Chapter 9: Propositions and their ConversesAll 17 Questions on This Page

Class 9 Maths Chapter 9 Exercise Set 9.1 Solutions

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. Complete step-by-step solutions for Questions 1–17 of Exercise Set 9.1 from NCERT Class 9 Mathematics (Ganita Manjari).

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Key Concepts for Exercise Set 9.1

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Mathematical Proposition

A proposition is a declarative statement that is either definitively true or definitively false, but cannot be both at the same time.

Converse of a Proposition

For any conditional proposition in the form 'If P, then Q', its converse is formed by interchanging the hypothesis and conclusion: 'If Q, then P'. The truth of a proposition does not imply that its converse is also true.

Proposition: If P ⟹ Q | Converse: If Q ⟹ P

Justification vs. Counterexample

To establish that a statement is true, a general deductive justification or proof is required. To establish that a statement is false, a single valid counterexample satisfying the hypothesis but violating the conclusion is sufficient.

Corresponding Angles & Parallel Lines

When a transversal cuts two parallel lines, corresponding angles are equal. Conversely, if a transversal cuts two lines such that corresponding angles are equal, the two lines are parallel.

Lines are parallel ⟺ Corresponding angles are equal

Square vs. Rectangle (Equal Angles vs. Equal Sides)

All four interior angles of both a square and a rectangle are 90°. However, having four equal angles does not guarantee equal sides; hence, a quadrilateral with equal angles is not necessarily a square.

Incentre and Angle Bisectors in Triangles

The interior angle bisectors of a triangle meet at its incentre (I). In an isosceles triangle with AB = AC, the angle bisector of ∠A is a line of symmetry. However, the converse does not hold in general: equal extended bisector segments (IE = IF) do not guarantee AB = AC.

AB = AC ⟹ IE = IF (True) | IE = IF ⟹ AB = AC (False)

Primes, Factors and Divisibility

A prime number has exactly two factors, 1 and itself; a number with more factors is composite. A number divisible by 8 is divisible by each of its factors, such as 2 and 4, but being divisible by 2 and 4 does not guarantee divisibility by 8.

Divisible by 8 ⟹ divisible by 2 and 4 (True) | Divisible by 2 and 4 ⟹ divisible by 8 (False)

Statement and Converse Both True

When a statement and its converse are both true, the relationship can be written as two If-then sentences. The divisibility rule for 3 is an example.

Divisible by 3 ⟺ Sum of digits is a multiple of 3

Question 1

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

Solution

Given Proposition

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

The proposition is true because corresponding angles formed by a transversal with two parallel lines are equal.

Two parallel lines cut by a transversal showing equal corresponding angles 1 and 2

Converse

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

The converse is true because equal corresponding angles imply that the two lines are parallel.

Proposition: TrueConverse: True

Answer:Proposition: True • Converse: True

Question 2

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

Solution

Proposition

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

The proposition is true because all four angles of a square are 90°.

90° = 90° = 90° = 90°

Converse

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

The converse is false.

Counterexample

Square versus Rectangle: Both have four 90-degree angles, but only the square has four equal sides

A rectangle of length 6 cm and breadth 4 cm has all four angles equal to 90°, but all its sides are not equal. Therefore, it is not a square.

Proposition: TrueConverse: False

Answer:Proposition: True • Converse: False (Counterexample: 6 cm × 4 cm rectangle)

Question 3

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

Fig. 9.1: Triangle ABC with Angle Bisectors BE and CFFig. 9.1

Fig. 9.1: Triangle ABC with angle bisectors at B and C meeting at incentre I and extended to E on AC and F on AB

Fig. 9.1: Angle bisectors BE and CF meeting at incentre I of △ABC.

Solution

Converse

If IE = IF, then AB = AC.

Proposition

Given

AB = AC

Therefore, △ABC is isosceles and is symmetric about the angle bisector of ∠A.

Since I lies on this line of symmetry, E and F are symmetric points.

