Class 9 Maths · Ganita Manjari Part 2 · Exercise 9.1
NCERT Solutions for Class 9 Maths Chapter 9 — Propositions and their Converses
All 17 questions of Exercise 9.1, solved with the converse, a justification or a counterexample — in the exact format you can write in your notebook and in the exam.
Exercise 9.1 · Proposition and converse
Q1If two lines are parallel, then the corresponding angles formed by a transversal are equal.
First, understand the question
The proposition is about two parallel lines cut by a transversal. To get the converse, we swap the “if” part and the “then” part. Then we check whether each statement is always true.
Given
- Proposition: If two lines are parallel, then the corresponding angles formed by a transversal are equal
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
Converse: If the corresponding angles formed by a transversal are equal, then the two lines are parallel.
Proposition – True. When a transversal cuts two parallel lines, the corresponding angles are equal (corresponding angles property):
∠1 = ∠5, ∠2 = ∠6, ∠3 = ∠7, ∠4 = ∠8
Converse – True. If a pair of corresponding angles is equal, the lines are parallel (converse of the corresponding angles property).
∴ Proposition – True, Converse – True.
Converse: If the corresponding angles formed by a transversal are equal, then the two lines are parallel.
Proposition – True. When a transversal cuts two parallel lines, the corresponding angles are equal (corresponding angles property):
Converse – True. If a pair of corresponding angles is equal, the lines are parallel (converse of the corresponding angles property).
∴ Proposition – True, Converse – True.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If the corresponding angles formed by a transversal are equal, then the two lines are parallel.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: When a transversal cuts two parallel lines, the angles formed at the same relative position (corresponding angles) are always equal: ∠1 = ∠5, ∠2 = ∠6, ∠3 = ∠7 and ∠4 = ∠8. This is the corresponding angles property of parallel lines.
Check the converse
Checking the Converse: The converse is True.
Justification: If a pair of corresponding angles formed by a transversal is equal, then the two lines are parallel. This is the converse of the corresponding angles property, which is used as a test for parallelism.
Conclude
Hence, the proposition is True and the converse is True.
Answer: Proposition – True | Converse – True
Exercise 9.1 · Proposition and converse
Q2If a quadrilateral is a square, then all its angles are equal.
First, understand the question
We swap the two parts to get the converse. A square has all its angles equal, but we must check whether every quadrilateral with equal angles is a square.
Given
- Proposition: If a quadrilateral is a square, then all its angles are equal
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
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 its angles are equal.
Converse – False. Counterexample: a rectangle of 6 cm × 4 cm has all its angles equal to 90°, but its sides are not equal, so it is not a square.
∴ Proposition – True, Converse – False.
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 its angles are equal.
Converse – False. Counterexample: a rectangle of 6 cm × 4 cm has all its angles equal to 90°, but its sides are not equal, so it is not a square.
∴ Proposition – True, Converse – False.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If all the angles of a quadrilateral are equal, then it is a square.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: Every angle of a square measures 90°. Hence, all its angles are equal.
Check the converse
Checking the Converse: The converse is False.
Counterexample: Consider a rectangle of length 6 cm and breadth 4 cm. All four of its angles are 90°, so all its angles are equal. But its four sides are not equal, so it is not a square.
Conclude
Hence, the proposition is True and the converse is False.
Answer: Proposition – True | Converse – False (Counterexample: Rectangle)
Exercise 9.1 · Proposition and converse ★
Q3In triangle ABC, the bisectors of angles B and C meet at the incentre I and are extended to meet the opposite sides at E and F. Proposition: If AB = AC, then IE = IF. Full question: 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 in Fig. 3.
First, understand the question
AB = AC means the triangle is isosceles. We check whether this always gives IE = IF, and whether IE = IF always forces AB = AC.
Given
- In △ABC, the bisectors of ∠B and ∠C meet at I and meet the opposite sides at E and F
- Proposition: If AB = AC, then IE = IF
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
Converse: If IE = IF, then AB = AC.
Proposition – True.
AB = AC ⇒ ∠B = ∠C\ ⇒ ∠IBC = ∠ICB\ ⇒ BI = CI
In △CIE and △BIF: ∠ICE = ∠IBF, ∠CIE = ∠BIF (vertically opposite angles) and CI = BI.
