Key Concepts at a Glance
- A proposition is a statement that is either true or false (never both).
- Many propositions have the form "if X then Y", also written "X implies Y", or "Y when X" — all three mean the same thing.
- The converse of "if X then Y" is "if Y then X" — the two parts swap places.
- A proposition being true tells you nothing about whether its converse is true. Any combination can happen: both true, only one true, or both false.
- A counterexample is a single example that satisfies X but not Y — one is enough to prove "if X then Y" is false.
- To prove a proposition true, you need an argument that works for every case — checking a few examples is never enough.
- When a proposition and its converse are both true, X and Y "imply each other" (e.g. n is a perfect square ⟺ n has an odd number of factors).
This chapter opens with a familiar geometry result — if two sides of a triangle are equal, the angles opposite them are equal — and asks whether flipping it around still gives a true statement. From there it builds the vocabulary of propositions, converses and counterexamples, with examples from everyday life (wet roads and rain), number theory (multiples, perfect squares, Fermat's numbers \(2^{2^n}+1\)) and geometry (triangle areas, the Baudhāyana–Pythagoras Theorem).
A good habit for every question in this chapter: first write the proposition and its converse clearly as "if … then …" sentences, then deal with each one separately — prove it if it's true, or give one clean counterexample if it's false.
In-Text Examples & Think and Reflect
Answers to the questions the chapter poses inside its worked examples, along with all three Think and Reflect boxes, in the order they appear in the textbook.
Think and Reflect
TRWe have proved the first statement in an earlier grade. Is the second statement true? Can you prove it?
(Statement 2: If two angles of a triangle are equal, then the sides opposite the equal angles have equal lengths.)
∠B = ∠C, with altitude AD drawn from the third vertex A, perpendicular to BC.
Yes, Statement 2 is true — it is the converse of Statement 1, and in this case both the proposition and its converse hold.
Given: △ABC with \(\angle B = \angle C\). To prove: \(AB = AC\).
Construction: Following the hint, draw the altitude AD from vertex A (the vertex holding the third angle) to BC, so that \(AD \perp BC\).
Proof: Compare △ABD and △ACD.
\(\angle ABD = \angle ACD\) (given, \(\angle B = \angle C\))
\(\angle ADB = \angle ADC = 90^\circ\) (AD is an altitude)
\(AD = AD\) (common side)
So \(\triangle ABD \cong \triangle ACD\) by the AAS congruence criterion, and hence the corresponding sides are equal: \(AB = AC\).
E2Example 2 — Proposition P: If a number is a multiple of 6, then it is a multiple of 3. Converse Q: If a number is a multiple of 3, then it is a multiple of 6.
Justify why P is true. Determine whether Q is true, and if not, give a counterexample.
P is true. If n is a multiple of 6, then \(n = 6k\) for some whole number k. So \(n = 3 \times (2k)\), which is 3 times a whole number — so n is a multiple of 3.
Q is false. Counterexample: 9 is a multiple of 3 (\(9 = 3 \times 3\)) but it is not a multiple of 6. (Any odd multiple of 3, such as 3, 15 or 21, works just as well.)
E3Example 3 — P: If n is a perfect square, then it has an odd number of factors. Q: If n has an odd number of factors, then it is a perfect square.
Which of them are true? The factor-partner argument in the Discussion box — does it prove P or Q?
The argument proves Q. It starts from "n has an odd number of factors" and ends with "n is a perfect square" — that is exactly the direction of Q. Factors come in factor-partner pairs; if no pair repeated a number, the factors would split neatly into pairs and their count would be even. An odd count forces some pair \((f, f)\), so \(n = f \times f\).
P is also true. If \(n = f \times f\), then \((f, f)\) is a factor-partner pair with a repeated number. There cannot be a second such pair \((g, g)\), because \(g \times g = f \times f\) forces \(g = f\) for positive numbers. So exactly one factor is "unpaired" and all the others come in distinct pairs — giving an odd total. For example, 36 has factor pairs (1, 36), (2, 18), (3, 12), (4, 9), (6, 6): that's 4 × 2 + 1 = 9 factors.
