NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 09 Propositions and their Converses Exercise 9.1

Exercise Set 9.1

Question 1. Frame the converse of the following proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
If two lines are parallel, then the corresponding angles formed by a transversal are equal.
Answer:
Proposition (P): If two lines are parallel, then the corresponding angles formed by a transversal are equal.
Converse (Q): If a transversal of two lines makes equal corresponding angles, then the two lines are parallel.

P is true. This is the corresponding angles property of parallel lines: if l ∥ m and the transversal t cuts them at P and Q, then ∠1 = ∠2.
Q is true. Suppose ∠1 = ∠2 but l and m are not parallel. Then they meet at some point R, and P, Q, R form a triangle.

∠1 is an exterior angle of △PQR at P, and ∠2 = ∠PQR is an interior opposite angle. By the exterior angle property,
∠1 = ∠2 + ∠PRQ, so ∠PRQ = ∠1 − ∠2 = 0°.
A triangle cannot have a 0° angle, so this is a contradiction. The lines cannot meet, i.e., l ∥ m.
Conclusion: P is true; Q is true.

Teacher's Note:
1. The converse is proved by contradiction: assume the lines meet and reach an impossible triangle.
2. Make sure ∠1 is the exterior angle and ∠2 the interior opposite angle of the same triangle.

Question 2. Frame the converse of the following proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
If a quadrilateral is a square, then all its angles are equal.
Answer:
Proposition (P): If a quadrilateral is a square, then all its angles are equal.
Converse (Q): If all the angles of a quadrilateral are equal, then it is a square.
P is true. By definition each angle of a square is 90°, so all four angles are equal.
Q is false. If all four angles are equal, each is \( \frac{360°}{4} \) = 90°. That makes the quadrilateral a rectangle, but not necessarily a square.
Counterexample: rectangle PQRS with PQ = 6 cm and QR = 3 cm has all four angles equal to 90°, but its sides are not all equal, so it is not a square.

Conclusion: P is true; Q is false.

Question 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.
Proposition: If AB = AC, then IE = IF. Frame the converse and decide whether each statement is true. Justify the true statements and give a counterexample for each false one.
Answer:
Proposition (P): If AB = AC, then IE = IF.
Converse (Q): If IE = IF, then AB = AC.
Here BE bisects ∠B (E lies on AC) and CF bisects ∠C (F lies on AB).
Part 1: P is true.

Given AB = AC, the angles opposite these sides are equal: ∠C = ∠B (Statement 1).
So ∠IBC = \( \frac{1}{2} \)∠B = \( \frac{1}{2} \)∠C = ∠ICB, and in △IBC the sides opposite these equal angles are equal: IB = IC (Statement 2).
Now compare △IBF and △ICE:
1. ∠IBF = ∠ICE (halves of the equal angles B and C)
2. IB = IC (shown above)
3. ∠BIF = ∠CIE (vertically opposite angles)
4. So △IBF ≅ △ICE (ASA), and hence IF = IE (CPCT).
P is true.
Part 2: Q is false.
Counterexample: take △ABC with ∠A = 60°, ∠B = 90°, ∠C = 30°. Here AC (opposite 90°) is longer than AB (opposite 30°), so AB ≠ AC. We show that IE = IF anyway.

Step 1: ∠ABE = ∠IBC = 45° and ∠ACF = ∠ICB = 15° (half angles).
Step 2: In △AFC, ∠AFC = 180° − 60° − 15° = 105°, so ∠IFB = 180° − 105° = 75°.
Step 3: In △ABE, ∠AEB = 180° − 60° − 45° = 75°, so ∠IEA = 75°.
Step 4: I is on the bisectors of ∠B and ∠C, so it is at the same distance from all three sides. Draw IP ⊥ AB and IQ ⊥ AC; then IP = IQ.
Step 5: In right triangles △IPF and △IQE: ∠IPF = ∠IQE = 90°, ∠IFP = ∠IEQ = 75° (Steps 2 and 3), IP = IQ (Step 4). So △IPF ≅ △IQE (AAS), and IF = IE (CPCT).
Numerical check with BC = 5 cm: AB ≈ 2.89 cm, AC ≈ 5.77 cm, while IE = IF ≈ 1.09 cm.
So IE = IF even though AB ≠ AC. Q is false.
Remark: the same reasoning works for any triangle with ∠A = 60°, because then ∠AFC and ∠AEB always add up to 180°. In fact IE = IF happens exactly when AB = AC or ∠A = 60°.
Conclusion: P is true; Q is false.