△CIE ≅ △BIF (ASA) ⇒ IE = IF
Converse – False. Counterexample: a triangle with ∠A = 60°, ∠B = 90°, ∠C = 30° has IE = IF, but AB ≠ AC.
∴ Proposition – True, Converse – False.
Converse: If IE = IF, then AB = AC.
Proposition – True.
In △CIE and △BIF: ∠ICE = ∠IBF, ∠CIE = ∠BIF (vertically opposite angles) and CI = BI.
Converse – False. Counterexample: a triangle with ∠A = 60°, ∠B = 90°, ∠C = 30° has IE = IF, but AB ≠ AC.
∴ Proposition – True, Converse – False.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If IE = IF, then AB = AC.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: Since AB = AC, the angles opposite to these sides are equal, i.e., ∠ABC = ∠ACB. Since BI and CI bisect these equal angles, ∠IBC = ∠ICB. Hence, in △BIC, the sides opposite to these equal angles are equal, so BI = CI. Now, in △CIE and △BIF: ∠ICE = ∠IBF (halves of equal angles), ∠CIE = ∠BIF (vertically opposite angles) and CI = BI (proved above). Therefore, △CIE ≅ △BIF by the ASA congruence rule. Hence, IE = IF (corresponding parts of congruent triangles).
Check the converse
Checking the Converse: The converse is False.
Counterexample: Consider a triangle with ∠A = 60°, ∠B = 90° and ∠C = 30°, with I, E and F constructed as above. In this triangle, IE = IF, but AB ≠ AC (since ∠B ≠ ∠C, the sides opposite to them cannot be equal).
Conclude
Hence, the proposition is True and the converse is False.
Exam Note: This is a starred (difficult) question in NCERT. In the exam, writing the counterexample with the three angles (60°, 90°, 30°) is sufficient — its detailed verification is not expected at Class 9 level.
Answer: Proposition – True | Converse – False
Exercise 9.1 · Proposition and converse
Q4If 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.
First, understand the question
This is about adding the same number a to both sides of an equation. We swap the two parts to get the converse, and check both.
Given
- Proposition: If x = y, then a + x = a + y
- x, y and a are any three numbers
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
Converse: If a + x = a + y, then x = y.
Proposition – True. Adding the same number to equal numbers gives equal results:
x = y ⇒ a + x = a + y
Converse – True. Subtract a from both sides:
a + x = a + y\ a + x – a = a + y – a\ x = y
∴ Proposition – True, Converse – True.
Converse: If a + x = a + y, then x = y.
Proposition – True. Adding the same number to equal numbers gives equal results:
Converse – True. Subtract a from both sides:
∴ Proposition – True, Converse – True.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If a + x = a + y, then x = y.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: Adding the same number a to two equal numbers gives equal results. This is called the addition property of equality. For example, if x = y = 4 and a = 3, then a + x = 7 and a + y = 7.
Check the converse
Checking the Converse: The converse is True.
Justification: If a + x = a + y, then subtracting a from both sides, we get a + x − a = a + y − a, i.e., x = y. This is the cancellation (subtraction) property of equality.
Conclude
Hence, the proposition is True and the converse is True.
Answer: Proposition – True | Converse – True
Exercise 9.1 · Proposition and converse
Q5If a and b are perfect squares, then ab is a perfect square.
First, understand the question
A perfect square is a number like 4, 9 or 16 (= m²). We check whether the product of two perfect squares is always a perfect square, and whether the reverse is always true.
Given
- Proposition: If a and b are perfect squares, then ab is a perfect square
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
Converse: If ab is a perfect square, then a and b are perfect squares.
Proposition – True. Let a = m² and b = n².
ab = m² × n² = (mn)²
So ab is a perfect square.
Converse – False. Counterexample: a = 2, b = 8.
ab = 2 × 8 = 16 = 4²
16 is a perfect square, but 2 and 8 are not.
∴ Proposition – True, Converse – False.
Converse: If ab is a perfect square, then a and b are perfect squares.
Proposition – True. Let a = m² and b = n².
So ab is a perfect square.
Converse – False. Counterexample: a = 2, b = 8.
16 is a perfect square, but 2 and 8 are not.
∴ Proposition – True, Converse – False.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If ab is a perfect square, then a and b are perfect squares.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: Let a = m² and b = n², where m and n are integers. Then ab = m² × n² = (mn)². Since mn is an integer, ab is a perfect square. For example, a = 4 and b = 9 gives ab = 36 = 6².