Hence,

IE = IF

Therefore, the proposition is true.

Converse

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

The converse is false.

Counterexample

Take a scalene triangle such that

∠A = 60°

∠B = 80°

∠C = 40°

Since

∠B ≠ ∠C

their opposite sides are unequal:

AB ≠ AC

But BI and CI bisect ∠B and ∠C, so

∠IBC = 40°

∠ICB = 20°

Thus, in △BIC,

∠BIC = 180° − 40° − 20°

= 120°

Since BE and CF are straight lines,

∠FIE = 120°

Also, AI bisects ∠A, so

∠FAI = ∠IAE = 30°

Hence, in quadrilateral AFIE,

∠FAE + ∠FIE = 60° + 120°

= 180°

Therefore, AFIE is cyclic.

So,

∠IEF = ∠IAF = 30°

and

∠EFI = ∠EAI = 30°

Therefore,

∠IEF = ∠EFI

and hence,

IE = IF

But

AB ≠ AC

Therefore, the converse is false.

Proposition: TrueConverse: False

Answer:Proposition — True • Converse — False (Counterexample: A scalene triangle with ∠A = 60°, ∠B = 80°, ∠C = 40° has IE = IF, but AB ≠ AC)

Question 4

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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

Given,

x = y

Adding a to both sides,

a + x = a + y

Therefore, the proposition is true.

Converse

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

Subtracting a from both sides,

x = y

Therefore, the converse is true.

Proposition: TrueConverse: True

Answer:Proposition: True • Converse: True

Question 5

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

Solution

Proposition

If a and b are perfect squares, we can write

a = m², b = n²

for some whole numbers m and n.

Then,

ab = m²n²

ab = (mn)²

Therefore, ab is a perfect square.

Hence, the proposition is true.

Converse

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

The converse is false.

Counterexample

Take

a = 2, b = 8

Then,

ab = 2 × 8 = 16 = 4²

So ab is a perfect square.

But 2 is not a perfect square and 8 is not a perfect square.

Therefore, the converse is false.

Proposition: TrueConverse: False

Answer:Proposition: True • Converse: False (Counterexample: a = 2, b = 8)

Question 6

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If x = y, then x² = y².

Solution

Proposition

Given,

x = y

Squaring both sides,

x² = y²

Therefore, the proposition is true.

Converse

Statement: If x² = y², then x = y.

The converse is false.

Counterexample

Take

x = 2, y = −2

Then,

x² = 2² = 4

and

y² = (−2)² = 4

Thus,

x² = y²

but

x ≠ y

Therefore, the converse is false.

Proposition: TrueConverse: False

Answer:Proposition: True • Converse: False (Counterexample: x = 2, y = −2)

Question 7

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If x = y, then x³ = y³.

Solution

Proposition

Given,

x = y

Cubing both sides,

x³ = y³

Therefore, the proposition is true.

Converse

Statement: If x³ = y³, then x = y.

Since x and y are real numbers, taking the cube root of both sides gives

x = y

Therefore, the converse is true.

Proposition: TrueConverse: True

Answer:Proposition: True • Converse: True

Question 8

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

Solution

Proposition

Given,

24 = 4 × 6

So, if n is divisible by 24, then n can be written as

n = 24k

for some positive integer k.

Then,

n = 4(6k)

so n is divisible by 4.

Also,

n = 6(4k)

so n is divisible by 6.

Therefore, the proposition is true.

Converse

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

The converse is false.

Counterexample

Take

n = 12

Then,

12 = 4 × 3

and

12 = 6 × 2

So 12 is divisible by both 4 and 6.

But 12 is not divisible by 24.

Therefore, the converse is false.

Proposition: TrueConverse: False

Answer:Proposition: True • Converse: False (Counterexample: n = 12)

Question 9

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

Solution

Proposition

Given,

60 = 5 × 12

If n is divisible by 60, then

n = 60k

for some positive integer k.