Teacher's Note:
1. Part 1 uses Statement 1 and then Statement 2 of this chapter, one after the other.
2. A counterexample must satisfy the ‘if’ part (IE = IF) but fail the ‘then’ part (AB ≠ AC).
3. Any triangle with ∠A = 60° and AB ≠ AC works as a counterexample.

Question 4. Frame the converse of the following proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
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.
Answer:
Proposition (P): If x = y, then a + x = a + y.
Converse (Q): If a + x = a + y, then x = y.
P is true. If x = y, they are the same number. Adding the same number a to the same number gives the same result, so a + x = a + y.
Q is true. Start from a + x = a + y and subtract a from both sides (that is, add −a):
(a + x) − a = (a + y) − a
x = y
These are exactly the steps behind “adding the same number to both sides” and “cancelling the same term from both sides” when we solve an equation such as x + 5 = 12.
Conclusion: P is true; Q is true.

Question 5. Frame the converse of the following proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
If a and b are perfect squares, then ab is a perfect square.
Answer:
Proposition (P): If a and b are perfect squares, then ab is a perfect square.
Converse (Q): If ab is a perfect square, then a and b are perfect squares.
P is true. Write a = m² and b = k² for whole numbers m and k.
ab = m² × k² = (m × k) × (m × k) = (mk)²
So ab is a perfect square. For example, 4 × 9 = 36 = 6².
Q is false. Counterexample: a = 2 and b = 8. Then ab = 16 = 4², a perfect square, but neither 2 nor 8 is a perfect square.
Conclusion: P is true; Q is false.

Question 6. Frame the converse of the following proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
(In Questions 6 and 7, x and y are real numbers.) If x = y, then x² = y².
Answer:
Proposition (P): If x = y, then x² = y².
Converse (Q): If x² = y², then x = y.
P is true. If x = y, then x² = x × x = y × y = y².
Q is false. From x² = y²:
x² − y² = 0
(x − y)(x + y) = 0
So x = y or x = −y. The second case gives counterexamples.
Counterexample: x = 2, y = −2. Then x² = 4 = y², but x ≠ y.
Conclusion: P is true; Q is false.

Question 7. Frame the converse of the following proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
If x = y, then x³ = y³.
Answer:
Proposition (P): If x = y, then x³ = y³.
Converse (Q): If x³ = y³, then x = y.
P is true. If x = y, then x³ = x × x × x = y × y × y = y³.
Q is true. Suppose x³ = y³. Using x³ − y³ = (x − y)(x² + xy + y²),
(x − y)(x² + xy + y²) = 0
So x − y = 0 or x² + xy + y² = 0.
Now \( x^2 + xy + y^2 = \left(x + \frac{y}{2}\right)^2 + \frac{3y^2}{4} \), a sum of two squares.
It equals 0 only when y = 0 and x + \( \frac{y}{2} \) = 0, i.e., x = y = 0.
So in every case x = y.
(Squares lose the sign, but cubes keep it: (−2)³ = −8 while 2³ = 8. That is why the counterexample of Question 6 does not work here.)
Conclusion: P is true; Q is true.

Teacher's Note:
1. Compare with Question 6: cubes keep the sign, squares do not.
2. Completing the square shows x² + xy + y² is never negative and is zero only at x = y = 0.

Question 8. Frame the converse of the following proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
(In Questions 8–12, n is a positive integer.) If n is divisible by 24, then it is divisible by both 4 and 6.
Answer:
Proposition (P): If n is divisible by 24, then it is divisible by both 4 and 6.
Converse (Q): If n is divisible by both 4 and 6, then it is divisible by 24.
P is true. Let n = 24k. Then n = 4 × (6k) and n = 6 × (4k), so n is divisible by 4 and by 6.
Q is false. Counterexample: n = 12. We have 12 = 4 × 3 and 12 = 6 × 2, so 12 is divisible by both 4 and 6, but 12 is not divisible by 24.
(The smallest number divisible by both 4 and 6 is LCM(4, 6) = 12, not 24.)
Conclusion: P is true; Q is false.

Question 9. Frame the converse of the following proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
If n is divisible by 60, then it is divisible by both 5 and 12.
Answer:
Proposition (P): If n is divisible by 60, then it is divisible by both 5 and 12.
Converse (Q): If n is divisible by both 5 and 12, then it is divisible by 60.
P is true. Let n = 60k. Then n = 5 × (12k) and n = 12 × (5k).
Q is true. As 12 = 2 × 2 × 3 divides n, the prime factorisation of n contains 2 × 2 × 3. As 5 divides n, the prime 5 also appears, and 5 is different from 2 and 3.
So n = 2 × 2 × 3 × 5 × (other factors) = 60 × (other factors), i.e., n is divisible by 60.
(5 and 12 have no common factor except 1, so LCM(5, 12) = 5 × 12 = 60. Compare Question 8, where 4 and 6 share the factor 2.)
Conclusion: P is true; Q is true.