Check the converse
Checking the Converse: The converse is False.
Counterexample: Take a = 2 and b = 8. Then ab = 2 × 8 = 16 = 4², which is a perfect square. But neither 2 nor 8 is a perfect square.
Conclude
Hence, the proposition is True and the converse is False.
Answer: Proposition – True | Converse – False (Counterexample: a = 2, b = 8)
Exercise 9.1 · Proposition and converse
Q6If x = y, then x² = y². In Questions 6 and 7, x and y are real numbers.
First, understand the question
Squaring removes the sign of a number, so two different numbers can have the same square. Keep this in mind while checking the converse.
Given
- Proposition: If x = y, then x² = y²
- x and y are real numbers
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
Converse: If x² = y², then x = y.
Proposition – True. Squaring both sides of x = y gives x² = y².
Converse – False. Counterexample: x = 2, y = −2.
x² = 2² = 4, y² = (-2)² = 4
So x² = y², but x ≠ y.
∴ Proposition – True, Converse – False.
Converse: If x² = y², then x = y.
Proposition – True. Squaring both sides of x = y gives x² = y².
Converse – False. Counterexample: x = 2, y = −2.
So x² = y², but x ≠ y.
∴ Proposition – True, Converse – False.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If x² = y², then x = y.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: Equal numbers have equal squares. If x = y, then squaring both sides gives x² = y². For example, x = y = 3 gives x² = y² = 9.
Check the converse
Checking the Converse: The converse is False.
Counterexample: Take x = 2 and y = −2. Then x² = 4 and y² = (−2)² = 4, so x² = y². But x ≠ y, since 2 ≠ −2.
Conclude
Hence, the proposition is True and the converse is False.
Answer: Proposition – True | Converse – False (Counterexample: x = 2, y = −2)
Exercise 9.1 · Proposition and converse
Q7If x = y, then x³ = y³. x and y are real numbers.
First, understand the question
Cubing keeps the sign of a number (a negative number has a negative cube). So, unlike squares, equal cubes come only from equal numbers.
Given
- Proposition: If x = y, then x³ = y³
- x and y are real numbers
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
Converse: If x³ = y³, then x = y.
Proposition – True. Cubing both sides of x = y gives x³ = y³.
Converse – True. Every real number has exactly one real cube root, so
x³ = y³ ⇒ ∛(x³) = ∛(y³) ⇒ x = y
∴ Proposition – True, Converse – True.
Converse: If x³ = y³, then x = y.
Proposition – True. Cubing both sides of x = y gives x³ = y³.
Converse – True. Every real number has exactly one real cube root, so
∴ Proposition – True, Converse – True.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If x³ = y³, then x = y.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: Equal numbers have equal cubes. If x = y, then cubing both sides gives x³ = y³. For example, x = y = 2 gives x³ = y³ = 8.
Check the converse
Checking the Converse: The converse is True.
Justification: Taking the real cube root of both sides of x³ = y³ gives x = y, because every real number has exactly one real cube root. For example, x³ = y³ = 8 gives x = y = 2.
Conclude
Hence, the proposition is True and the converse is True.
Answer: Proposition – True | Converse – True
Exercise 9.1 · Divisibility
Q8If n is divisible by 24, then it is divisible by both 4 and 6. In Questions 8 to 12, n is a positive integer.
First, understand the question
A number divisible by 24 can be written as 4 × (6k) or 6 × (4k). For the converse, we look for a number that is divisible by 4 and 6 but not by 24.
Given
- Proposition: If n is divisible by 24, then it is divisible by both 4 and 6
- n is a positive integer
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
Converse: If n is divisible by both 4 and 6, then n is divisible by 24.
Proposition – True. Let n = 24k.
n = 24k = 4 × (6k) = 6 × (4k)
So n is divisible by 4 and by 6.
Converse – False. Counterexample: n = 12.
12 = 4 × 3 = 6 × 2
But 12 is not divisible by 24.
∴ Proposition – True, Converse – False.
Converse: If n is divisible by both 4 and 6, then n is divisible by 24.
Proposition – True. Let n = 24k.
So n is divisible by 4 and by 6.
Converse – False. Counterexample: n = 12.
But 12 is not divisible by 24.
