Solutions to Exercise Set 9.1 (Mathematical Reasoning / Proofs)

Instructions (Questions 1 to 12)

Frame the converse for each of the propositions in Questions 1 to 12. Then, determine if each of the two statements (Original Proposition and Converse) is true or not. Justify the true statements and give a counterexample for each false statement.

Question 1

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

Truth Value: True
Justification: By the Corresponding Angles Postulate, when two parallel lines are cut by a transversal, the corresponding angles are equal.

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

Truth Value: True
Justification: By the Converse of the Corresponding Angles Postulate, if corresponding angles are equal, the lines must be parallel.

Question 2

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

Truth Value: True
Justification: By definition, a square is a regular quadrilateral with four equal interior angles (each measuring 90 degrees).

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

Truth Value: False
Counterexample: A rectangle that is not a square (for example, sides of length 3 cm and 5 cm). All four angles are equal to 90 degrees, but it is not a square because its adjacent sides are not equal.

Question 3

Given: Any Triangle ABC, let us bisect the angles at B and C. The bisectors meet at the incentre I of the triangle. Extend the bisectors beyond I till they meet the opposite sides at E and F respectively.

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

Truth Value: True
Justification:
1. 1. In Triangle ABC, since AB = AC, Triangle ABC is isosceles, so Angle B = Angle C.
2. 2. BI and CI bisect Angle B and Angle C, so Angle IBC = half of Angle B = half of Angle C = Angle ICB. Hence, IB = IC.
3. 3. Also, Angle ABI = Angle ACI = half of Angle B.
4. 4. In Triangle FBI and Triangle ECI:
  - – Angle FBI = Angle ECI
  - – IB = IC
  - – Angle FIB = Angle EIC (Vertically opposite angles)
5. 5. By ASA Congruence Criterion, Triangle FBI is congruent to Triangle ECI.
6. 6. Therefore, by CPCTC, IF = IE (or IE = IF).

Converse: In Triangle ABC, if IE = IF, then AB = AC.

Truth Value: True
Justification:
1. 1. Let Angle B = 2 * beta and Angle C = 2 * gamma. Then Angle IBC = beta and Angle ICB = gamma.
2. 2. In Triangle IBC, Angle BIC = 180 degrees – (beta + gamma). Thus, vertically opposite angles Angle FIE = Angle BIC = 180 degrees – (beta + gamma), and Angle FIB = Angle EIC = beta + gamma.
3. 3. Using the sine rule in Triangle FBI and Triangle ECI:
  - – IF / sin(beta) = IB / sin(Angle BFI)
  - – IE / sin(gamma) = IC / sin(Angle CEI)
4. 4. Given IE = IF and using the symmetry of angle bisector segments from the incentre to opposite sides, IE = IF holds if and only if beta = gamma (i.e., Angle B = Angle C), which implies AB = AC.

Question 4

Original Statement: If x = y, then a + x = a + y, where x, y, a are any three numbers.

Truth Value: True
Justification: Addition property of equality. Adding the same number a to equal quantities preserves equality.

Converse: If a + x = a + y, then x = y, where x, y, a are any three numbers.

Truth Value: True
Justification: Cancellation property of addition (subtracting a from both sides yields x = y).

Question 5

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

Truth Value: True
Justification: Let a = m squared and b = n squared for integers m, n. Then ab = m squared * n squared = (m * n) squared, which is an integer squared, hence a perfect square.

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

Truth Value: False
Counterexample: Let a = 2 and b = 8. Neither 2 nor 8 is a perfect square, but ab = 2 * 8 = 16 = 4 squared, which is a perfect square.

Question 6 (where x, y belong to Real Numbers)

Original Statement: If x = y, then x squared = y squared.

Truth Value: True
Justification: Squaring both sides of an equality yields equal results (x * x = y * y).

Converse: If x squared = y squared, then x = y.

Truth Value: False
Counterexample: Let x = 2 and y = -2. x squared = 4 and y squared = 4, so x squared = y squared, but 2 is not equal to -2.

Question 7 (where x, y belong to Real Numbers)

Original Statement: If x = y, then x cubed = y cubed.

Truth Value: True
Justification: Cubing both sides of an equality yields equal results.

Converse: If x cubed = y cubed, then x = y.

Truth Value: True
Justification: The function f(t) = t cubed is strictly increasing on Real Numbers, so it is one-to-one. Hence x cubed = y cubed implies x = y.

Question 8 (where n is a positive integer)

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

Truth Value: True
Justification: Since 24 = 4 * 6, any multiple of 24 is also a multiple of both 4 and 6.

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

Truth Value: False
Counterexample: n = 12. 12 is divisible by 4 (12 = 4 * 3) and by 6 (12 = 6 * 2), but 12 is not divisible by 24.

Question 9 (where n is a positive integer)

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

Truth Value: True
Justification: 60 = 5 * 12. Any number divisible by 60 must be divisible by its factors 5 and 12.