Therefore,

n = 5(12k)

so n is divisible by 5.

Also,

n = 12(5k)

so n is divisible by 12.

Therefore, the proposition is true.

Converse

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

Since 5 and 12 have no common factor other than 1,

LCM(5, 12) = 5 × 12 = 60

Therefore, any number divisible by both 5 and 12 is divisible by 60.

Hence, the converse is true.

Proposition: TrueConverse: True

Answer:Proposition: True • Converse: True

Question 10

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

Solution

Proposition

Let p be a prime number.

Given,

n = p²

The factors of p² are

1, p, p²

There are exactly 3 factors.

Therefore, the proposition is true.

Converse

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

Suppose the three factors of n are

1, m, n.

Since m divides n, write

n = mk

If m ≠ k, then both m and k would be different factors of n, giving more than three factors.

Therefore,

m = k

Hence,

n = m²

Also, m must be prime. Otherwise, m would have a factor other than 1 and m, which would give another factor of n.

Therefore, n is the square of a prime number.

Hence, the converse is true.

Proposition: TrueConverse: True

Answer:Proposition: True • Converse: True

Question 11

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

Solution

Proposition

Let p and q be two unequal prime numbers.

Given,

n = pq

Since p and q are prime and p ≠ q, the divisors of n are

1, p, q, pq

Thus, n has exactly 4 divisors.

Therefore, the proposition is true.

Converse

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

The converse is false.

Counterexample

Take

n = 8

Its divisors are

1, 2, 4, 8

So 8 has exactly 4 divisors.

But

8 = 2 × 2 × 2

It is not the product of two unequal prime numbers.

Therefore, the converse is false.

Proposition: TrueConverse: False

Answer:Proposition: True • Converse: False (Counterexample: n = 8)

Question 12

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

Note

Here, “no factors in common” means that n and n+3 have no common factor other than 1.

Solution

Proposition

Suppose n is a multiple of 3.

Then,

n = 3k

for some positive integer k.

Therefore,

n + 3 = 3k + 3 = 3(k + 1)

So both n and n + 3 have 3 as a common factor.

Therefore, if n and n + 3 have no common factor other than 1, n cannot be a multiple of 3.

Hence, the proposition is true.

Converse

Statement: If n is not a multiple of 3, then n and n + 3 have no common factor other than 1.

Suppose d is a common factor of n and n + 3.

Then d also divides their difference:

(n + 3) − n = 3

Therefore, d must be a factor of 3.

The only positive factors of 3 are

1, 3

But n is not a multiple of 3, so 3 cannot be a factor of n.

Therefore, the only common factor is

Hence, the converse is true.

Proposition: TrueConverse: True

Answer:Proposition: True • Converse: True

Question 13

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There 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.

Solution

(i) 4n² + 1

Take

n = 4

Then,

4n² + 1 = 4(4²) + 1

= 4(16) + 1

= 65

But,

65 = 5 × 13

So 65 is not prime.

Therefore, the claim is false.

Counterexample:

n = 4

(ii) n² + n + 11

Take

n = 10

Then,

n² + n + 11 = 10² + 10 + 11

= 100 + 10 + 11

= 121

But,

121 = 11 × 11

So 121 is not prime.

Therefore, the claim is false.

Counterexample:

n = 10

(iii) 4ⁿ + 3

Take

n = 4

Then,

4ⁿ + 3 = 4⁴ + 3

= 256 + 3

= 259

But,

259 = 7 × 37

So 259 is not prime.

Therefore, the claim is false.

Counterexample:

n = 4

Answer:Counterexamples: (i) n = 4 • (ii) n = 10 • (iii) n = 4

Question 14

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Find 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.

Solution

(i) 2ⁿ − 1

Take the prime number

n = 11

Then,

2ⁿ − 1 = 2¹¹ − 1

= 2048 − 1

= 2047

But,

2047 = 23 × 89

So 2047 is not prime.