∴ Proposition – True, Converse – False.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If n is divisible by both 4 and 6, then n is divisible by 24.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: Let n = 24k for some positive integer k. Then n = 4 × (6k) and n = 6 × (4k). Hence, n is divisible by both 4 and 6. For example, 24 is divisible by 4 (24 = 4 × 6) and by 6 (24 = 6 × 4).
Check the converse
Checking the Converse: The converse is False.
Counterexample: Take n = 12. Now 12 is divisible by 4 (12 = 4 × 3) and also by 6 (12 = 6 × 2), but 12 is not divisible by 24.
Conclude
Hence, the proposition is True and the converse is False.
Answer: Proposition – True | Converse – False (Counterexample: n = 12)
Exercise 9.1 · Divisibility
Q9If n is divisible by 60, then it is divisible by both 5 and 12.
First, understand the question
This is like Q8, with 60, 5 and 12. This time 5 and 12 have no common factor, and that decides whether the converse is true.
Given
- Proposition: If n is divisible by 60, then it is divisible by both 5 and 12
- n is a positive integer
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
Converse: If n is divisible by both 5 and 12, then n is divisible by 60.
Proposition – True. Let n = 60k.
n = 60k = 5 × (12k) = 12 × (5k)
Converse – True. 5 and 12 are co-prime, so
LCM(5, 12) = 5 × 12 = 60
A number divisible by both 5 and 12 is divisible by their LCM, 60.
∴ Proposition – True, Converse – True.
Converse: If n is divisible by both 5 and 12, then n is divisible by 60.
Proposition – True. Let n = 60k.
Converse – True. 5 and 12 are co-prime, so
A number divisible by both 5 and 12 is divisible by their LCM, 60.
∴ Proposition – True, Converse – True.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If n is divisible by both 5 and 12, then n is divisible by 60.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: Let n = 60k for some positive integer k. Then n = 5 × (12k) and n = 12 × (5k). Hence, n is divisible by both 5 and 12. For example, 60 is divisible by 5 and by 12.
Check the converse
Checking the Converse: The converse is True.
Justification: 5 and 12 have no common factor except 1 (they are co-prime), so LCM(5, 12) = 5 × 12 = 60. Every number divisible by both 5 and 12 is a multiple of their LCM, 60. Hence, n is divisible by 60.
Conclude
Hence, the proposition is True and the converse is True.
Answer: Proposition – True | Converse – True
Exercise 9.1 · Factors
Q10If n is the square of a prime number, then it has exactly 3 factors.
First, understand the question
We count the factors of n. The square of a prime, like 25 = 5², has only 1, 5 and 25 as factors. For the converse, we start from a number that has exactly 3 factors.
Given
- Proposition: If n is the square of a prime number, then it has exactly 3 factors
- n is a positive integer
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
Converse: If n has exactly 3 factors, then n is the square of a prime number.
Proposition – True. Let n = p², where p is prime. Its only factors are 1, p and p² (3 factors).
Converse – True. Let the factors be 1, m and n (1 < m < n). Then n ÷ m is also a factor between 1 and n, so
(n)/(m) = m ⇒ n = m²
If m were composite, n would have a fourth factor. So m is prime, and n = m² is the square of a prime.
∴ Proposition – True, Converse – True.
Converse: If n has exactly 3 factors, then n is the square of a prime number.
Proposition – True. Let n = p², where p is prime. Its only factors are 1, p and p² (3 factors).
Converse – True. Let the factors be 1, m and n (1 < m < n). Then n ÷ m is also a factor between 1 and n, so
If m were composite, n would have a fourth factor. So m is prime, and n = m² is the square of a prime.
∴ Proposition – True, Converse – True.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If n has exactly 3 factors, then n is the square of a prime number.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: Let n = p², where p is a prime number. The only positive factors of n are 1, p and p². Hence, n has exactly 3 factors. For example, for p = 5, n = 25 has factors 1, 5 and 25 — exactly three.
Check the converse
Checking the Converse: The converse is True.
Justification: Suppose n has exactly three factors, say 1, m and n, where 1 < m < n. Since m divides n, n ÷ m is also a factor of n lying between 1 and n, so it must equal m. Hence n = m². Also, m must be prime: if m were composite, a factor of m between 1 and m would give a fourth factor of n, which is impossible. Therefore, n is the square of a prime number.
Conclude
Hence, the proposition is True and the converse is True.
Answer: Proposition – True | Converse – True
Exercise 9.1 · Divisors
Q11If n is a product of two unequal prime numbers, then it has exactly 4 divisors.
