Chapter 9: Key concepts
A proposition is a statement that can be either true or false. If a statement says “If X, then Y”, its converse is “If Y, then X”. Another way to say “If X, then Y” is “Y when X”. In both forms, X is the condition.
The original is True; the converse is False, because 9 is divisible by 3 but not by 6.
| Possible situation | Simple example |
|---|---|
| Original true; converse false | Divisible by 6 ⇒ divisible by 3; reverse is false. |
| Both true | For a positive integer: perfect square ⇔ odd number of positive factors. |
| Original false; converse true | Equal-area triangles ⇒ congruent (false); congruent triangles ⇒ equal area (true). |
| Both false | Even ⇒ multiple of 3 (false); multiple of 3 ⇒ even (false). |
In-text questions, worked examples & reflections
Covers the questions and discussion prompts on PDF pages 1–5, including Examples 1–5 and the textbook reflections.
Equal angles imply equal opposite sides. If two angles of a triangle are equal, are their opposite sides also equal? Prove it by drawing an altitude from the third vertex.
Yes. In △ABC, suppose ∠B = ∠C. They are equal and their sum is less than 180°, so each is acute. Draw the altitude from A to BC, meeting BC at D between B and C.
- ∠ADB = ∠ADC = 90°.
- ∠BAD = 90° − ∠B and ∠CAD = 90° − ∠C. Since ∠B = ∠C, ∠BAD = ∠CAD.
- In each triangle the equal angles are at A and D. Their included side is the same segment AD.
- Therefore △ABD ≅ △ACD by ASA, so AB = AC.
This converse of the isosceles triangle theorem is true.
Wet road and rain. Proposition: If it rains, the road becomes wet. Converse: If the road is wet, it must have rained. Must both be true?
In the textbook’s ordinary situation, the original proposition is treated as True. The converse is False.
Counterexample: A road can be wet because a water tanker spilled water on it, even when there has been no rain.
Fermat’s claim and Euler’s counterexample. Why does one composite number disprove the claim that every number of the form 22n + 1 is prime?
For n = 5, the expression is:
Since the value has factors other than 1 and itself, it is composite. A single counterexample is enough to reject an “all numbers” claim.
Multiples of 6 and 3. If a number is a multiple of 6, is it a multiple of 3? Is the converse true?
Original true Every multiple of 6 is 6k = 3(2k), so it is divisible by 3.
Converse false A multiple of 3 need not be a multiple of 6. For example, 9 is divisible by 3 but not by 6.
Perfect squares and odd numbers of factors. Let n be a positive integer. P: If n is a perfect square, it has an odd number of positive factors. Q: A positive integer with an odd number of positive factors is a perfect square. Which are true, and which does the book’s partner-factor argument prove?
P true Q true
Factors usually come in pairs: for example, 12 has (1, 12), (2, 6), and (3, 4), so it has six factors.
- If n has an odd number of positive factors, one factor must be paired with itself: n = d × d. Thus n is a perfect square. This is the textbook chain (i) ⇒ (ii) ⇒ (iii), and it proves Q only.
- The same chain does not yet prove P. A repeated pair gives a square, but two different repeated pairs, (f, f) and (g, g), would leave an even number of factors. That case cannot occur: if f and g are positive and f × f = g × g, then f = g. So a square n = d2 has exactly one repeated pair. Every other factor pairs with a different factor, and the number of positive factors is odd. This proves P.
Equal-area versus congruent triangles. P: Triangles with the same area are congruent. Q: Congruent triangles have the same area. Determine both truth values.
P false Take two right triangles with perpendicular legs (3, 4) and (2, 6). Both have area 6 square units, because ½ × 3 × 4 = ½ × 2 × 6 = 6; their side lengths differ, so the triangles are not congruent.
Q true Congruent triangles have corresponding equal side lengths and equal angles; their bases and corresponding heights are equal, so their areas must be equal.
Can a proposition and its converse both be false? Give one example in which the original statement and its converse are both false.
Original: If a number is even, then it is a multiple of 3. False Counterexample: 2.
Converse: If a number is a multiple of 3, then it is even. False Counterexample: 3.
Converse of the Baudhāyana–Pythagoras theorem. If a triangle with sides a, b and c satisfies a² + b² = c², must it be right-angled?
Converse: If a2 + b2 = c2, then the triangle is right-angled, with side c opposite the right angle.
Let △ABC be that triangle, with BC = a, CA = b and AB = c.
- Construct right triangle △XYZ with the right angle at Z, and with perpendicular legs YZ = a and XZ = b.
- By the Pythagoras theorem, the hypotenuse satisfies XY2 = a2 + b2 = c2. Lengths are positive, so XY = c.
- △ABC and △XYZ then have the same three side lengths. By SSS, △ABC ≅ △XYZ, matching BC with YZ, CA with XZ, and AB with XY.
