Ch 9 Propositions And Their Converses Solutions Class 9 NCERT

P = Two lines are parallel

Q = Corresponding angles are equal

P implies Q  (P ⇒ Q)

In the figure,

L ∥ M,  T is a transversal

∴  ∠1 = ∠5 ;  ∠4 = ∠8

    ∠2 = ∠6 ;  ∠3 = ∠7

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

If corresponding angles are equal, then the lines are parallel.

This is true by the Converse of Corresponding Angles Theorem.

P = A quadrilateral is a square

Q = All its angles are equal

P implies Q  (P ⇒ Q)

By definition, a square has all four angles equal to 90°.

So whenever a quadrilateral is a square, all its angles are equal.

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

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

Counterexample: A rectangle has all four angles equal (each 90°), but its adjacent sides are not equal, so it is not a square.

∴  Converse, Q ⇒ P is false.

In △ABC, the bisectorsⓘ (tap to see meaning) of ∠B and ∠C meet at the incentreⓘ (tap to see meaning)  I.

These bisectors are extended beyond I to meet the opposite sides, the B-bisector meets AC at E, and the C-bisector meets AB at F.

P = AB = AC

Q = IE = IF

P implies Q  (P ⇒ Q)

Since AB = AC, △ABC is isosceles, so it has a line of symmetry through A and the midpoint of BC.

Reflecting the triangle across this axis swaps B ↔ C, and hence swaps side AC ↔ AB.

The incentre I lies on this axis of symmetry, so I maps to itself under the reflection.

The bisector of ∠B maps to the bisector of ∠C, so point E (on AC) maps to point F (on AB).

Since reflection preserves distances, and I → I, E → F,

    IE = IF

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

If IE = IF, then AB = AC.

This is also true ,it is closely related to a famous result in triangle geometry (the Steiner–Lehmus theorem), which shows that equal bisector-related lengths force a triangle to be isosceles.

Unlike the forward direction, this cannot be proved by a simple symmetry argument ,it needs an indirect proof (by contradiction), which is beyond this exercise.

∴  Converse, Q ⇒ P is true.

P = x = y

Q = a + x = a + y

(where x, y and a are any three numbers)

P implies Q  (P ⇒ Q)

Since x = y, adding the same number a to both sides,

    a + x = a + y

This follows from the Addition Property of Equality.

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

If a + x = a + y, subtracting a from both sides,

    x = y

This is true by the Subtraction Property of Equality (Cancellation Law of Addition).

P = a and b are perfect squares

Q = ab is a perfect square

P implies Q  (P ⇒ Q)

Let a = m² and b = n² for some integers m, n.

Then,  ab = m²n² = (mn)²

So ab is also a perfect square.

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

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

Counterexample: Let a = 2 and b = 8. Then ab = 16 = 4², a perfect square ,but neither 2 nor 8 is itself a perfect square.

∴  Converse, Q ⇒ P is false.

P = x = y

Q = x² = y²

(where x, y are any real numbers)

P implies Q  (P ⇒ Q)

Since x = y, squaring both sides,

    x² = y²

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

If x² = y², then x = y.

Counterexample: Let x = 2 and y = −2. Then x² = 4 = y², but x ≠ y.

∴  Converse, Q ⇒ P is false.

P = x = y

Q = x³ = y³

(where x, y are any real numbers)

P implies Q  (P ⇒ Q)

Since x = y, cubing both sides,

    x³ = y³

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

If x³ = y³, then x = y.

Unlike squaring, the cube of a real number is one-to-one ,unequal real numbers always give unequal cubes (a negative number cubed stays negative, so no sign-cancellation like x² is possible).

So x³ = y³ forces x = y.

∴  Converse, Q ⇒ P is true.

P = n is divisible by 24

Q = n is divisible by both 4 and 6

(where n is a positive integer)

P implies Q  (P ⇒ Q)

24 = 4 × 6, and both 4 and 6 divide 24.

So if n is divisible by 24, i.e. n = 24k,

    n = 4(6k)  and  n = 6(4k)

Hence n is divisible by both 4 and 6.

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

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

The LCM(4, 6) = 12, not 24 ,so n need only be a multiple of 12.

Counterexample: Let n = 12. It is divisible by 4 (12÷4 = 3) and by 6 (12÷6 = 2), but 12 is not divisible by 24.

∴  Converse, Q ⇒ P is false.

P = n is divisible by 60

Q = n is divisible by both 5 and 12

(where n is a positive integer)

P implies Q  (P ⇒ Q)

60 = 5 × 12, and both 5 and 12 divide 60.

So if n is divisible by 60, i.e. n = 60k,

    n = 5(12k)  and  n = 12(5k)

Hence n is divisible by both 5 and 12.

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

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

Since HCF(5, 12) = 1 (they share no common factor), their LCM(5, 12) = 5 × 12 = 60.

So any n divisible by both 5 and 12 must be divisible by their LCM, which is 60.