First, understand the question
A product of two different primes, like 6 = 2 × 3, has the divisors 1, 2, 3 and 6. For the converse, we look for a number with 4 divisors that is not of this form.
Given
- Proposition: If n is a product of two unequal prime numbers, then it has exactly 4 divisors
- n is a positive integer
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
Converse: If n has exactly 4 divisors, then n is a product of two unequal prime numbers.
Proposition – True. Let n = pq (p ≠ q, both prime). Its divisors are 1, p, q and pq — exactly 4.
Converse – False. Counterexample: n = 8.
Divisors of 8: 1, 2, 4, 8, 8 = 2³
8 has 4 divisors, but it is not a product of two unequal primes.
∴ Proposition – True, Converse – False.
Converse: If n has exactly 4 divisors, then n is a product of two unequal prime numbers.
Proposition – True. Let n = pq (p ≠ q, both prime). Its divisors are 1, p, q and pq — exactly 4.
Converse – False. Counterexample: n = 8.
8 has 4 divisors, but it is not a product of two unequal primes.
∴ Proposition – True, Converse – False.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If n has exactly 4 divisors, then n is a product of two unequal prime numbers.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: Let n = pq, where p and q are unequal prime numbers. The positive divisors of n are 1, p, q and pq. These are four distinct divisors, and there are no others. Hence, n has exactly 4 divisors. For example, 6 = 2 × 3 has divisors 1, 2, 3 and 6.
Check the converse
Checking the Converse: The converse is False.
Counterexample: Take n = 8. Its divisors are 1, 2, 4 and 8, so it has exactly 4 divisors. But 8 = 2 × 2 × 2 = 2³, which is not a product of two unequal prime numbers.
Conclude
Hence, the proposition is True and the converse is False.
Answer: Proposition – True | Converse – False (Counterexample: n = 8)
Exercise 9.1 · Common factors
Q12If 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.
First, understand the question
“No factors in common” means the only common factor is 1. We use the fact that any common factor of n and n + 3 must also divide their difference, which is 3.
Given
- Proposition: If n and n + 3 have no factors in common, then n is not a multiple of 3
- n is a positive integer
To find
- The converse of the proposition
- Whether the proposition and the converse are true or false
Write this in your exam
Converse: If n is not a multiple of 3, then n and n + 3 have no factors in common.
Proposition – True. If n = 3k, then
n + 3 = 3k + 3 = 3(k + 1)
so 3 would be a common factor, which is not possible. Hence n is not a multiple of 3.
Converse – True. A common factor of n and n + 3 divides
(n + 3) – n = 3
so it can only be 1 or 3. Since n is not a multiple of 3, it must be 1.
∴ Proposition – True, Converse – True.
Converse: If n is not a multiple of 3, then n and n + 3 have no factors in common.
Proposition – True. If n = 3k, then
so 3 would be a common factor, which is not possible. Hence n is not a multiple of 3.
Converse – True. A common factor of n and n + 3 divides
so it can only be 1 or 3. Since n is not a multiple of 3, it must be 1.
∴ Proposition – True, Converse – True.
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Write the converse
Interchange the “if” part and the “then” part of the proposition.
Converse: If n is not a multiple of 3, then n and n + 3 have no factors in common.
Check the proposition
Checking the Proposition: The proposition is True.
Justification: Suppose n were a multiple of 3, say n = 3k. Then n + 3 = 3k + 3 = 3(k + 1) would also be a multiple of 3, so 3 would be a common factor of n and n + 3. This contradicts the given condition. Hence, n is not a multiple of 3.
Check the converse
Checking the Converse: The converse is True.
Justification: Any common factor of n and n + 3 must also divide their difference, (n + 3) − n = 3. Hence, the only possible common factors are 1 and 3. Since n is not a multiple of 3, the common factor cannot be 3. Therefore, n and n + 3 have no common factor other than 1.
Conclude
Hence, the proposition is True and the converse is True.
Answer: Proposition – True | Converse – True
Exercise 9.1 · Counterexamples
Q13Find counterexamples to the 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. There are no known ‘neat’ expressions that generate only primes!
First, understand the question
A claim like “all numbers of this form are prime” is false if we find even one value of n that gives a composite number. That value is the counterexample.