- The angle opposite XY is the right angle at Z. It matches ∠C, which sits between sides a and b and opposite side c. So ∠C = 90°, and the converse is true.
Examples of true propositions with false converses. Give everyday, geometrical and number-theory examples where the proposition is true but the converse fails.
- Daily life: If an object is a bicycle, it has wheels. Converse false: a car also has wheels.
- Geometry: If a quadrilateral is a square, it is a rectangle. Converse false: a non-square rectangle is not a square.
- Number theory: If n is divisible by 12, it is divisible by 3. Converse false: 9 is divisible by 3 but not by 12.
These are sample responses; other correct examples are possible.
Exercise Set 9.1 – Complete solutions
Questions 1–12 require the converse, the truth value of both statements, and a proof or counterexample. In Q8–Q12, n is a positive integer. Questions 13–17 focus on counterexamples, divisibility and logical conditions.
Questions 1–6
Parallel lines and corresponding angles. If two lines are parallel, the corresponding angles formed by a transversal are equal. Write the converse; test both statements.
- Original proposition True – Corresponding-angles theorem.
- Converse True – Converse of that theorem.
Converse: If a transversal cuts two lines and a pair of corresponding angles are equal, then the two lines are parallel.
Original: When a transversal cuts two parallel lines, each pair of corresponding angles is equal. This is the corresponding-angles theorem from earlier grades.
Converse: If a transversal cuts two distinct coplanar lines and one pair of corresponding angles is equal, the lines are parallel. To see why, draw the parallel to the first line through the point where the transversal meets the second line. By the original theorem, this parallel makes the required corresponding angle. On the specified side of the transversal, only one ray makes that angle, so the second line coincides with the parallel.
Squares and equal angles. If a quadrilateral is a square, all its angles are equal. Write the converse; determine if each direction is true.
- Original proposition True – Proved below.
- Converse False – Disproved by a counterexample below.
Converse: If all four angles of a quadrilateral are equal, then it is a square.
Original: Every square has four 90° angles.
Counterexample to converse: A rectangle measuring 2 cm × 5 cm has four equal angles but is not a square, because its adjacent sides are unequal.
Incentre and equal segments IE and IF. In △ABC, the internal angle bisectors of B and C meet at I. They meet AC and AB at E and F. Proposition: If AB = AC, then IE = IF. Frame the converse and test both.
- Original proposition True – Proved below.
- Converse False – Disproved by a counterexample below.
Converse: If IE = IF, then AB = AC.
Original (true): If AB = AC, the triangle is isosceles with ∠B = ∠C, and the bisector of ∠A is its axis of symmetry. Reflection in that bisector swaps B with C and swaps the B-bisector with the C-bisector, so E on AC swaps with F on AB. The incentre I lies on the bisector, so it stays fixed. Hence IE = IF.
Counterexample to the converse: Take ∠A = 60°, ∠B = 30°, ∠C = 90°. Then ∠B ≠ ∠C, so the opposite sides are unequal: AB ≠ AC. Even so, IE = IF.
- BE bisects ∠B, so ∠EBC = 15°. In △BEC the angle at C is still ∠C = 90°, because E lies on AC. Thus ∠BEC = 180° − (15° + 90°) = 75°.
- CF bisects ∠C, so ∠BCF = 45°. In △BCF the angle at B is still ∠B = 30°, because F lies on AB. Thus ∠BFC = 180° − (30° + 45°) = 105°. A, F and B are collinear in that order, so the adjacent angle ∠AFC = 75°.
- Drop perpendiculars from I to lines AC and AB, meeting them at D and H. The incentre is equidistant from the sides, so ID = IH.
- Ray EI makes 75° with ray EC. That angle is acute, so the foot D falls on ray EC, and the angle at E in right triangle △IED is 75° rather than the 105° angle on the other side of EI. Likewise, ray FI makes 75° with ray FA, so H falls on ray FA and the angle at F in right triangle △IFH is 75°.
- Each small triangle then has angles 90°, 75° and 15°. Pair ∠IDE with ∠IHF (both 90°) and ∠DIE with ∠HIF (both 15°). The included sides are ID and IH, which are equal. By ASA, △IDE ≅ △IHF, so the hypotenuses are equal: IE = IF.
Adding the same number to equal numbers. For any numbers a, x and y: If x = y, then a + x = a + y. Write the converse and decide both directions.
- Original proposition True – Proved below.
- Converse True – Proved below.
Converse: If a + x = a + y, then x = y.
Original: Adding a to both sides of x = y preserves equality.
Converse: From a + x = a + y, subtract a from both sides to obtain x = y.
Product of perfect squares. If a and b are perfect squares, ab is a perfect square. Write and test the converse.
- Original proposition True – Proved below.
