NCERT Solutions Class 9 Maths Ch 9: Propositions & Converses

No, a true proposition does not always have a true converse. But Statement 2 is true, and it can be proved.

1. GivenIn △ABC, ∠ABC = ∠BCA. We must show AB = AC.
2. ConstructionDraw AD ⊥ BC, meeting BC at D.
3. Compare the angles∠ADB = ∠ADC = 90°. In the two right triangles, ∠BAD = 90° − ∠ABC and ∠CAD = 90° − ∠BCA. Since ∠ABC = ∠BCA, we get ∠BAD = ∠CAD.
4. CongruenceAD is common. So △ABD ≅ △ACD by ASA.
5. ConclusionAB = AC (corresponding sides of congruent triangles). So the sides opposite equal angles are equal.

Why not always? The proposition “if x = y then x² = y²” is true, but its converse is false (2² = (−2)² but 2 ≠ −2). See Q10.

Yes. For example: “If a positive integer is even, then it is a multiple of 3.”

1. PropositionIf a positive integer is even, then it is a multiple of 3. This is false: 2 is even but not a multiple of 3.
2. ConverseIf a positive integer is a multiple of 3, then it is even. This is also false: 3 is a multiple of 3 but not even.

Example: “If it rains on an open road, the road gets wet” is true, but “If the road is wet, then it has rained” is false.

1. Example 1: rain and a wet roadProposition: If it rains on an uncovered road, the road becomes wet (true). Converse: If the road is wet, then it has rained (false). A tanker may have spilled water.
2. Example 2: buying notebooksA shop sells each notebook for ₹10. Proposition: If I buy three notebooks, I spend ₹30 (true). Converse: If I spend ₹30, I bought three notebooks (false). I could have bought a ₹30 pen.

Four examples are given below, two from geometry and two from number theory.

Both the proposition and its converse are true.

1. PropositionIf two lines are parallel, the corresponding angles made by a transversal are equal. True (the corresponding angles property).
2. ConverseIf the corresponding angles made by a transversal with two lines are equal, then the two lines are parallel.
3. Converse: TrueEqual corresponding angles are exactly the condition for two lines to be parallel.

So l ∥ m ⇒ α = β, and α = β ⇒ l ∥ m: the lines are parallel if and only if the corresponding angles are equal.

The proposition is true. Its converse is false.

1. Proposition: TrueEvery angle of a square is 90°, so all its angles are equal.
2. ConverseIf all the angles of a quadrilateral are equal, then it is a square.
3. Converse: FalseA rectangle of length 6 cm and breadth 4 cm has four equal angles (90° each), but its sides are not all equal, so it is not a square. This one counterexample is enough.

The proposition is true. Its converse (if IE = IF then AB = AC) is false.

1. Proposition: proofSince AB = AC, ∠ABC = ∠BCA. The bisectors BI and CI cut these equal angles in half, so ∠IBC = ∠ICB, and hence IB = IC (by Q1).
2. Compare △CIE and △BIF∠ICE = ∠IBF (each is half of an equal base angle), ∠CIE = ∠BIF (vertically opposite angles), and CI = BI. So △CIE ≅ △BIF by ASA, and IE = IF. The proposition is true.
3. ConverseIf IE = IF, then AB = AC. This is false.
4. CounterexampleTake △ABC with ∠A = 60°, ∠B = 90° and ∠C = 30°, so AB ≠ AC. Take AB = 1, BC = √3, AC = 2, with B at the origin, A = (1, 0) and C = (0, √3). Then the incentre is I = (√3 − 12, √3 − 12), and calculation givesIE = IF = √2(2 − √3) ≈ 0.379So IE = IF although AB ≠ AC.
5. Why it works (angle chase)∠AEB = 180° − 60° − 45° = 75° and ∠AFC = 180° − 60° − 15° = 105°. Mark G on AB with AG = AE. Then △AEI ≅ △AGI (SAS), so IG = IE and ∠AGI = 75°. In △IFG, ∠IFG = 180° − 105° = 75° = ∠IGF, so IF = IG. Hence IE = IG = IF.

The proposition is true. Its converse is true.

1. Proposition: TrueAdding the same number a to equal numbers gives equal results.
2. ConverseIf a + x = a + y, then x = y.
3. Converse: TrueSubtract a from both sides: (a + x) − a = (a + y) − a, so x = y.

The proposition is true. Its converse is false.

1. Proposition: TrueLet a = m² and b = n². Then ab = m² × n² = (mn)², and mn is an integer, so ab is a perfect square.
2. ConverseIf ab is a perfect square, then a and b are perfect squares.
3. Converse: FalseTake a = 2 and b = 8. Then ab = 16 = 4² is a perfect square, but neither 2 nor 8 is a perfect square.

The proposition is true. Its converse is false.

1. Proposition: TrueEqual numbers have equal squares.
2. ConverseIf x² = y², then x = y.
3. Converse: FalseTake x = 2 and y = −2. Then x² = 4 = y², but 2 ≠ −2. (In general x² = y² only tells us x = y or x = −y.)

The proposition is true. Its converse is true.

