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

Truth Value: Both the proposition and its converse are True (These are the fundamental parallel line postulates/theorems in Euclidean geometry).

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

Truth Value: The proposition is True (all interior angles of a square are equal to $90^\circ$). The converse is False.

Counterexample: A non-square rectangle has all four angles equal ($90^\circ$ each), but it is not a square because its adjacent sides are not equal.

Converse: If $IE = IF$, then $AB = AC$.

Truth Value: Both the proposition and its converse are True.

Justification (Proposition): If $AB = AC$, then $\angle B = \angle C$. Since $BE$ and $CF$ are angle bisectors, $\angle IBC = \frac{1}{2}\angle B = \frac{1}{2}\angle C = \angle ICB$, making $\Delta IBC$ isosceles with $IB = IC$.
In $\Delta IBF$ and $\Delta ICE$:

• $\angle IBF = \frac{1}{2}\angle B = \frac{1}{2}\angle C = \angle ICE$
• $IB = IC$
• $\angle FIB = \angle EIC$ (vertically opposite angles)

Converse: If $a + x = a + y$, then $x = y$.

Truth Value: Both the proposition and its converse are True.

Justification: The proposition relies on the fundamental additive property of equality (adding the same number to equal quantities results in equal quantities). The converse is justified by the subtraction property of equality (subtracting $a$ from both sides: $(a + x) - a = (a + y) - a \implies x = y$).

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

Truth Value: The proposition is True. The converse is False.

Counterexample: Let $a = 2$ and $b = 8$. Their product is $ab = 2 \times 8 = 16 = 4^2$, which is a perfect square. However, neither $2$ nor $8$ are perfect squares.

Converse: If $x^2 = y^2$, then $x = y$.

Truth Value: The proposition is True. The converse is False.

Counterexample: Let $x = 3$ and $y = -3$. Here, $x^2 = 3^2 = 9$ and $y^2 = (-3)^2 = 9$, so $x^2 = y^2$. However, $x \neq y$ because $3 \neq -3$.

Converse: If $x^3 = y^3$, then $x = y$.

Truth Value: Both the proposition and its converse are True.

Justification: The cubing function $f(t) = t^3$ is strictly increasing across all real numbers. Hence, each real number has a unique real cube root, so $x^3 = y^3 \implies x = y$.

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

Truth Value: The proposition is True. The converse is False.

Counterexample: Let $n = 12$. The number 12 is divisible by both 4 ($12 = 4 \times 3$) and 6 ($12 = 6 \times 2$), but 12 is not divisible by 24.
Explanation: For divisibility by both 4 and 6, $n$ only needs to be a multiple of their least common multiple, $\text{lcm}(4, 6) = 12$, not necessarily $4 \times 6 = 24$.

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

Truth Value: Both the proposition and its converse are True.

Justification for Converse: The numbers 5 and 12 are coprime, meaning they share no common factors other than 1 ($\gcd(5, 12) = 1$). Therefore, any number divisible by both 5 and 12 must be divisible by their lowest common multiple: $$\text{lcm}(5, 12) = 5 \times 12 = 60$$

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

Truth Value: Both the proposition and its converse are True.

Justification: If $n = p^2$ where $p$ is prime, its only divisors are $1$, $p$, and $p^2$ (exactly 3 distinct positive factors).
Conversely, if an integer has an odd number of factors, it must be a perfect square. If it has exactly 3 factors, those factors must be $1$, $\sqrt{n}$, and $n$. For there to be no other factors, $\sqrt{n}$ cannot have any divisors other than 1 and itself, which forces $\sqrt{n}$ to be a prime number $p$. Thus, $n = p^2$.

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

Truth Value: The proposition is True. The converse is False.

Counterexample: Let $n = 8$. The divisors of 8 are $1, 2, 4,$ and $8$ (exactly 4 divisors). However, 8 is not the product of two unequal prime numbers; it is the cube of a single prime ($8 = 2^3$). (In general, any number of the form $p^3$ has $(3+1) = 4$ divisors).

Converse: If $n$ is not a multiple of 3, then $n$ and $n+3$ have no factors in common (other than 1).

Truth Value: Both the proposition and its converse are True.

Justification: Any common factor $d$ of $n$ and $n+3$ must also divide their difference: $$(n + 3) - n = 3$$ The only positive divisors of 3 are 1 and 3. Therefore, if $n$ is not a multiple of 3, then 3 cannot be a factor of $n$, meaning $n$ and $n+3$ share no factors greater than 1.

• (i) Let $n = 4$. Then: $$4(4^2) + 1 = 4(16) + 1 = 64 + 1 = 65$$ 65 is composite because $65 = 5 \times 13$.
• (ii) Let $n = 11$. Then: $$11^2 + 11 + 11 = 121 + 22 = 143$$ 143 is composite because $143 = 11 \times 13$.
  (Alternatively, for $n = 10$, $10^2 + 10 + 11 = 121 = 11^2$).
• (iii) Let $n = 4$. Then: $$4^4 + 3 = 256 + 3 = 259$$ 259 is composite because $259 = 7 \times 37$.

• (i) Let $n = 11$ (11 is a prime number). Then: $$2^{11} - 1 = 2048 - 1 = 2047$$ 2047 is composite because $2047 = 23 \times 89$.
• (ii) Let $n = 6$ (6 is an even number). Then: $$2^6 + 1 = 64 + 1 = 65$$ 65 is composite because $65 = 5 \times 13$.
  (Another example: for $n = 10$, $2^{10} + 1 = 1025 = 5^2 \times 41$).

• (i) Justification: Since 8 is a multiple of both 2 and 4 ($8 = 2 \times 4$), any number that is divisible by 8 has 8 as a factor. Therefore, it inherently contains both 2 and 4 as factors, making it divisible by both 2 and 4.
• (ii) Shortcut Check: No, checking whether a number is divisible by 2 and 4 is not enough to determine if it is divisible by 8.
  
  Reason: The converse statement ("If a number is divisible by both 2 and 4, then it is divisible by 8") is False.
  Counterexample: Consider 12. 12 is divisible by 2 ($12 = 2 \times 6$) and divisible by 4 ($12 = 4 \times 3$), but 12 is not divisible by 8. (Any number divisible by both 2 and 4 only needs to be a multiple of $\text{lcm}(2, 4) = 4$).

The relationship consists of two true 'If-then' statements (a proposition and its converse):

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

2. Converse: "If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3."

Conclusion: Because both the conditional proposition and its converse are true, the statement is an "if and only if" (biconditional) equivalence.

• (i) Property: "If a quadrilateral is of type $Q$, then it has equal-length diagonals."
  • (a) Should the two sticks be of equal length?
    Yes. If the final quadrilateral is of type $Q$, the property guarantees that its diagonals must be equal. Therefore, using sticks of unequal length would make it impossible to obtain a quadrilateral of type $Q$.
  • (b) Will it matter how the two sticks are put together?
    Yes. Having equal-length diagonals alone is a necessary condition, but not always a sufficient one (the converse is not necessarily true). For instance, if type $Q$ represents a rectangle, the sticks must not only be equal in length but must also bisect each other.
• (ii) Property: "If a quadrilateral has equal diagonals, then it is of type $Q$."
  • (a) Should the two sticks be of equal length?
    Yes. In order to satisfy the 'if' condition (equal diagonals) which guarantees obtaining type $Q$, the two sticks must be of equal length.
  • (b) Will it matter how the two sticks are put together?
    No, it will not matter. The statement guarantees that any quadrilateral having equal-length diagonals is automatically of type $Q$, regardless of the angle or position at which the two sticks intersect.