Propositions and their Converses

1. Given: A triangle $\triangle ABC$ with sidelengths $BC = a$, $AC = b$, $AB = c$ such that: $$a^2 + b^2 = c^2 \quad \text{--- (1)}$$
2. To Prove: $\triangle ABC$ is a right-angled triangle with $\angle C = 90^\circ$.
3. Construction: Construct another triangle $\triangle XYZ$ such that:
   • $\angle Z = 90^\circ$ (a right angle)
   • Side $YZ = a$ (equal to $BC$)
   • Side $XZ = b$ (equal to $AC$)
4. Applying Pythagoras on $\triangle XYZ$: Since $\triangle XYZ$ is right-angled at $Z$, by the Baudhāyana–Pythagoras Theorem: $$XY^2 = YZ^2 + XZ^2 = a^2 + b^2$$ Using equation (1), $a^2 + b^2 = c^2$, so: $$XY^2 = c^2 \implies XY = c = AB$$
5. Congruence Criterion: Compare $\triangle ABC$ and $\triangle XYZ$:
   • $BC = YZ = a$ (by construction)
   • $AC = XZ = b$ (by construction)
   • $AB = XY = c$ (just proved)
   Therefore, by the SSS Congruence Criterion: $$\triangle ABC \cong \triangle XYZ$$
6. Conclusion: By Corresponding Parts of Congruent Triangles (CPCT): $$\angle C = \angle Z$$ Since $\angle Z = 90^\circ$, it follows that: $$\angle C = 90^\circ$$ Thus, $\triangle ABC$ is indeed a right-angled triangle! Q.E.D.

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

• Truth of Proposition: TRUE. This is the fundamental Corresponding Angles Axiom of Euclidean geometry.
• Converse Statement: "If a transversal intersects two lines such that the corresponding angles are equal, then the two lines are parallel."
• Truth of Converse: TRUE. This is the Converse of the Corresponding Angles Axiom, universally used to prove lines are parallel.
• Biconditional status: Both statements are true ($X \iff Y$).

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

• Truth of Proposition: TRUE. Every square has four interior angles each measuring exactly $90^\circ$.
• Converse Statement: "If all the angles of a quadrilateral are equal, then it is a square."
• Truth of Converse: FALSE!
• Counterexample: A rectangle with length $6\text{ cm}$ and breadth $4\text{ cm}$ has all four interior angles equal to $90^\circ$ ($\frac{360^\circ}{4} = 90^\circ$), but its adjacent sides are unequal ($6 \neq 4$), so it is definitely NOT a square!

Given: In $\triangle ABC$, bisectors of $\angle B$ and $\angle C$ meet at incenter $I$, and are extended to meet opposite sides $AC$ and $AB$ at $E$ and $F$ respectively.

Proposition: "If $AB = AC$, then $IE = IF$."

• Truth of Proposition: TRUE.
  Proof: If $AB = AC$, then $\angle B = \angle C$ (angles opp. equal sides).
  Their half angles are equal: $\angle IBC = \angle ICB = \frac{1}{2}\angle B \implies IB = IC$.
  Now in $\triangle IBF$ and $\triangle ICE$:
  • $\angle FIB = \angle EIC$ (vertically opposite angles)
  • $IB = IC$
  • $\angle IBF = \angle ICE = \frac{1}{2}\angle B$
  By ASA congruence criterion, $\triangle IBF \cong \triangle ICE \implies IE = IF$ (by CPCT).
• Converse Statement: "In $\triangle ABC$, if the angle bisectors of $B$ and $C$ meet the opposite sides at $E$ and $F$ through incenter $I$ such that $IE = IF$, then $AB = AC$."
• Truth of Converse: TRUE. This is directly related to the celebrated Steiner-Lehmus Theorem. If $IE = IF$, the symmetry forces $\angle B = \angle C$, establishing that $\triangle ABC$ must be isosceles ($AB = AC$).

Original Proposition: "If $x = y$, then $a + x = a + y$, where $x, y$ and $a$ are any three real numbers."

• Truth of Proposition: TRUE. Adding the same quantity $a$ to equal quantities preserves equality (Euclid's Axiom 2: If equals are added to equals, the wholes are equal).
• Converse Statement: "If $a + x = a + y$, then $x = y$, where $x, y$ and $a$ are any three real numbers."
• Truth of Converse: TRUE. Subtracting $a$ from both sides yields $(a + x) - a = (a + y) - a \implies x = y$ (Additive cancellation property of real numbers).
• Application: This biconditional relation is routinely used in solving algebraic equations.

Original Proposition: "If $a$ and $b$ are perfect squares, then $ab$ is a perfect square."

