3.
Checking the converse:
The converse is true.
Justification: If n is not a multiple of 3, then n and n + 3 cannot have 3 as a common factor. Since any common factor of n and n + 3 must divide their difference,
(n + 3) − n = 3, their only possible common factor greater than 1 is 3. Therefore, they have no common factor.

Educational Resources

13. There are no known ‘neat’ expressions that generate only primes! Find counterexamples to the following claims.
(i) AII numbers of the form 4n² + 1 are prime.
(ii) AII numbers of the form n² + n + 11 are prime.
(iii) AII numbers of the form 4ⁿ + 3 are prime.
Solution:
(i) Take n = 4, then
4n² + 1 = 4(4)² + 1 = 64 + 1 = 65
But 65 = 5 × 13
Therefore, 65 is not prime.
Counterexample: n = 4.

(ii) Take n = 10, then
n² + n + 11 = (10)² + 10 + 11 = 100 + 10 + 11 = 121
But 121 = 11 × 11
Therefore, 121 is not prime.
Counterexample: n = 10.

(iii) Take n = 4, then
4ⁿ + 3 = 4⁴ + 3 = 256 + 3 = 259
But 259 = 7 × 37
Therefore, 259 is not prime.
Counterexample: n = 4.

14. Find counterexamples to the fouowing statements.
(i) If n is a prime number then 2ⁿ – 1is a prime number.
(ii) If n is an even number, then 2ⁿ + 1 is a prime number.
Solution:
(i) Take n = 11, then
2ⁿ – 1 = 2¹¹ – 1 = 2048 – 1 = 2047
But 2047 = 23 × 89
Therefore, 2047 is not a prime number.
Counterexample: n = 11.

(ii) Take n = 6, then
2ⁿ + 1 = 2⁶ + 1 = 64 + 1 = 65
But 65 = 5 × 13
Therefore, 65 is not a prime number.
Counterexample: n = 6.

15. Consider the statement: ‘If a number is divisible by 8, then it is divisible by both Z and 4’.
(i) Justify the statement.
(ii) RecaII the divisibility shortcuts that we have studied for di-fferent numbers. To check whether a given number is divisible by 8, is it enough to check whether it is divisille by 2 and 4? Why or why not?
Solution:
(i) Justification:
Since 8 = 2 × 4,
So, every number divisible by 8 is also divisible by both 2 and 4.
For example, 24 is divisible by 8, and
24 ÷ 2 = 12
24 ÷ 4 = 6
Hence, the statement is true.

Philosophy

(ii) No, it is not enough.
For example, 12 is divisible by both 2 and 4:
12 ÷ 2 = 6 and 12 ÷ 4 = 3
But
12 ÷ 8 = 1.5
Thus, 12 is not divisible by 8.
Therefore, being divisible by both 2 and 4 does not guarantee that a number is divisible by 8.

16. Recall that a shortcut to check whether a given number is divisible by 3 is to add the digits of the number and check if the sum is a multiple of 3. Express the relationship betlveen ‘a number is divisible by 3’ and ‘sum of the digits is a multiple of 3’ using ‘If – then’ sentences.
Solution:
The relationship can be expressed using the following If–then statements:
1. If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.
2. If a number is divisible by 3, then the sum of its digits of a number is a multiple of 3.
Thus, both statements are true, and are converses of each other.

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17. We have identified different types of quadrilaterals – squares, rectangles, parallelograms, rhombi, kites and trapezia. One can identify more types (e.g., we could create a category of quadrilaterals that have equal-length opposite sides).
Suppose we have identified a category of quadrilaterals called Q, and we have to construct a quadrilateral of this type. For this, we are to use two thin sticks, put them together as diagonals so that the quadrilateral obtained by joining their endpoints is of type Q (see the Fig. 9.2).

(i) Suppose Q satisfies the following property.
If a quadrilateral is of type Q, then it has equal-length diagonals.
(a) Should the two sticks be of equal length? Why or why not?
(b) WiIl it matter how the two sticks are put together?
(ii) Instead of the property mentioned above, suppose Q satisfies the following property.
If a quadrilateral has equal diagonals, then it is of type Q. What will be your answers to (a) and (b) now?
Solution:
(i) If a quadrilateral is of type Q, then it has equal-length diagonals.
(a) Yes. The two sticks represent the diagonals of the quadrilateral. Since every quadrilateral of type Q has equal-length diagonals, the two sticks must be of equal length.
(b) No. As long as the two sticks intersect to form the diagonals of a quadrilateral, their position does not affect the fact that their lengths are equal.

(ii) If a quadrilateral has equal-length diagonals, then it is of type Q.
(a) Yes. The two sticks represent the diagonals. If they are of equal length, the quadrilateral formed will have equal diagonals and hence, by the given property, will be of type Q.
(b) No. The property depends only on the diagonals being equal in length. Therefore, their arrangement does not matter, provided they form a quadrilateral.

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