Question 10. Frame the converse of the following proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
If n is the square of a prime number, then it has exactly 3 factors.
Answer:
Proposition (P): If n is the square of a prime number, then it has exactly 3 factors.
Converse (Q): If n has exactly 3 factors, then it is the square of a prime number.
P is true. Let n = p² with p prime. Its only prime factor is p (taken twice), so its factors are 1, p and p², three different numbers. Examples: 4 → 1, 2, 4; 9 → 1, 3, 9; 25 → 1, 5, 25.
Q is true. Suppose n has exactly 3 factors.
1. 3 is odd, so by Example 3 (Statement Q) n is a perfect square: n = m².
2. n ≠ 1 (1 has only one factor), so m ≥ 2 and 1, m, m² are three different factors of n. These must be all of them.
3. If m were not prime, it would have a factor d with 1 < d < m. Then d would also divide n = m², giving a fourth factor, which is impossible.
So m is prime and n = m² is the square of a prime.
Conclusion: P is true; Q is true.

Teacher's Note:
1. The converse uses Example 3: an odd number of factors means a perfect square.
2. Then check that m must be prime, otherwise a fourth factor appears.

Question 11. Frame the converse of the following proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
If n is a product of two unequal prime numbers, then it has exactly 4 divisors.
Answer:
Proposition (P): If n is a product of two unequal prime numbers, then it has exactly 4 divisors.
Converse (Q): If n has exactly 4 divisors, then it is a product of two unequal prime numbers.
P is true. Let n = pq with p and q different primes. Every divisor is built from these prime factors, so the divisors are 1, p, q and pq, four different numbers. Example: 15 = 3 × 5 has divisors 1, 3, 5, 15.
Q is false. Counterexample: n = 8 has exactly 4 divisors, 1, 2, 4, 8, but 8 = 2 × 2 × 2 is not a product of two unequal primes. (27 = 3³, with divisors 1, 3, 9, 27, works too.)
Conclusion: P is true; Q is false.

Question 12. Frame the converse of the following proposition. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
If n and n + 3 have no factors in common, then n is not a multiple of 3.
Answer:
(“No factors in common” means no common factor other than 1.)
Proposition (P): If n and n + 3 have no factors in common, then n is not a multiple of 3.
Converse (Q): If n is not a multiple of 3, then n and n + 3 have no factors in common.
P is true. Suppose instead that n is a multiple of 3, say n = 3k. Then n + 3 = 3k + 3 = 3(k + 1) is also a multiple of 3, so 3 is a common factor of n and n + 3. This contradicts the given condition, so n is not a multiple of 3.
Q is 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, so d = 1.
Hence n and n + 3 have no common factor other than 1. Check: 4 and 7, 5 and 8, 10 and 13 have no common factor.
Conclusion: P is true; Q is true.

Teacher's Note:
1. A common factor of two numbers also divides their difference.
2. P is easiest to prove by assuming the opposite (n = 3k).

Question 13. 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.
Answer:
Substitute n = 1, 2, 3, … until a value that is not prime appears.
(i) 4n² + 1
n = 1: 5 (prime); n = 2: 17 (prime); n = 3: 37 (prime); n = 4: 65 (not prime)
n = 4: 4 × 4² + 1 = 64 + 1 = 65 = 5 × 13
Counterexample: n = 4 gives 65, which is composite.
(ii) n² + n + 11
n = 1 to 9 give 13, 17, 23, 31, 41, 53, 67, 83, 101, all prime.
n = 10: 10² + 10 + 11 = 100 + 10 + 11 = 121 = 11 × 11
Counterexample: n = 10 gives 121, which is composite. (n = 11 also works: 121 + 11 + 11 = 143 = 11 × 13.)
(iii) 4ⁿ + 3
n = 1: 7 (prime); n = 2: 19 (prime); n = 3: 67 (prime); n = 4: 259 (not prime)
n = 4: 4⁴ + 3 = 256 + 3 = 259 = 7 × 37
Counterexample: n = 4 gives 259, which is composite.