• Truth of Proposition: TRUE.
  Justification: Since $a$ and $b$ are squares, $a = m^2$ and $b = n^2$ for integers $m, n$. Their product is $ab = m^2 n^2 = (mn)^2$, which is the square of the integer $mn$.
• Converse Statement: "If the product $ab$ is a perfect square, then both $a$ and $b$ are perfect squares."
• Truth of Converse: FALSE!
• Counterexample: Let $a = 2$ and $b = 8$.
  Product $ab = 2 \times 8 = 16 = 4^2$ (a perfect square!).
  However, neither $a = 2$ nor $b = 8$ is a perfect square! (Another easy counterexample: $a = 3, b = 12 \implies ab = 36$).

Original Proposition: "If $x = y$, then $x^2 = y^2$ (for real numbers $x, y$)."

• Truth of Proposition: TRUE. Multiplying equal numbers by equal numbers produces equal results ($x \times x = y \times y$).
• Converse Statement: "If $x^2 = y^2$, then $x = y$ (for real numbers $x, y$)."
• Truth of Converse: FALSE!
• Counterexample: Let $x = 5$ and $y = -5$.
  Here $x^2 = 5^2 = 25$ and $y^2 = (-5)^2 = 25$, so $x^2 = y^2$.
  However, $5 \neq -5$. In general, $x^2 = y^2 \implies x = \pm y$, which does not guarantee $x = y$.

Original Proposition: "If $x = y$, then $x^3 = y^3$ (for real numbers $x, y$)."

• Truth of Proposition: TRUE. Cubing identical real numbers yields identical values.
• Converse Statement: "If $x^3 = y^3$, then $x = y$ (for real numbers $x, y$)."
• Truth of Converse: TRUE!
• Justification: $$x^3 - y^3 = 0 \implies (x - y)(x^2 + xy + y^2) = 0$$ The quadratic factor can be rewritten by completing the square: $$x^2 + xy + y^2 = \left(x + \frac{y}{2}\right)^2 + \frac{3y^2}{4}$$ This expression is strictly positive for any non-zero real numbers ($>0$). The only way the product can equal zero is if $x - y = 0 \implies x = y$. Unlike squares, odd powers preserve negative signs ($(-2)^3 = -8 \neq 2^3 = 8$), making the cubing operation one-to-one (injective) over the real numbers!

Original Proposition: "If a positive integer $n$ is divisible by 24, then it is divisible by both 4 and 6."

• Truth of Proposition: TRUE. Since $24 = 4 \times 6$, any integer $n = 24k = 4(6k) = 6(4k)$ is automatically a multiple of both 4 and 6.
• Converse Statement: "If a positive integer $n$ is divisible by both 4 and 6, then it is divisible by 24."
• Truth of Converse: FALSE!
• Counterexample: Take $n = 12$ (or $n = 36, 60$).
  $12$ is divisible by $4$ ($12 = 4 \times 3$) and divisible by $6$ ($12 = 6 \times 2$).
  However, $12$ is definitely NOT divisible by 24!
  Why this happens: $4$ and $6$ share a common factor ($\gcd(4, 6) = 2$). The least common multiple is $\text{LCM}(4, 6) = 12$, not $24$.

Original Proposition: "If a positive integer $n$ is divisible by 60, then it is divisible by both 5 and 12."

• Truth of Proposition: TRUE. Since $60 = 5 \times 12$, if $n = 60k$, then $n = 5(12k)$ and $n = 12(5k)$.
• Converse Statement: "If a positive integer $n$ is divisible by both 5 and 12, then it is divisible by 60."
• Truth of Converse: TRUE!
• Justification: Unlike Question 8, here $5$ and $12$ are co-prime integers ($\gcd(5, 12) = 1$).
  When two numbers are co-prime, their least common multiple is simply their product: $$\text{LCM}(5, 12) = 5 \times 12 = 60$$ Therefore, any number divisible by both 5 and 12 must be a multiple of their LCM ($60$), proving the converse true!

Original Proposition: "If a positive integer $n$ is the square of a prime number, then it has exactly 3 factors."

• Truth of Proposition: TRUE.
  Justification: Let $n = p^2$, where $p$ is a prime. Since $p$ has no divisors other than $1$ and $p$, the only divisors of $p^2$ are $1, p,$ and $p^2$. Total factors $= 3$. (Examples: $4 = 2^2$ has $\{1, 2, 4\}$; $9 = 3^2$ has $\{1, 3, 9\}$; $25 = 5^2$ has $\{1, 5, 25\}$).
• Converse Statement: "If a positive integer $n$ has exactly 3 factors, then $n$ is the square of a prime number."
• Truth of Converse: TRUE!
• Justification: Any integer $n > 1$ always has $1$ and $n$ as factors. If it has exactly one additional factor $d$, then $d$ must pair with itself ($d \times d = n \implies n = d^2$). If $d$ were composite, $d$ would possess other sub-factors, causing $n$ to have more than 3 factors. Hence, $d$ must be prime, proving $n$ is the square of a prime!