1. Proposition: TrueEqual numbers have equal cubes.
2. ConverseIf x³ = y³, then x = y.
3. Converse: TrueEvery real number has exactly one real cube root. Taking the real cube root of both sides of x³ = y³ gives x = y.

The proposition is true. Its converse is false.

1. Proposition: TrueLet n = 24k. Then n = 4 × 6k and n = 6 × 4k, so n is divisible by both 4 and 6.
2. ConverseIf n is divisible by both 4 and 6, then n is divisible by 24.
3. Converse: FalseTake n = 12. It is divisible by 4 and by 6, but not by 24. (Divisible by both 4 and 6 only guarantees divisible by their LCM, 12.)

The proposition is true. Its converse is true.

1. Proposition: TrueLet n = 60k. Then n = 5 × 12k and n = 12 × 5k, so n is divisible by both 5 and 12.
2. ConverseIf n is divisible by both 5 and 12, then n is divisible by 60.
3. Converse: True5 and 12 have no common factor, so their LCM is 5 × 12 = 60. Every number divisible by both 5 and 12 is a multiple of their LCM, which is 60.

The proposition is true. Its converse is true.

1. Proposition: TrueLet n = p², where p is prime. Its only positive factors are 1, p and p². So it has exactly 3 factors.
2. ConverseIf n has exactly 3 factors, then n is the square of a prime number.
3. Converse: TrueLet the three factors be 1, p and n with 1 < p < n. Since p divides n, n ÷ p is also a factor, and it lies between 1 and n, so n ÷ p = p. Hence n = p². If p were not prime, it would have a factor between 1 and p, giving n an extra factor, so p is prime.

The proposition is true. Its converse is false.

1. Proposition: TrueLet n = pq with p ≠ q primes. Its divisors are 1, p, q and pq: exactly four.
2. ConverseIf n has exactly 4 divisors, then n is a product of two unequal prime numbers.
3. Converse: FalseTake n = 8. Its divisors are 1, 2, 4 and 8 (exactly four), but 8 = 2 × 2 × 2 is not a product of two unequal primes.

The proposition is true. Its converse is true.

1. Proposition: TrueIf n were a multiple of 3, then n + 3 would also be a multiple of 3, so 3 would be a common factor. This contradicts the condition, so n is not a multiple of 3.
2. ConverseIf n is not a multiple of 3, then n and n + 3 have no factors in common (other than 1).
3. Converse: TrueAny common factor of n and n + 3 divides their difference (n + 3) − n = 3. The only positive factors of 3 are 1 and 3, and 3 does not divide n. So the only common factor is 1.

Counterexamples: n = 4 gives 65 = 5 × 13; n = 10 gives 121 = 11 × 11; n = 4 gives 259 = 7 × 37.

One counterexample is enough to show that each claim is false.

(i) n = 11 gives 2047 = 23 × 89. (ii) n = 6 gives 65 = 5 × 13.

1. (i)Take n = 11 (a prime): 2¹¹ − 1 = 2048 − 1 = 2047 = 23 × 89, which is not prime.
2. (ii)Take n = 6 (an even number): 2⁶ + 1 = 64 + 1 = 65 = 5 × 13, which is not prime.

If n = 8k, then n = 2 × 4k and n = 4 × 2k, so n is divisible by both 2 and 4.

1. Let the number be n = 8kwhere k is an integer.
2. Rewriten = 2 × 4k   and   n = 4 × 2k

So the statement is true.

No. A number divisible by both 2 and 4 need not be divisible by 8.

1. A counterexample12 is divisible by 2 and by 4, but 12 is not divisible by 8. So the converse is false.
2. WhyA number divisible by both 2 and 4 is divisible by their LCM, which is 4, not 8.
3. The right shortcutCheck whether the number formed by the last three digits is divisible by 8. For a number with fewer than three digits, check the whole number.

Both “if-then” statements are true, and each is the converse of the other.

1. Statement 1If a number is divisible by 3, then the sum of its digits is a multiple of 3.
2. Statement 2 (the converse)If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.
3. ExampleFor 123: the digit sum is 1 + 2 + 3 = 6, a multiple of 3, and 123 = 3 × 41 is divisible by 3.

So a number is divisible by 3 if and only if the sum of its digits is a multiple of 3.

(i) The sticks must be equal, and the way they are put together may matter. (ii) The sticks need not be equal, but if they are equal, how they are put together does not matter.

1. (i)(a) Should the sticks be equal?Yes. Every quadrilateral of type Q has equal diagonals, so the two sticks must be of equal length. Unequal sticks can never give a quadrilateral of type Q.
2. (i)(b) Does it matter how they are put together?Yes, it may. The property says Q has equal diagonals, but not that every quadrilateral with equal diagonals is of type Q. For example, if Q means rectangles, then both diagonals are equal, but the sticks must also be crossed at their midpoints to get a rectangle.
3. (ii)(a) Should the sticks be equal?Not necessarily. The property says nothing about what the diagonals of a type-Q quadrilateral are like, so Q may contain quadrilaterals with unequal diagonals. To be sure of getting type Q, use equal sticks.
4. (ii)(b) Does it matter how they are put together?No. If the sticks are equal, then any way of crossing them gives a quadrilateral with equal diagonals, and by the property it is of type Q.