- Converse False – Disproved by a counterexample below.
Converse: If ab is a perfect square, then both a and b are perfect squares.
Proof of original: Write a = m2 and b = n2 for integers m,n. Then ab = (mn)2, a perfect square.
Counterexample: a = 2 and b = 8. Their product is 16 = 42, but neither 2 nor 8 is a perfect square.
Equal real numbers and their squares. For real x,y, if x = y then x² = y². Is the converse also true?
- Original proposition True – Proved below.
- Converse False – Disproved by a counterexample below.
Converse: If x2 = y2, then x = y.
Original: Equal numbers have equal squares.
Counterexample: x = 2 and y = −2 give x2 = y2 = 4, although x ≠ y.
Indeed, x2 = y2 implies x = y or x = −y, not necessarily x = y.
Questions 7–12
Equal real numbers and their cubes. For real x,y, if x = y then x³ = y³. Is the converse true?
- Original proposition True – Proved below.
- Converse True – Proved below.
Converse: If x3 = y3, then x = y.
Proof: Equal numbers have equal cubes. Conversely, the function t ↦ t3 is strictly increasing on the real numbers, so two distinct real numbers cannot have the same cube. Alternatively, taking the real cube root of both sides gives x = y.
Divisibility by 24, 4 and 6. If a positive integer n is divisible by 24, then it is divisible by both 4 and 6. Write the converse and test it.
- Original proposition True – Proved below.
- Converse False – Disproved by a counterexample below.
Converse: If n is divisible by both 4 and 6, then it is divisible by 24.
Original: 4 divides 24 and 6 divides 24, so both divide every multiple of 24. The product 4 × 6 = 24 does not by itself make the converse true, because 4 and 6 are not coprime.
Counterexample: 12 is divisible by 4 and 6 but not by 24. The least common multiple of 4 and 6 is 12, not 24.
Divisibility by 60, 5 and 12. If n is divisible by 60, it is divisible by both 5 and 12. Write and test the converse.
- Original proposition True – Proved below.
- Converse True – Proved below.
Converse: If n is divisible by both 5 and 12, then it is divisible by 60.
Original: Because 60 is a multiple of 5 and 12, any multiple of 60 is divisible by both.
Converse: 5 and 12 have no common prime factor, so LCM(5, 12) = 5 × 12 = 60. An integer divisible by both must be a multiple of 60.
Square of a prime and exactly three factors. If n is the square of a prime, n has exactly 3 positive factors. Write and test the converse.
- Original proposition True – Proved below.
- Converse True – Proved below.
Converse: If n has exactly three positive factors, then n is the square of a prime number.
Original: If n = p2 and p is prime, its only positive factors are 1, p, p2.
Converse: With three factors, 1 and n make one factor-partner pair, and the remaining factor must pair with itself. Thus n = d2. If d were composite, its proper divisor would produce extra factors of n, contradicting “exactly 3”. Therefore d is prime.
Four divisors and distinct prime factors. If n is the product of two unequal prime numbers, it has exactly 4 positive divisors. Is the converse true?
- Original proposition True – Proved below.
- Converse False – Disproved by a counterexample below.
Converse: If n has exactly four positive divisors, then it is a product of two unequal prime numbers.
Original: For distinct primes p,q, the only positive divisors of pq are 1, p, q and pq.
Counterexample: 8 = 23 has exactly four divisors: 1, 2, 4 and 8. It is not a product of two unequal primes.
A number with exactly four positive divisors is either p³ (a prime cube) or pq (a product of two distinct primes).
Common factors of n and n + 3. If n and n + 3 have no common factor greater than 1, then n is not a multiple of 3. Write the converse and test both.
- Original proposition True – Proved below.
- Converse True – Proved below.
Converse: If n is not a multiple of 3, then n and n + 3 have no common factor greater than 1.
Original: If n were divisible by 3, then n + 3 would also be divisible by 3, so they would share the factor 3. This contradicts the condition. Therefore n is not divisible by 3.
Converse: Any common divisor d of n and n + 3 must divide their difference, which is 3. So d is either 1 or 3. If n is not divisible by 3, d cannot be 3; their greatest common divisor is 1.
Questions 13–17
Find counterexamples to three prime-generating claims. Find counterexamples to: (i) 4n² + 1 is always prime; (ii) n² + n + 11 is always prime; (iii) 4ⁿ + 3 is always prime.
- (i) Let n = 4: 4n2 + 1 = 4(16) + 1 = 65 = 5 × 13, not prime.
- (ii) Let n = 10: n2 + n + 11 = 100 + 10 + 11 = 121 = 11 × 11, not prime.
- (iii) Let n = 4: 4n + 3 = 256 + 3 = 259 = 7 × 37, not prime.
Each expression fails to produce a prime for at least one positive integer n, so all three universal claims are false.