E4Example 4 — P: If two triangles have the same area, then they are congruent. Q: If two triangles are congruent, then they have the same area.
Determine if these statements are true or not. Justify the true statements and give a counterexample for each false statement.
Two triangles with equal area (6 square units each) that are clearly not congruent.
P is false. Counterexample: a right triangle with legs 3 and 4 (sides 3, 4, 5) has area \(\frac{1}{2} \times 4 \times 3 = 6\). A triangle with base 6 and height 2 has area \(\frac{1}{2} \times 6 \times 2 = 6\) too. The areas are equal, but the second triangle has a side of length 6, which the first doesn't have — so they are not congruent.
Q is true. Congruent triangles can be placed exactly on top of each other, so they cover exactly the same region of the plane — and hence have the same area. (Equivalently: corresponding sides and altitudes are equal, so \(\frac{1}{2} \times \text{base} \times \text{height}\) gives the same value for both.)
Think and Reflect
TRIt can also happen that both the proposition and converse are false! Can you give an example?
Number example. P: If a number is a multiple of 4, then it is a multiple of 6. This is false — 4 itself is a multiple of 4 but not of 6.
Converse Q: If a number is a multiple of 6, then it is a multiple of 4. This is also false — 6 is a multiple of 6 but not of 4.
Geometry example. P: If a quadrilateral is a rhombus, then it is a rectangle. False — a rhombus with angles 60° and 120° is not a rectangle.
Converse Q: If a quadrilateral is a rectangle, then it is a rhombus. False — a 4 cm × 2 cm rectangle does not have all sides equal.
Everyday example. P: If a person is a cricketer, then they are a teacher. Q: If a person is a teacher, then they are a cricketer. Both are clearly false — each needs just one person who is one but not the other.
E5Example 5 — Converse of the Baudhāyana–Pythagoras Theorem: Let a, b, c be the sidelengths of △ABC with \(a^2 + b^2 = c^2\). Construct a right triangle XYZ whose perpendicular sides YZ and XZ have lengths a and b. What can we say about the length of XY? What can we say about △ABC and △XYZ?
Length of XY. △XYZ is right-angled at Z, so by the Baudhāyana–Pythagoras Theorem: \(XY^2 = YZ^2 + XZ^2 = a^2 + b^2\). But we are given \(a^2 + b^2 = c^2\), so \(XY^2 = c^2\), giving \(XY = c\).
Comparing the triangles. Now the three sides match:
\(BC = YZ = a\), \(\quad CA = ZX = b\), \(\quad AB = XY = c\)
So \(\triangle ABC \cong \triangle XYZ\) by the SSS congruence criterion. Corresponding angles of congruent triangles are equal, so \(\angle C = \angle Z = 90^\circ\).
Hence △ABC is right-angled (at C, the vertex opposite the longest side c), and the converse of the Baudhāyana–Pythagoras Theorem is true.
Think and Reflect
TR1. Identify real-life examples in which a proposition is true but not the converse. Give nice counterexamples!
2. Give more examples from geometry as well as number theory in which a proposition is true but not its converse. Give appropriate counterexamples.
1. Real-life examples
(a) P: If a fruit is a mango, then it is a fruit that grows on a tree. (True.) Converse: If a fruit grows on a tree, then it is a mango. False — an apple grows on a tree too.
(b) P: If a city is in Kerala, then it is in India. (True.) Converse: If a city is in India, then it is in Kerala. False — Jaipur is in India but not in Kerala.
(c) P: If it is Sunday, then schools are closed. (True for regular schools.) Converse: If schools are closed, then it is Sunday. False — schools are also closed on public holidays such as 15 August.
2. Geometry examples
(a) P: If a quadrilateral is a square, then it is a rhombus. (True — all sides are equal.) Converse: If a quadrilateral is a rhombus, then it is a square. False — a rhombus with angles 60° and 120°.