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

Truth Value: True
Justification: Greates Common Divisor gcd(5, 12) = 1 (they are coprime). Therefore, if a number is divisible by both 5 and 12, it must be divisible by their product 5 * 12 = 60.

Question 10 (where n is a positive integer)

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

Truth Value: True
Justification: Let n = p squared, where p is prime. The factors (positive divisors) of p squared are 1, p, and p squared, which are exactly 3 factors.

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

Truth Value: True
Justification: If a number n has prime factorization n = p1^a1 * p2^a2 …, its number of factors is (a1 + 1)(a2 + 1)… = 3. Since 3 is prime, we must have a1 + 1 = 3 implies a1 = 2, so n = p squared for some prime p.

Question 11 (where n is a positive integer)

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

Truth Value: True
Justification: Let n = p * q where p, q are distinct primes. The divisors of n are 1, p, q, and pq, making a total of 4 divisors.

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

Truth Value: False
Counterexample: n = 8 = 2 cubed. The divisors of 8 are 1, 2, 4, 8 (exactly 4 divisors), but 8 is not the product of two unequal prime numbers; it is the cube of a prime.

Question 12 (where n is a positive integer)

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

Truth Value: True
Justification: Any common factor d of n and n + 3 must also divide (n + 3) – n = 3. If n were a multiple of 3, then 3 would be a common factor of n and n + 3. Since they have no common factors (other than 1), n cannot be a multiple of 3.

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

Truth Value: True
Justification: The only possible common factor greater than 1 for n and n + 3 is 3. If n is not a multiple of 3, then 3 does not divide n, so n and n + 3 share no factors in common.

Prime Forms and Counterexamples

Question 13

Task: Find counterexamples to the following claims that all numbers of a given form are prime.

(i) All numbers of the form 4 * n squared + 1 are prime.

Counterexample: Let n = 4.
4 * (4 squared) + 1 = 4 * 16 + 1 = 65.
Since 65 = 5 * 13, it is composite. Thus, n = 4 is a counterexample.

(ii) All numbers of the form n squared + n + 11 are prime.

Counterexample: Let n = 11.
11 squared + 11 + 11 = 11 * (11 + 1 + 1) = 11 * 13 = 143.
Since 143 is divisible by 11 and 13, it is composite. Thus, n = 11 is a counterexample.

(iii) All numbers of the form 4^n + 3 are prime.

Counterexample: Let n = 4.
4 to the power of 4 + 3 = 256 + 3 = 259.
Since 259 = 7 * 37, it is composite. Thus, n = 4 is a counterexample.

Question 14

Task: Find counterexamples to the following statements.

(i) If n is a prime number, then 2^n – 1 is a prime number.

Counterexample: Let n = 11 (which is a prime number).
2 to the power of 11 – 1 = 2048 – 1 = 2047.
Since 2047 = 23 * 89, it is composite. Thus, n = 11 is a counterexample.

(ii) If n is an even number, then 2^n + 1 is a prime number.

Counterexample: Let n = 6 (which is an even number).
2 to the power of 6 + 1 = 64 + 1 = 65.
Since 65 = 5 * 13, it is composite. Thus, n = 6 is a counterexample.

Divisibility and Logic

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.

Justification: 8 = 2 * 4. If a number n is divisible by 8, then n = 8k = 2(4k) = 4(2k) for some integer k. Therefore, n is divisible by both 2 and 4.

(ii) Recall the divisibility shortcuts. 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: No, it is not enough.
Reason: For a divisibility rule combining two factors a and b to imply divisibility by a * b, a and b must be coprime (i.e., gcd(a, b) = 1). Here, gcd(2, 4) = 2, which is not 1.
Counterexample: n = 4. It is divisible by 2 and by 4, but it is not divisible by 8.

Question 16

Task: 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.

Statement (If-then form): If a number is divisible by 3, then the sum of its digits is a multiple of 3.
Converse (If-then form): If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.
Combined (If and only if form): A number is divisible by 3 if and only if the sum of its digits is a multiple of 3.

Question 17

Suppose we have identified a category of quadrilaterals called Q, and we construct a quadrilateral of this type using two thin sticks put together as diagonals so that the quadrilateral obtained by joining their endpoints is of type Q.

(i) Suppose Q satisfies the 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?

Answer: Yes, the two sticks MUST be of equal length.
Reason: Since type Q quadrilaterals have equal-length diagonals, and the sticks form the diagonals of the constructed quadrilateral, the sticks must be equal in length to satisfy this necessary condition.

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

Answer: No, it does not matter how they are put together (as long as they cross each other).
Reason: The proposition only states a necessary condition (having equal diagonals), not a sufficient one. Any quadrilateral formed by joining the endpoints of two equal-length sticks will automatically have equal diagonals, regardless of their intersection point or angle.

(ii) Instead of the property mentioned above, suppose Q satisfies the property: “If a quadrilateral has equal diagonals, then it is of type Q.”

What will be your answers to (a) and (b) now?