Original Proposition: "If a positive integer $n$ is a product of two unequal prime numbers, then it has exactly 4 divisors."

• Truth of Proposition: TRUE.
  Justification: Let $n = p \times q$ with distinct primes $p \neq q$. The only divisors are $1, p, q,$ and $pq$. Total count $= 4$. (e.g., $15 = 3 \times 5$ has divisors $\{1, 3, 5, 15\}$).
• Converse Statement: "If a positive integer $n$ has exactly 4 divisors, then $n$ is a product of two unequal prime numbers."
• Truth of Converse: FALSE!
• Counterexample: Take $n = p^3$, the cube of any prime!
  For instance, let $n = 8 = 2^3$ (or $n = 27 = 3^3$).
  The divisors of $8$ are: $\{1, 2, 4, 8\}$ — exactly 4 divisors!
  However, $8$ is NOT the product of two unequal prime numbers! ($8 = 2 \times 2 \times 2$). Hence, the converse is false.

Original Proposition: "If positive integers $n$ and $n + 3$ have no factors in common, then $n$ is not a multiple of 3."

• Truth of Proposition: TRUE.
  Proof: Let $d$ be any common factor of $n$ and $n+3$. Then $d$ must divide their difference: $$(n + 3) - n = 3$$ Since $3$ is prime, its only positive divisors are $1$ and $3$. Thus, the only possible common factor greater than $1$ is $3$. If $n$ were a multiple of $3$, then $n = 3k \implies n + 3 = 3(k+1)$, meaning both would share $3$ as a common factor. Therefore, having NO common factors guarantees that $n$ cannot be a multiple of $3$.
• Converse Statement: "If positive integer $n$ is not a multiple of 3, then $n$ and $n + 3$ have no factors in common."
• Truth of Converse: TRUE!
• Proof of Converse: If $n$ is not a multiple of $3$, then $3 \nmid n$. Since the only possible common factor between $n$ and $n+3$ is $3$ (as shown above), and $3$ does not divide $n$, they cannot share any factor $> 1$. Hence, $\gcd(n, n+3) = 1$. Both statements are true ($X \iff Y$).

Context: There are no known simple algebraic polynomials that generate exclusively prime numbers. Disprove each claim by providing a counterexample:

(i) Claim: "All numbers of the form $4n^2 + 1$ are prime (for positive integer $n$)."

• Testing $n = 1: 4(1)^2 + 1 = 5$ (Prime)
• Testing $n = 2: 4(4) + 1 = 17$ (Prime)
• Testing $n = 3: 4(9) + 1 = 37$ (Prime)
• Counterexample at $n = 4$: $$4(4)^2 + 1 = 4(16) + 1 = 64 + 1 = 65 = 5 \times 13 \quad \text{(Composite!)}$$ (Another counterexample is $n=6: 4(36)+1 = 145 = 5 \times 29$).

(ii) Claim: "All numbers of the form $n^2 + n + 11$ are prime."

• Testing $n = 1: 1 + 1 + 11 = 13$ (Prime)
• Testing $n = 2: 4 + 2 + 11 = 17$ (Prime)
• Counterexample at $n = 10$: $$10^2 + 10 + 11 = 100 + 10 + 11 = 121 = 11 \times 11 = 11^2 \quad \text{(Composite!)}$$ (Also at $n = 11$: $11^2 + 11 + 11 = 11(11 + 1 + 1) = 11 \times 13 = 143$, which is composite).

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

• Testing $n = 1: 4^1 + 3 = 7$ (Prime)
• Testing $n = 2: 4^2 + 3 = 16 + 3 = 19$ (Prime)
• Testing $n = 3: 4^3 + 3 = 64 + 3 = 67$ (Prime)
• Counterexample at $n = 4$: $$4^4 + 3 = 256 + 3 = 259 = 7 \times 37 \quad \text{(Composite!)}$$

(i) Claim: "If $n$ is a prime number, then $2^n - 1$ is a prime number." (Mersenne Conjecture)

• Testing primes: $n=2 \implies 2^2 - 1 = 3$ (Prime); $n=3 \implies 2^3 - 1 = 7$ (Prime); $n=5 \implies 2^5 - 1 = 31$ (Prime); $n=7 \implies 2^7 - 1 = 127$ (Prime).
• Counterexample at prime $n = 11$: $$2^{11} - 1 = 2048 - 1 = 2047 = 23 \times 89$$ Here $n=11$ is prime, but $2^{11} - 1 = 2047$ is a composite number!