Therefore, the statement is false.

Counterexample:

n = 11

(ii) 2ⁿ + 1

Take the even number

n = 6

Then,

2ⁿ + 1 = 2⁶ + 1

= 64 + 1

= 65

But,

65 = 5 × 13

So 65 is not prime.

Therefore, the statement is false.

Counterexample:

n = 6

Answer:Counterexamples: (i) n = 11 • (ii) n = 6

Question 15

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Consider 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?

Solution

(i) Justify the statement

Suppose a number is divisible by 8.

Then it can be written as

for some whole number k.

Since

8 = 2 × 4,

we have

8k = 2(4k)

Therefore, the number is divisible by 2.

Also,

8k = 4(2k)

Therefore, the number is divisible by 4.

Hence, the statement is true.

(ii) Is checking 2 and 4 enough?

No, it is not enough.

Checking whether a number is divisible by both 2 and 4 considers the converse of the given statement.

The converse is:

If a number is divisible by both 2 and 4, then it is divisible by 8.

This is false.

For example,

is divisible by 2:

12 = 2 × 6

and by 4:

12 = 4 × 3

But 12 is not divisible by 8.

Therefore, checking divisibility by 2 and 4 is not enough.

Answer:(i) True • (ii) No (Counterexample: 12)

Question 16

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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 between ‘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 these two If-then statements:

(i) Statement

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

(ii) Converse

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

Thus, both the statement and its converse are true.

Answer:Both If-then statements are true.

Question 17

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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) 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?

Fig. 9.2: Quadrilateral with Two Sticks as DiagonalsFig. 9.2

Fig. 9.2: Two thin sticks put together as diagonals with endpoints joined to form a quadrilateral

Fig. 9.2: Two thin sticks used as diagonals to form a quadrilateral.

Solution

The two sticks are used as the diagonals of the quadrilateral.

(i)

Given property:

If a quadrilateral is of type Q, then it has equal-length diagonals.

(a) Should the two sticks be of equal length?

Yes.

Since every quadrilateral of type Q has equal-length diagonals, the two sticks used as the diagonals should be of equal length.

Yes, the two sticks should be of equal length.

(b) Will it matter how the two sticks are put together?

Yes, it may matter.

The given statement only tells us that a quadrilateral of type Q must have equal diagonals.

It does not say that every quadrilateral with equal diagonals is of type Q.

For example, suppose Q represents rectangles. A rectangle has equal diagonals, but simply using two equal sticks does not guarantee that the quadrilateral formed will be a rectangle. Their arrangement can affect the shape.

Therefore,

Yes, the arrangement may matter.

(ii)

Now the property is:

If a quadrilateral has equal diagonals, then it is of type Q.

(a) Should the two sticks be of equal length?

Yes.

The sticks are the diagonals of the quadrilateral.

Therefore, to make the diagonals equal, the two sticks must have equal lengths.

Yes, the two sticks should be of equal length.

(b) Will it matter how the two sticks are put together?

No.

The given statement says that whenever a quadrilateral has equal diagonals, it is of type Q.

So, once the two sticks have equal lengths, the quadrilateral has equal diagonals and therefore must be of type Q.

Thus, the particular arrangement of the two equal sticks does not matter, as long as they form the required quadrilateral.

No, the arrangement does not matter.

Answer:(i) (a) Yes, the two sticks should be of equal length. (b) Yes, the arrangement may matter. • (ii) (a) Yes, the two sticks should be of equal length. (b) No, the arrangement does not matter.

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💡Key Concepts for Exercise Set 9.1

Mathematical Proposition

Converse of a Proposition

Justification vs. Counterexample

Corresponding Angles & Parallel Lines

Square vs. Rectangle (Equal Angles vs. Equal Sides)

Incentre and Angle Bisectors in Triangles

Primes, Factors and Divisibility

Statement and Converse Both True

Key Concepts for Exercise Set 9.1