(b) P: If two angles are vertically opposite, then they are equal. (True.) Converse: If two angles are equal, then they are vertically opposite. False — the three 60° angles of an equilateral triangle are equal but not vertically opposite.
2. Number theory examples
(a) P: If a number is divisible by 9, then it is divisible by 3. (True: \(9k = 3 \times 3k\).) Converse: If a number is divisible by 3, then it is divisible by 9. False — 6.
(b) P: If p is a prime greater than 2, then p is odd. (True — an even number greater than 2 has 2 as a factor.) Converse: If a number greater than 2 is odd, then it is prime. False — 9 = 3 × 3.
Exercise Set 9.1
For Questions 1–12: frame the converse of each proposition, then decide whether each of the two statements is true. Justify the true statements and give a counterexample for each false one.
1If two lines are parallel, then the corresponding angles formed by a transversal are equal.
Lines l and m cut by transversal t; ∠1 and ∠2 are corresponding angles.
Converse: If the corresponding angles formed by a transversal on two lines are equal, then the two lines are parallel.
Proposition — true. This is the corresponding-angles property of parallel lines, established in earlier grades: when \(l \parallel m\), a transversal t makes \(\angle 1 = \angle 2\).
Converse — true. Suppose \(\angle 1 = \angle 2\) but l and m are not parallel. Then they meet at some point R, and together with t they form a triangle. In that triangle, \(\angle 1\) is an exterior angle and \(\angle 2\) is one of its interior opposite angles. An exterior angle equals the sum of the two interior opposite angles, so \(\angle 1 = \angle 2 + \angle R\). With \(\angle 1 = \angle 2\), this forces \(\angle R = 0^\circ\) — impossible for a triangle. So l and m cannot meet: they are parallel.
2If a quadrilateral is a square, then all its angles are equal.
Converse: If all the angles of a quadrilateral are equal, then it is a square.
Proposition — true. Every angle of a square is 90°, so all four are equal.
Converse — false. Counterexample: a rectangle measuring 4 cm × 2 cm. All four of its angles are 90° (equal), but its sides are not all equal, so it is not a square.
3*Given any △ABC, let us bisect the angles at B and C. The bisectors meet at the incentre I of the triangle. Now extend the bisectors beyond I till they meet the opposite sides at E and F respectively, as shown (Fig. 9.1).
Proposition: If AB = AC, then IE = IF.
Converse: If IE = IF, then AB = AC.
Proposition — true. Here E lies on AC (on the bisector BE) and F lies on AB (on the bisector CF).
Since \(AB = AC\), \(\angle B = \angle C\), so their halves are equal: \(\angle IBC = \angle ICB\) and \(\angle IBF = \angle ICE\).
In △IBC, \(\angle IBC = \angle ICB\), so \(IB = IC\) (sides opposite equal angles — the TR result above).
Now compare △IBF and △ICE:
\(\angle IBF = \angle ICE\) (each is half of the equal angles B and C)
\(IB = IC\) (shown above)
\(\angle BIF = \angle CIE\) (vertically opposite angles, since BE and CF cross at I)
So \(\triangle IBF \cong \triangle ICE\) (ASA), and therefore \(IF = IE\).
Converse — false. Surprisingly, IE = IF can happen even when AB ≠ AC. It happens whenever \(\angle A = 60^\circ\). Counterexample: the 30°–60°–90° triangle with \(\angle A = 60^\circ\), \(\angle B = 90^\circ\), \(\angle C = 30^\circ\).
B = (0, 0), A = (0, 1), C = (√3, 0): ∠A = 60°, ∠B = 90°, ∠C = 30°. The highlighted segments IE and IF are equal, yet AB = 1 ≠ AC = 2.
Place the triangle on a coordinate plane: \(B = (0,0)\), \(A = (0,1)\), \(C = (\sqrt3, 0)\). Then \(AB = 1\), \(BC = \sqrt3\), \(AC = 2\), so \(AB \ne AC\).