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

• Testing evens: $n=2 \implies 2^2 + 1 = 5$ (Prime); $n=4 \implies 2^4 + 1 = 17$ (Prime).
• Counterexample at even $n = 6$: $$2^6 + 1 = 64 + 1 = 65 = 5 \times 13$$ Here $n = 6$ is an even number, but $2^6 + 1 = 65$ is composite! (Also at $n = 8: 2^8 + 1 = 257$ is prime, but at $n=10: 2^{10}+1 = 1025 = 25 \times 41$).

Statement: "If a number is divisible by 8, then it is divisible by both 2 and 4."

• (i) Justification: Let $n$ be divisible by 8 $\implies n = 8k = 2(4k)$ and $n = 4(2k)$. Since 2 and 4 are both factors of 8, every multiple of 8 must be divisible by both 2 and 4. The statement is universally True.
• (ii) Testing the shortcut: "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?"
  NO, IT IS NOT ENOUGH!
  Mathematical Reason: $4$ is already a multiple of $2$. Therefore, any number divisible by $4$ is automatically divisible by $2$. Checking both $2$ and $4$ gives no extra information beyond divisibility by $4$.
  $\text{LCM}(2, 4) = 4$, which is strictly less than $8$.
  Counterexample: The number $12$ (or $20, 28, 36$).
  $12$ is divisible by $2$ ($12 \div 2 = 6$) and divisible by $4$ ($12 \div 4 = 3$).
  Yet $12$ is NOT divisible by 8 ($12 \div 8 = 1.5$).
  Official Divisibility Test for 8: Because $1000 = 8 \times 125$ is divisible by 8, a number is divisible by 8 if and only if its last three digits form a number divisible by 8!

Task: Express the relationship between "a number is divisible by 3" and "sum of the digits is a multiple of 3" using 'If–then' propositions.

• Proposition 1 ($X \implies Y$):
  "If a positive integer is divisible by 3, then the sum of its digits is a multiple of 3." (TRUE)
• Proposition 2 (Converse $Y \implies X$):
  "If the sum of the digits of a positive integer is a multiple of 3, then the number is divisible by 3." (TRUE)
• Unified Biconditional Formulation:
  "A positive integer is divisible by 3 if and only if the sum of its digits is a multiple of 3."
  Algebraic Proof: Any number $N = a_n 10^n + \dots + a_1 10 + a_0$ can be rewritten as: $$N = [a_n(10^n - 1) + \dots + a_1(9)] + (a_n + \dots + a_1 + a_0)$$ Since $10^k - 1 = 99\dots9$ is always a multiple of 3, $N$ and its digit sum leave the exact same remainder upon division by 3.

Setup: You are given two thin sticks to be joined together as diagonals to construct a category of quadrilaterals called $Q$ by connecting their endpoints (Fig. 9.2 in textbook).

Part (i): Category $Q$ satisfies: "If a quadrilateral is of type $Q$, then it has equal-length diagonals." ($Q \implies \text{Equal Diagonals}$)

• (a) Should the two sticks be of equal length? Why or why not?
  YES. Since the proposition guarantees that every quadrilateral of type $Q$ has equal-length diagonals, using sticks of unequal lengths will produce a quadrilateral whose diagonals are unequal, making it impossible to be of type $Q$. Equal sticks are a necessary condition.
• (b) Will it matter how the two sticks are put together?
  YES. Having equal sticks is necessary, but not sufficient to guarantee category $Q$. How they intersect matters completely! For example:
  • If they cross at their midpoints at $90^\circ$, they form a Square.
  • If they cross at their midpoints at an oblique angle, they form a Rectangle.
  • If they do not bisect each other, they might form an Isosceles Trapezium.
  Therefore, the angle and point of intersection are critical.

Part (ii): Category $Q$ satisfies: "If a quadrilateral has equal diagonals, then it is of type $Q$." ($\text{Equal Diagonals} \implies Q$)

• (a) Should the two sticks be of equal length? Why or why not?
  YES. To invoke the condition "has equal diagonals", you must select two sticks of equal length.
• (b) Will it matter how the two sticks are put together?
  NO! Under this new rule, having equal diagonals is sufficient by itself to qualify for category $Q$. No matter how you position the sticks (at any angle or crossing point), as long as their lengths are equal, the resulting quadrilateral automatically satisfies the premise and thus belongs to type $Q$!