Given
- (i) 4n² + 1
- (ii) n² + n + 11
- (iii) 4ⁿ + 3
To find
- A counterexample for each claim
Write this in your exam
(i) Take n = 4:
4n² + 1 = 4 × 4² + 1\ = 64 + 1\ = 65 = 5 × 13
(ii) Take n = 10:
n² + n + 11 = 10² + 10 + 11\ = 121 = 11 × 11
(iii) Take n = 4:
4ⁿ + 3 = 4⁴ + 3\ = 256 + 3\ = 259 = 7 × 37
∴ 65, 121 and 259 are composite, so all three claims are false.
(i) Take n = 4:
(ii) Take n = 10:
(iii) Take n = 4:
∴ 65, 121 and 259 are composite, so all three claims are false.
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(i) Find a counterexample for 4n² + 1
(i) The claim is False.
Counterexample: Take n = 4.
4n² + 1 = 4 × (4)² + 1 = 4 × 16 + 1 = 65 = 5 × 13
Since 65 is composite, not all numbers of the form 4n² + 1 are prime.
(ii) Find a counterexample for n² + n + 11
(ii) The claim is False.
Counterexample: Take n = 10.
n² + n + 11 = (10)² + 10 + 11 = 100 + 10 + 11 = 121 = 11 × 11
Since 121 is composite, not all numbers of the form n² + n + 11 are prime.
(iii) Find a counterexample for 4ⁿ + 3
(iii) The claim is False.
Counterexample: Take n = 4.
4ⁿ + 3 = 4⁴ + 3 = 256 + 3 = 259 = 7 × 37
Since 259 is composite, not all numbers of the form 4ⁿ + 3 are prime.
Conclude
Hence, all three claims are False.
Answer: (i) n = 4 → 65 | (ii) n = 10 → 121 | (iii) n = 4 → 259 — All Claims False
Exercise 9.1 · Counterexamples
Q14Find counterexamples: (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.
First, understand the question
Each statement is false if we can find one value of n that satisfies the “if” part but gives a composite number.
Given
- (i) If n is prime, then 2ⁿ − 1 is prime
- (ii) If n is even, then 2ⁿ + 1 is prime
To find
- A counterexample for each statement
Write this in your exam
(i) Take n = 11 (a prime):
2¹¹ – 1 = 2048 – 1\ = 2047 = 23 × 89
(ii) Take n = 6 (an even number):
2⁶ + 1 = 64 + 1\ = 65 = 5 × 13
∴ 2047 and 65 are composite, so both statements are false.
(i) Take n = 11 (a prime):
(ii) Take n = 6 (an even number):
∴ 2047 and 65 are composite, so both statements are false.
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(i) Find a counterexample for 2ⁿ − 1
(i) The statement is False.
Counterexample: Take n = 11, which is a prime number.
2ⁿ − 1 = 2¹¹ − 1 = 2048 − 1 = 2047 = 23 × 89
Since 2047 is composite, 2ⁿ − 1 is not prime even though n = 11 is prime.
(ii) Find a counterexample for 2ⁿ + 1
(ii) The statement is False.
Counterexample: Take n = 6, which is an even number.
2ⁿ + 1 = 2⁶ + 1 = 64 + 1 = 65 = 5 × 13
Since 65 is composite, 2ⁿ + 1 is not prime even though n = 6 is even.
Conclude
Hence, both statements are False.
Answer: (i) n = 11 → 2047 = 23 × 89 | (ii) n = 6 → 65 = 5 × 13 — Both False
Exercise 9.1 · Divisibility test
Q15If a number is divisible by 8, then it is divisible by both 2 and 4 — justify the statement, and say whether checking divisibility by 2 and 4 is enough to test divisibility by 8. Full question: 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?
First, understand the question
Part (i) asks us to prove the statement. Part (ii) asks whether its converse (divisible by 2 and 4 ⇒ divisible by 8) can be used as a test for 8.
Given
- Statement: If a number is divisible by 8, then it is divisible by both 2 and 4
To find
- (i) A justification of the statement
- (ii) Is checking 2 and 4 enough to test divisibility by 8?
Write this in your exam
(i) True. Let n = 8k.
n = 8k = 2 × (4k) = 4 × (2k)
So n is divisible by both 2 and 4.
(ii) No. 12 is divisible by 2 and by 4, but not by 8. In fact,
LCM(2, 4) = 4
so divisibility by 2 and 4 only guarantees divisibility by 4.