For a right triangle, the inradius is \(r = \dfrac{1 + \sqrt3 - 2}{2} = \dfrac{\sqrt3 - 1}{2}\), so \(I = (r, r)\).
E (bisector of ∠B, the line \(y = x\), meets AC, the line \(\dfrac{x}{\sqrt3} + y = 1\)): \(E = \left(\dfrac{3-\sqrt3}{2}, \dfrac{3-\sqrt3}{2}\right)\).
F (bisector of ∠C meets AB, the y-axis): the bisector makes a 15° angle with BC, and \(\tan 15^\circ = 2 - \sqrt3\), so \(F = \left(0,\ \sqrt3(2-\sqrt3)\right) = (0,\ 2\sqrt3 - 3)\).
\(IE^2 = 2\left(\dfrac{3-\sqrt3}{2} - \dfrac{\sqrt3-1}{2}\right)^2 = 2(2-\sqrt3)^2 = 14 - 8\sqrt3\)
\(IF^2 = \left(\dfrac{\sqrt3-1}{2}\right)^2 + \left(2\sqrt3 - 3 - \dfrac{\sqrt3-1}{2}\right)^2 = \dfrac{(4-2\sqrt3) + (52-30\sqrt3)}{4} = 14 - 8\sqrt3\)
So \(IE = IF \approx 0.379\), even though \(AB \ne AC\).
Why ∠A = 60° works: \(\angle FIE = \angle BIC = 90^\circ + \dfrac{\angle A}{2} = 120^\circ\), so \(\angle FAE + \angle FIE = 180^\circ\) and A, F, I, E lie on one circle. In that circle, chords IF and IE subtend equal angles \(\dfrac{\angle A}{2}\) at A (AI bisects ∠A), so IF = IE — no matter what B and C are.
4If x = y, then a + x = a + y, where x, y and a are any three numbers. This proposition and its converse are routinely used while solving equations.
Converse: If \(a + x = a + y\), then \(x = y\).
Proposition — true. If x and y are the same number, then adding the same number a to each gives the same result: \(a + x = a + y\).
Converse — true. Start from \(a + x = a + y\) and add \(-a\) to both sides (i.e. subtract a): \(a + x - a = a + y - a\), which gives \(x = y\).
This is exactly why, while solving equations, we can both add the same number to both sides (proposition) and cancel the same number from both sides (converse).
5If a and b are perfect squares, then ab is a perfect square.
Converse: If ab is a perfect square, then a and b are perfect squares.
Proposition — true. If \(a = p^2\) and \(b = q^2\), then \(ab = p^2q^2 = (pq)^2\), a perfect square.
Converse — false. Counterexample: \(a = 2\), \(b = 8\). Then \(ab = 16 = 4^2\) is a perfect square, but neither 2 nor 8 is a perfect square.
In Questions 6 and 7, x and y are real numbers.
6If x = y, then \(x^2 = y^2\).
Converse: If \(x^2 = y^2\), then \(x = y\).
Proposition — true. If x and y are the same number, squaring gives the same result: \(x \times x = y \times y\).
Converse — false. Counterexample: \(x = 3\), \(y = -3\). Then \(x^2 = 9 = y^2\), but \(3 \ne -3\).
(From \(x^2 = y^2\) we only get \((x-y)(x+y) = 0\), i.e. \(x = y\) or \(x = -y\). The converse would be true if x and y were both restricted to non-negative numbers.)
7If x = y, then \(x^3 = y^3\).
Converse: If \(x^3 = y^3\), then \(x = y\).
Proposition — true. Cubing the same number gives the same result.
Converse — true. If \(x^3 = y^3\), then \(x^3 - y^3 = 0\), i.e.
\((x - y)(x^2 + xy + y^2) = 0\)
So either \(x - y = 0\), or \(x^2 + xy + y^2 = 0\). Now
\(x^2 + xy + y^2 = \left(x + \dfrac{y}{2}\right)^2 + \dfrac{3y^2}{4}\)
is a sum of two squares, so it is zero only when both squares are zero: \(y = 0\) and then \(x = 0\) — in which case \(x = y\) anyway. Either way, \(x = y\).