Question 14. 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.
Answer:
(i) Try prime values of n:
n = 2: 3; n = 3: 7; n = 5: 31; n = 7: 127 (all prime)
n = 11: 2¹¹ − 1 = 2048 − 1 = 2047 = 23 × 89
Counterexample: n = 11 is prime, but 2¹¹ − 1 = 2047 = 23 × 89 is not prime.
(ii) Try even values of n:
n = 2: 5; n = 4: 17 (both prime)
n = 6: 2⁶ + 1 = 64 + 1 = 65 = 5 × 13
Counterexample: n = 6 is even, but 2⁶ + 1 = 65 = 5 × 13 is not prime.

Question 15. 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?
Answer:
(i) Let the number be n = 8k. Then
n = 8k = 2 × (4k) and n = 8k = 4 × (2k).
So n is divisible by both 2 and 4. The statement is true.
(ii) No, it is not enough. Using “divisible by 2 and 4” as a test for 8 would need the converse: “If a number is divisible by both 2 and 4, then it is divisible by 8.” This converse is false.
Counterexample: 12 is divisible by 2 and by 4, but 12 ÷ 8 leaves remainder 4, so 12 is not divisible by 8. (20, 28, 1012, … also work.)
Reason: any number divisible by 4 is already divisible by 2, so checking 2 and 4 is the same as checking only 4, and LCM(2, 4) = 4, not 8.
The correct shortcut: a number is divisible by 8 if the number formed by its last three digits is divisible by 8.

Teacher's Note:
1. A divisibility test needs the converse direction: ‘passes the test ⇒ divisible’.
2. Checking 2 and 4 together adds nothing beyond checking 4.

Question 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 between ‘a number is divisible by 3’ and ‘sum of the digits is a multiple of 3’ using ‘If-then’ sentences.
Answer:
The relationship is a proposition together with its converse, and both are true:
1. If a number is divisible by 3, then the sum of its digits is a multiple of 3.
2. If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.
So the two statements imply each other: a number is divisible by 3 exactly when its digit sum is a multiple of 3.
Why both hold (for a 3-digit number with digits a, b, c):
100a + 10b + c = (99a + 9b) + (a + b + c) = 9(11a + b) + (a + b + c)
9(11a + b) is always a multiple of 3, so the number and its digit sum differ by a multiple of 3. Hence one is a multiple of 3 exactly when the other is. The same idea works for any number of digits.
Examples: 462 → 4 + 6 + 2 = 12 (a multiple of 3), and 462 = 3 × 154. 247 → 2 + 4 + 7 = 13 (not a multiple of 3), and 247 is not divisible by 3.

Question 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 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?
Answer:
The sticks form the diagonals, so their lengths are the lengths of the diagonals.
(i) Property: “If a quadrilateral is of type Q, then it has equal-length diagonals.”
(a) Yes, the sticks must be of equal length. Every type-Q quadrilateral has equal diagonals. With unequal sticks the diagonals would be unequal, and the quadrilateral could not be of type Q. Equal sticks are necessary.
(b) Yes, it may matter. The property only describes something every type-Q quadrilateral has; it does not say that every quadrilateral with equal diagonals is of type Q. So equal sticks alone may not be enough. For instance, if type Q were ‘rectangles’, the equal sticks would also have to bisect each other; if they cross elsewhere, the quadrilateral has equal diagonals but is not a rectangle.

So, without more information about Q, we cannot be sure that every arrangement of two equal sticks gives a type-Q quadrilateral.
(ii) Property: “If a quadrilateral has equal diagonals, then it is of type Q.”
(a) Equal sticks are now enough (sufficient): any quadrilateral with equal diagonals is of type Q, so equal sticks guarantee success. The property does not say that type-Q quadrilaterals must have equal diagonals, so equal sticks are not strictly necessary; but with unequal sticks we cannot be sure. The safe choice is two sticks of equal length.
(b) No, it will not matter how the equal sticks are put together (as long as they cross and form a quadrilateral). In both arrangements (a) and (b) above, the diagonals are equal, so the quadrilateral is of type Q.
Note: in (i) the property is “Q ⇒ equal diagonals”, which makes equal sticks necessary; in (ii) it is the converse, “equal diagonals ⇒ Q”, which makes equal sticks sufficient. A proposition and its converse give different information.

Teacher's Note:
1. ‘Q ⇒ equal diagonals’ tells us what is necessary; ‘equal diagonals ⇒ Q’ tells us what is sufficient.
2. Use rectangles as a concrete type Q to see why placement matters in (i).
3. Figure (b) shows equal diagonals without a rectangle.