Chapter 9: Propositions and their Converses (Propositions)

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

• Proposition: true. This is the corresponding angles property of parallel lines (studied in earlier grades).
• Converse: true. If the lines were not parallel, they would meet on one side of the transversal and form a triangle; then one of the "corresponding" angles would be an exterior angle of that triangle and would be greater than the other, contradicting equality. So equal corresponding angles force the lines to be parallel.

Both the proposition and its converse are true.

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

• Proposition: true. All angles of a square are 90°.
• Converse: false. Counterexample: a rectangle with sides 4 cm and 2 cm has all four angles equal to 90°, but it is not a square. (Equal angles force each angle to be 360° ÷ 4 = 90°, so the quadrilateral is a rectangle, but its sides need not be equal.)

Proposition true; converse false (a non-square rectangle).

Here E is on AC (on the bisector from B) and F is on AB (on the bisector from C).

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

Proposition: true. If AB = AC then ∠B = ∠C, so their halves are equal: $\angle IBC=\angle ICB$, and therefore IB = IC. Now compare △IBF and △ICE:

• $\angle IBF=\frac{1}{2}\angle B=\frac{1}{2}\angle C=\angle ICE$
• $IB=IC$
• $\angle BIF=\angle CIE$ (vertically opposite angles, since BIE and CIF are straight lines)

So $△IBF\cong △ICE$ (ASA), and hence IF = IE.

Converse: false. Write $\angle B=2\beta$, $\angle C=2\gamma$ and let r be the inradius. In △ICE, $\angle ICE=\gamma$ and $\angle CIE=\beta +\gamma$ (the exterior angle of △IBC at I), so $\angle IEC=180°-(\beta +2\gamma )$. Dropping the perpendicular from I to AC (length r) gives $IE=\frac{r}{\sin (\beta +2\gamma )}$; in the same way $IF=\frac{r}{\sin (2\beta +\gamma )}$. These are equal when $\beta =\gamma$, but also when $(\beta +2\gamma )+(2\beta +\gamma )=180°$, i.e. $\beta +\gamma =60°$, i.e. $\angle A=60°$.

Counterexample: take ∠A = 60°, ∠B = 80°, ∠C = 40°. Then IE = IF (measuring, or by the formula: $\sin 100°=\sin 80°$), but AB ≠ AC, because ∠B ≠ ∠C.

Proposition true (ASA congruence of △IBF and △ICE). Converse false: any triangle with ∠A = 60° and ∠B ≠ ∠C (e.g. 60°, 80°, 40°) has IE = IF but AB ≠ AC.

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

• Proposition: true. Adding the same number a to two equal numbers gives equal results.
• Converse: true. Add −a to both sides (i.e. subtract a): $(a+x)-a=(a+y)-a$, so x = y.

Both are true.

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

• Proposition: true. If $a=m^2$ and $b=n^2$, then $ab=m^2{n}^2=(mn{)}^2$.
• Converse: false. Counterexample: a = 2, b = 8. Then ab = 16 = 4², but neither 2 nor 8 is a perfect square.

Proposition true; converse false (2 × 8 = 16).

Converse: If x² = y², then x = y.

• Proposition: true. Equal numbers have equal squares.
• Converse: false. Counterexample: x = 2, y = −2. Then x² = y² = 4, but x ≠ y. (In fact x² = y² only gives x = y or x = −y.)

Proposition true; converse false (2² = (−2)²).

Converse: If x³ = y³, then x = y.

• Proposition: true. Equal numbers have equal cubes.
• Converse: true. $x^3-y^3=(x-y)(x^2+xy+y^2)$, and $x^2+xy+y^2={\left(x+\frac{y}{2}\right)}^2+\frac{3y^2}{4}$, which is positive unless x = y = 0. So if $x^3=y^3$, the product is 0 and therefore x − y = 0. (Unlike squares, cubes keep the sign: there is no "−2" problem.)

Both are true.

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

• Proposition: true. If n = 24k, then n = 4(6k) = 6(4k).
• Converse: false. Counterexample: n = 12 is divisible by 4 and by 6, but not by 24. (Being divisible by 4 and 6 only means being divisible by their LCM, 12.)

Proposition true; converse false (n = 12).

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

• Proposition: true. If n = 60k, then n = 5(12k) = 12(5k).
• Converse: true. If 12 divides n, then n = 12k. Since 5 divides 12k and 5 has no common factor with 12, 5 must divide k, say k = 5m. Then n = 60m. (The LCM of 5 and 12 is 60.)