Unlike squaring, cubing keeps the sign of a number (\((-3)^3 = -27\), \(3^3 = 27\)), so two different real numbers never have the same cube.
In Questions 8–12, n is a positive integer.
8If n is divisible by 24, then it is divisible by both 4 and 6.
Converse: If n is divisible by both 4 and 6, then it is divisible by 24.
Proposition — true. If \(n = 24k\), then \(n = 4 \times (6k)\) and \(n = 6 \times (4k)\), so n is divisible by both 4 and 6.
Converse — false. Counterexample: \(n = 12\). It is divisible by 4 (\(12 = 4 \times 3\)) and by 6 (\(12 = 6 \times 2\)), but not by 24. (4 and 6 share the factor 2, so being divisible by both only guarantees divisibility by their LCM, 12 — not by \(4 \times 6 = 24\).)
9If n is divisible by 60, then it is divisible by both 5 and 12.
Converse: If n is divisible by both 5 and 12, then it is divisible by 60.
Proposition — true. If \(n = 60k\), then \(n = 5 \times (12k)\) and \(n = 12 \times (5k)\).
Converse — true. Since 12 divides n, write \(n = 12m\). Since 5 also divides n, 5 divides \(12m = 2 \times 2 \times 3 \times m\). As 5 is prime and does not divide 2, 2 or 3, it must divide m: \(m = 5t\). Then \(n = 12 \times 5t = 60t\), so 60 divides n.
The key difference from Question 8: 5 and 12 have no common factor (other than 1), so their LCM is the full product \(5 \times 12 = 60\).
10If n is the square of a prime number, then it has exactly 3 factors.
Converse: If n has exactly 3 factors, then it is the square of a prime number.
Proposition — true. If \(n = p^2\) with p prime, any factor of n can only have p in its prime factorisation, at most twice. So the factors are exactly 1, p and \(p^2\) — three factors. (E.g. 49: 1, 7, 49.)
Converse — true. If n has exactly 3 factors, that is an odd number of factors, so by Example 3, n is a perfect square: \(n = m^2\). Since 1 has only one factor, \(m > 1\), and then 1, m and \(m^2\) are three different factors of n — so they are all its factors.
If m were not prime, it would have a factor d with \(1 < d < m\), and d would be a fourth factor of n — a contradiction. So m is prime, and n is the square of a prime.
11If n is a product of two unequal prime numbers, then it has exactly 4 divisors.
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 distinct primes, any divisor of n can only contain p and q in its prime factorisation, each at most once. So the divisors are exactly 1, p, q and pq — four divisors. (E.g. 15: 1, 3, 5, 15.)
Converse — false. Counterexample: \(n = 8\). Its divisors are 1, 2, 4, 8 — exactly four — but \(8 = 2^3\) is not a product of two unequal primes. (In general, the cube of any prime, such as 27, also has exactly 4 divisors.)
12If n and n + 3 have no factors in common, then n is not a multiple of 3.
("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. Suppose n were a multiple of 3, say \(n = 3k\). Then \(n + 3 = 3(k+1)\) is also a multiple of 3, so 3 would be a common factor — contradicting what we're given. So n is not a multiple of 3.
Converse — true. Let d be any common factor of n and n + 3. Then d also divides their difference, \((n+3) - n = 3\). So \(d = 1\) or \(d = 3\). If n is not a multiple of 3, then \(d = 3\) is impossible, leaving only \(d = 1\). So n and n + 3 have no common factor.
13There are no known "neat" expressions that generate only primes! Find counterexamples to the following claims.
(i) All numbers of the form \(4n^2 + 1\) are prime.
(ii) All numbers of the form \(n^2 + n + 11\) are prime.