∴ (i) True (ii) No, it is not enough.
(i) True. Let n = 8k.
So n is divisible by both 2 and 4.
(ii) No. 12 is divisible by 2 and by 4, but not by 8. In fact,
so divisibility by 2 and 4 only guarantees divisibility by 4.
∴ (i) True (ii) No, it is not enough.
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(i) Justify the statement
(i) The statement is True.
Justification: Let the number be n = 8k, where k is an integer. Then n = 2 × (4k), so n is divisible by 2; and n = 4 × (2k), so n is divisible by 4. Hence, every number divisible by 8 is divisible by both 2 and 4. For example, 24 is divisible by 8, and 24 ÷ 2 = 12 and 24 ÷ 4 = 6.
(ii) Check whether 2 and 4 are enough
Checking 2 and 4 is the converse of the statement, so we test the converse.
(ii) No, it is not enough.
Reason: Checking divisibility by 2 and 4 is actually the converse of the given statement, and the converse is false. For example, 12 is divisible by both 2 (12 ÷ 2 = 6) and 4 (12 ÷ 4 = 3), but 12 is not divisible by 8. In fact, LCM(2, 4) = 4, so divisibility by both 2 and 4 guarantees only divisibility by 4, not by 8.
Conclude
Hence, (i) the statement is True, and (ii) checking divisibility by 2 and 4 is not sufficient to test divisibility by 8.
Answer: (i) Statement – True | (ii) No, not enough (Counterexample: 12)
Exercise 9.1 · If–then statements
Q16Express the relationship between ‘a number is divisible by 3’ and ‘sum of the digits is a multiple of 3’ using ‘If–then’ sentences. 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.
First, understand the question
The divisibility rule for 3 works both ways. We write it as two “If–then” sentences, and each one is the converse of the other.
Given
- P: a number is divisible by 3
- Q: the sum of its digits is a multiple of 3
To find
- The relationship as “If–then” sentences
Write this in your exam
Statement 1: If a number is divisible by 3, then the sum of its digits is a multiple of 3.
Statement 2: If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.
Both are true, and each is the converse of the other. Example:
54: 5 + 4 = 9 and 54 = 3 × 18
∴ Both “If–then” statements are true (converses of each other).
Statement 1: If a number is divisible by 3, then the sum of its digits is a multiple of 3.
Statement 2: If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.
Both are true, and each is the converse of the other. Example:
∴ Both “If–then” statements are true (converses of each other).
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Write the two If–then sentences
The relationship is expressed by the following two ‘If–then’ statements:
Statement 1: If a number is divisible by 3, then the sum of its digits is a multiple of 3.
Statement 2: If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.
Check them with examples
Both statements are True, and each statement is the converse of the other.
Example: For the number 54, the sum of digits is 5 + 4 = 9, which is a multiple of 3. Hence, 54 is divisible by 3 (54 = 3 × 18). Similarly, for 123, the digit sum is 1 + 2 + 3 = 6, a multiple of 3, and 123 = 3 × 41.
Conclude
Hence, ‘a number is divisible by 3’ and ‘the sum of its digits is a multiple of 3’ imply each other.
Answer: Both ‘If–then’ Statements – True (Converses of Each Other)
Exercise 9.1 · Necessary and sufficient conditions
Q17Two thin sticks are put together as the diagonals of a quadrilateral of type Q — must they be of equal length, and does the arrangement matter? Full question: 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 Fig. 4).
First, understand the question
The two sticks become the diagonals of the quadrilateral. In (i), equal diagonals are something every type-Q shape has (necessary). In (ii), equal diagonals are enough to make type Q (sufficient).
Given
- (i) If a quadrilateral is of type Q, then it has equal diagonals
- (ii) If a quadrilateral has equal diagonals, then it is of type Q
To find
- (a) Must the sticks be of equal length?
- (b) Does it matter how they are put together?
Write this in your exam
(i) (a) Yes. The sticks are the diagonals, and every type-Q quadrilateral has equal diagonals.
(b) Yes, it may matter. The property does not say that every quadrilateral with equal diagonals is of type Q (that would be its converse).
(ii) (a) Yes. Equal sticks give equal diagonals, so the quadrilateral is of type Q.
(b) No. In every arrangement the diagonals stay equal, so it is always of type Q.