∴  Converse, Q ⇒ P is true.

P = n is the square of a prime number

Q = n has exactly 3 factors

(where n is a positive integer)

P implies Q  (P ⇒ Q)

Let n = p², where p is prime.

The factors of n are 1, p and p² ,exactly 3 factors.

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

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

For n = p^a (a single prime raised to a power), the number of factors is (a + 1).

For exactly 3 factors, a + 1 = 3, so a = 2, giving n = p².

If n had two or more distinct prime factors (e.g. n = pq), it would have at least 4 factors.

So exactly 3 factors forces n to be the square of a prime.

∴  Converse, Q ⇒ P is true.

P = n is a product of two unequal prime numbers

Q = n has exactly 4 divisors

(where n is a positive integer)

P implies Q  (P ⇒ Q)

Let n = p × q, where p and q are unequal primes.

The divisors of n are 1, p, q and pq ,exactly 4 divisors.

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

If n has exactly 4 divisors, then n is a product of two unequal primes.

Counterexample: Let n = 8 = 2³. Its divisors are 1, 2, 4, 8 , exactly 4 divisors, but 8 is not a product of two unequal primes; it is the cube of a single prime.

∴  Converse, Q ⇒ P is false.

P = n and n + 3 have no factors in common

Q = n is not a multiple of 3

(where n is a positive integer)

P implies Q  (P ⇒ Q)

Note that HCF(n, n + 3) = HCF(n, 3), since (n + 3) − n = 3.

If n were a multiple of 3, then HCF(n, 3) = 3, so n and n + 3 would share the common factor 3.

So if n and n + 3 have no factor in common, n cannot be a multiple of 3.

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

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

Since n is not a multiple of 3, HCF(n, 3) = 1.

So HCF(n, n + 3) = HCF(n, 3) = 1 ,i.e. n and n + 3 share no common factor other than 1.

∴  Converse, Q ⇒ P is true.

Claim: All numbers of the form 4n² + 1 are prime.

Counterexample: For n = 4,  4n² + 1 = 4(16) + 1 = 65 = 5 × 13, which is not prime.

Claim: All numbers of the form n² + n + 11 are prime.

This holds for n = 1 to 9 (giving 13, 17, 23, 31, 41, 53, 67, 83, 101 – all prime), but fails at n = 10.

Counterexample: For n = 10,  n² + n + 11 = 100 + 10 + 11 = 121 = 11², which is not prime.

Claim: All numbers of the form 4ⁿ + 3 are prime.

Counterexample: For n = 4,  4⁴ + 3 = 256 + 3 = 259 = 7 × 37, which is not prime.

Claim: If n is a prime number, then 2ⁿ − 1 is a prime number.

Counterexample: For n = 11 (prime),  2¹¹ − 1 = 2047 = 23 × 89, which is not prime.

Claim: If n is an even number, then 2ⁿ + 1 is a prime number.

Counterexample: For n = 6 (even),  2⁶ + 1 = 65 = 5 × 13, which is not prime.

P = n is divisible by 8

Q = n is divisible by both 2 and 4

(where n is a positive integer)

P implies Q  (P ⇒ Q)

8 = 2 × 4, and both 2 and 4 divide 8.

So if n is divisible by 8, i.e. n = 8k,

    n = 2(4k)  and  n = 4(2k)

Hence n is divisible by both 2 and 4.

∴  Statement, P ⇒ Q is true (justified).

This asks whether the converse holds:

Q implies P  (Q ⇒ P)

If n is divisible by both 2 and 4, then n is divisible by 8.

The LCM(2, 4) = 4, not 8 – so being divisible by 2 and 4 only guarantees divisibility by 4.

Counterexample: Let n = 12. It is divisible by 2 and by 4, but 12 is not divisible by 8.

∴  No, checking divisibility by 2 and 4 is not enough to conclude divisibility by 8. There is no shortcut for 8 built from the 2 and 4 tests – one must check divisibility by 8 directly (e.g. by testing if the last three digits form a number divisible by 8).

P = A number is divisible by 3

Q = The sum of the digits of the number is a multiple of 3

P implies Q  (P ⇒ Q)

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

Since 10 = 3 × 3 + 1, every power of 10 leaves remainder 1 when divided by 3.

So a number and the sum of its digits always leave the same remainder on division by 3.

Hence if the number is divisible by 3 (remainder 0), its digit sum is too.

∴  Proposition, P ⇒ Q is true.

Now,

Q implies P  (Q ⇒ P)

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

Using the same fact ,the number and its digit sum leave the same remainder on division by 3 ,

if the digit sum has remainder 0, the number itself must have remainder 0 too.

∴  Converse, Q ⇒ P is true.

Two thin sticks are placed as the diagonals of a quadrilateral. Joining the endpoints of the sticks gives a quadrilateral of category Q.

Type Q  ⇒  Equal-length diagonals

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

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

Equal-length diagonals  ⇒  Type Q

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

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