Find counterexamples involving 2ⁿ − 1 and 2ⁿ + 1. Disprove (i) if n is prime, then 2ⁿ − 1 is prime; (ii) if n is even, then 2ⁿ + 1 is prime.
- (i) Choose n = 11, which is prime. Then 211 − 1 = 2048 − 1 = 2047 = 23 × 89, composite.
- (ii) Choose n = 6, which is even. Then 26 + 1 = 64 + 1 = 65 = 5 × 13, composite.
Therefore, both claims are false.
Testing divisibility by 8. “If a number is divisible by 8, then it is divisible by both 2 and 4.” (i) Justify. (ii) Can checking divisibility by 2 and 4 alone establish divisibility by 8?
(i) True. If n = 8k, then n = 2(4k) and n = 4(2k). Hence it is divisible by both 2 and 4.
(ii) No. For example, 12 is divisible by 2 and by 4 but not by 8. In fact, checking 2 and 4 together gives no more than checking 4, because every multiple of 4 is already even.
Useful shortcut: A whole number is divisible by 8 exactly when the number formed by its last three digits is divisible by 8 (use all the digits if there are fewer than three).
Digit sum and divisibility by 3. Express the divisibility-by-3 shortcut in two “If–then” sentences.
Statement 1: If a number is divisible by 3, then the sum of its digits is divisible by 3.
Converse (Statement 2): If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.
Both true For example, 5,124 has digit sum 5 + 1 + 2 + 4 = 12. Since 12 is divisible by 3, so is 5,124.
Why this works: 10, 100, 1000, … each leave remainder 1 when divided by 3. So a number and its digit sum have the same remainder upon division by 3.
Quadrilateral Q and the lengths of its diagonals. Four endpoints of two crossing sticks form a quadrilateral. (i) If being type Q implies equal diagonals, (a) must the sticks be equal and (b) does arrangement matter? (ii) If equal diagonals imply type Q instead, answer (a) and (b) again.
(i) “If a quadrilateral is of type Q, then its diagonals are equal.”
(a) Yes. Every Q quadrilateral must have equal-length diagonals, so the two sticks used as its diagonals must be equal.
(b) The arrangement can matter, depending on the type Q. The given implication alone does not tell us which arrangements give Q. Equal diagonals are required, but the definition of Q may impose another condition. For instance, if Q means “rectangle”, the diagonals must be equal and bisect one another. Merely crossing two equal sticks at arbitrary points need not give a rectangle.
(ii) “If a quadrilateral has equal diagonals, then it is of type Q.”
(a) Using equal sticks is sufficient to guarantee Q, but they are not necessarily required. This implication does not say that every Q quadrilateral must have equal diagonals; Q may include other quadrilaterals as well.
(b) Once two equal sticks really are the diagonals of a non-degenerate quadrilateral, their manner of crossing does not matter for guaranteeing Q. The assumed statement says that every quadrilateral with equal diagonals belongs to Q.
Chapter summary and quick revision
For Questions 1–12, the exercise asks you to state the converse and decide which statements are true. Use this answer map for revision; the full reasons are in the worked answers above.
| Question | Original | Converse | Key reason |
|---|---|---|---|
| Q1 | True | True | Corresponding-angle criterion |
| Q2 | True | False | A rectangle need not be a square |
| Q3 | True | False | With ∠A = 60°, ∠B = 30° and ∠C = 90°, IE = IF but AB ≠ AC |
| Q4 | True | True | Add/subtract the same number |
| Q5 | True | False | 2 × 8 = 16 |
| Q6 | True | False | (−2)² = 2² |
| Q7 | True | True | Real cubes are one-to-one |
| Q8 | True | False | 12 is divisible by 4 and 6 |
| Q9 | True | True | LCM(5,12) = 60 |
| Q10 | True | True | Exactly 3 divisors ⇔ prime square |
| Q11 | True | False | 8 has 4 divisors, but equals 2³ |
| Q12 | True | True | gcd(n,n+3) = gcd(n,3) |
- Split a sentence into its “if” condition and “then” conclusion.
- Swap those two parts to form the converse; do not replace them with their negatives.
- Test the original and converse separately.
- For a true claim, write a valid proof. For a false claim, give one counterexample that satisfies the condition but breaks the conclusion.
- Distinguish necessary and sufficient conditions, especially in Question 17.
Frequently asked questions
What is a converse in mathematics?
The converse of “If P, then Q” is “If Q, then P”. It reverses the direction of the implication.
Is the converse of every true statement true?
No. For example, every square is a rectangle, but not every rectangle is a square.
How many counterexamples are needed to prove a statement false?
One valid counterexample is enough to disprove a statement that claims something is true for all cases.
Can both a statement and its converse be true?
Yes. A positive integer is a perfect square if and only if it has an odd number of positive factors.