∴ (i) Yes; Yes, it may matter (ii) Yes; No
(i) (a) Yes. The sticks are the diagonals, and every type-Q quadrilateral has equal diagonals.
(b) Yes, it may matter. The property does not say that every quadrilateral with equal diagonals is of type Q (that would be its converse).
(ii) (a) Yes. Equal sticks give equal diagonals, so the quadrilateral is of type Q.
(b) No. In every arrangement the diagonals stay equal, so it is always of type Q.
∴ (i) Yes; Yes, it may matter (ii) Yes; No
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(i) Property: If type Q, then equal diagonals
(a) Yes, the two sticks must be of equal length.
Reason: The two sticks themselves form the diagonals of the quadrilateral. Since every quadrilateral of type Q has equal-length diagonals, the two sticks (diagonals) must be of equal length. Unequal sticks would give unequal diagonals, and the quadrilateral could not be of type Q.
(b) Yes, it may matter how the two sticks are put together.
Reason: The given property only says that every type-Q quadrilateral has equal diagonals; it does not say that every quadrilateral with equal diagonals is of type Q (that would be its converse). Type Q may require some additional condition. For example, if Q meant “rectangles”, the two equal sticks would also have to bisect each other, i.e., cross at their midpoints. Hence, merely using two equal sticks in any position may not produce a quadrilateral of type Q.
(ii) Property: If equal diagonals, then type Q
(a) Yes — choosing two sticks of equal length will guarantee a quadrilateral of type Q.
Reason: If the sticks are equal, the diagonals of the quadrilateral formed will be equal, and by the given property the quadrilateral will be of type Q. (Note: the property does not claim that a type-Q quadrilateral cannot be formed in any other way; equal sticks are a sufficient way, not the only possible way.)
(b) No, it will not matter how the two sticks are put together.
Reason: With equal sticks, the diagonals will be equal in every arrangement (as long as the sticks cross each other so that joining the endpoints forms a proper quadrilateral). By the given property, the quadrilateral will be of type Q in every case. No extra condition on the angle or the midpoint is required.
Conclude
Hence, (i) (a) Yes (b) Yes, it may matter; (ii) (a) Yes, equal sticks guarantee type Q (b) No, it will not matter.
Note: Some solution websites give a different answer to (i)(b). The answer given here — “Yes, it may matter” — is the logically correct one and matches the textbook’s intention: from “Q ⇒ equal diagonals” alone, we cannot conclude that equal diagonals will always produce Q.
Answer: (i) (a) Yes (b) Yes, may matter | (ii) (a) Yes — sufficient (b) No
2-Minute Revision — Key Terms of Chapter 9
| Term | Meaning in One Line |
|---|---|
| Proposition | A statement that is either true or false. |
| Converse | The statement obtained by interchanging the ‘If’ and ‘Then’ parts of a proposition. |
| Counterexample | One example satisfying the ‘If’ part but not the ‘Then’ part; it proves a statement false. |
| Justification | A reason or property showing why a true statement is always true. |
| Necessary Condition | A condition without which the result is impossible (Q17(i): equal diagonals). |
| Sufficient Condition | A condition which alone guarantees the result (Q17(ii): equal diagonals). |
True Statement ➜ Justify • False Statement ➜ Counterexample • Converse = Interchange ‘If’ and ‘Then’
Frequently Asked Questions on Chapter 9
No. It may be true (Q1, Q4, Q7) or false (Q2, Q5, Q6). The proposition and its converse must always be checked separately.
Only one. A single valid counterexample is enough to prove a general statement false.
No. In Q13, n² + n + 11 gives primes for n = 1 to 9 but fails at n = 10. A proof must give a reason that works for every case.
Yes. Example: “If a number is even, then it is a multiple of 3” is false (counterexample: 2), and its converse “If a number is a multiple of 3, then it is even” is also false (counterexample: 3).
Squaring removes the sign, so x² = y² allows x = −y (converse false). Cubing preserves the sign, so x³ = y³ forces x = y (converse true).
For Q1–Q12, always write four things: the Converse, whether the Proposition is True/False with justification or counterexample, whether the Converse is True/False with justification or counterexample, and a final “Hence…” line — exactly as shown in the “Write this in your exam” box of each question.
Class 9 Maths • Ganita Manjari Part 2 • Chapter 9: Propositions and their Converses • Exercise 9.1 Solutions (Q1–Q17) — Prepared for students to understand and to write directly in exams.