Both are true (5 and 12 have no common factor, so LCM = 60).

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

• Proposition: true. If $n=p^2$ (p prime), its factors are exactly 1, p and $p^2$.
• Converse: true. Factors come in partner pairs (d and n/d). With an odd number of factors, one factor must be its own partner, so n is a perfect square, $n=d^2$, and the three factors are 1, d, $d^2$. If d were not prime, say d = ab with 1 < a < d, then a would be a fourth factor of n. So d is prime.

Both are true.

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

• Proposition: true. If n = pq with p ≠ q primes, its divisors are exactly 1, p, q and pq.
• Converse: false. Counterexample: n = 8 = 2³ has exactly 4 divisors (1, 2, 4, 8), but it is not a product of two unequal primes. (Every cube of a prime, $p^3$, is a counterexample.)

Proposition true; converse false (n = 8).

("No factors in common" means no common factor other than 1.)

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

• Proposition: true. If n were a multiple of 3, then n + 3 would also be a multiple of 3, and 3 would be a common factor.
• Converse: true. Any common factor d of n and n + 3 divides their difference, 3. So d = 1 or d = 3. If n is not a multiple of 3, d cannot be 3, so d = 1.

Both are true: n and n + 3 share a factor other than 1 exactly when 3 divides n.

(i) n = 1, 2, 3 give 5, 17, 37 (all prime), but n = 4 gives $4\times 16+1=65=5\times 13$.

(ii) n = 0 to 9 give 11, 13, 17, 23, 31, 41, 53, 67, 83, 101 (all prime), but n = 10 gives $100+10+11=121=11\times 11$.

(iii) n = 1, 2, 3 give 7, 19, 67 (all prime), but n = 4 gives $256+3=259=7\times 37$. (Also n = 0 gives 4.)

(i) n = 4: 65 = 5 × 13 (ii) n = 10: 121 = 11² (iii) n = 4: 259 = 7 × 37

(i) For n = 2, 3, 5, 7 we get 3, 7, 31, 127 (all prime). But n = 11 is prime and $2^{11}-1=2047=23\times 89$.

(ii) n = 2 gives 5 and n = 4 gives 17 (prime). But n = 6 is even and $2^6+1=65=5\times 13$.

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

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

(ii) No. That would be using the converse, which is false. Divisible by 2 and 4 means divisible by 4 (any multiple of 4 is already even), and not every multiple of 4 is a multiple of 8. Counterexample: 12 is divisible by 2 and 4 but not by 8. The correct shortcut for 8 is: check whether the number formed by the last three digits is divisible by 8.

(i) 8k = 2(4k) = 4(2k). (ii) No; for example 12 passes the test for 2 and 4 but is not divisible by 8. Use the last-three-digits test instead.

Both directions are true, so we need two sentences:

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

Together: a number is divisible by 3 if and only if the sum of its digits is a multiple of 3. (Reason: a number differs from its digit sum by a multiple of 9, e.g. $372-(3+7+2)=3\times 99+7\times 9$, so both are multiples of 3 or neither is.)

"If a number is divisible by 3, then its digit sum is a multiple of 3" and its converse "If the digit sum is a multiple of 3, then the number is divisible by 3" are both true.

(i) "Q ⇒ equal diagonals".

• (a) Yes. Every quadrilateral of type Q has equal diagonals, so with unequal sticks we can never get a Q.
• (b) It may matter. Equal diagonals are necessary, but this property does not say they are enough. For example, if Q were "rectangles", the sticks would also have to bisect each other; if Q were "squares", they would have to bisect each other at right angles. So we cannot just place equal sticks anyhow.

(ii) "Equal diagonals ⇒ Q".

• (a) Equal sticks are enough: any quadrilateral with equal diagonals is a Q, so using equal sticks guarantees success. (The property does not forbid other Q's with unequal diagonals, but equal sticks are the safe choice.)
• (b) It does not matter how the equal sticks are put together (as long as their ends form a quadrilateral): every such quadrilateral has equal diagonals and is therefore of type Q.

This question shows the difference between a proposition and its converse: in (i), equal diagonals are a necessary condition; in (ii), they are a sufficient condition.

(i) (a) Yes, they must be equal (b) Yes, the arrangement may matter. (ii) (a) Equal sticks guarantee a Q (b) The arrangement does not matter.