(iii) All numbers of the form \(4n + 3\) are prime.
| n | \(4n^2+1\) | \(n^2+n+11\) | \(4n+3\) |
|---|---|---|---|
| 1 | 5 | 13 | 7 |
| 2 | 17 | 17 | 11 |
| 3 | 37 | 23 | 15 = 3 × 5 |
| 4 | 65 = 5 × 13 | 31 | 19 |
| … | … | ||
| 10 | 121 = 11 × 11 |
(i) \(n = 4\): \(4(4^2) + 1 = 65 = 5 \times 13\), not prime.
(ii) \(n = 10\): \(10^2 + 10 + 11 = 121 = 11 \times 11\), not prime. (Values for n = 0 to 9 are all prime: 11, 13, 17, 23, 31, 41, 53, 67, 83, 101 — which is exactly why checking a few cases is never a proof! A quick way to spot this one: at \(n = 11\), every term is a multiple of 11, giving \(143 = 11 \times 13\).)
(iii) \(n = 3\): \(4(3) + 3 = 15 = 3 \times 5\), not prime.
14Find counterexamples to the following statements.
(i) If n is a prime number, then \(2^n - 1\) is a prime number.
(ii) If n is an even number, then \(2^n + 1\) is a prime number.
(i) Checking primes in order: \(2^2 - 1 = 3\), \(2^3 - 1 = 7\), \(2^5 - 1 = 31\), \(2^7 - 1 = 127\) — all prime. But for \(n = 11\) (prime):
\(2^{11} - 1 = 2048 - 1 = 2047 = 23 \times 89\), which is not prime.
(ii) \(2^2 + 1 = 5\) and \(2^4 + 1 = 17\) are prime, but for \(n = 6\) (even):
\(2^6 + 1 = 64 + 1 = 65 = 5 \times 13\), which is not prime.
15Consider the statement: "If a number is divisible by 8, then it is divisible by both 2 and 4".
(i) Justify the statement.
(ii) Recall the divisibility shortcuts that we have studied for different numbers. 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?
(i) If n is divisible by 8, then \(n = 8k\). So \(n = 2 \times (4k)\) and \(n = 4 \times (2k)\) — n is divisible by both 2 and 4.
(ii) No. Checking 2 and 4 is the converse of the statement — "if divisible by 2 and 4, then divisible by 8" — and that converse is false. Counterexample: 12 is divisible by 2 and by 4, but not by 8. (Also, every number divisible by 4 is already divisible by 2, so checking 2 adds nothing new; the LCM of 2 and 4 is only 4.)
The correct shortcut for 8 is to check whether the number formed by the last three digits is divisible by 8.
16Recall 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.
Proposition: If a number is divisible by 3, then the sum of its digits is a multiple of 3.
Converse: If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.
Both are true, so the two statements imply each other — which is exactly why the digit-sum test works in both directions. To see why, take a 3-digit number with digits a, b, c:
\(100a + 10b + c = (99a + 9b) + (a + b + c)\)
The part \(99a + 9b = 9(11a + b)\) is always a multiple of 3. So the number and its digit sum differ by a multiple of 3 — if either one is a multiple of 3, so is the other. The same idea works for any number of digits, since \(10, 100, 1000, \ldots\) are each 1 more than a multiple of 9.
17We have identified different types of quadrilaterals — squares, rectangles, parallelograms, rhombi, kites and trapezia. One can identify more types (e.g., we could create a category of quadrilaterals that have equal-length opposite sides). Suppose we have identified a category of quadrilaterals called Q, and we have to construct a quadrilateral of this type. For this, we are to use two thin sticks, put them together as diagonals so that the quadrilateral obtained by joining their endpoints is of type Q (see Fig. 9.2).
(i) Suppose Q satisfies the following 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? (b) Will it matter how the two sticks are put together?
(ii) Instead of the property mentioned above, suppose Q satisfies the following property. If a quadrilateral has equal diagonals, then it is of type Q. What will be your answers to (a) and (b) now?
(i) Property: "type Q ⇒ equal diagonals"
(a) Yes, the sticks must be of equal length. Every type-Q quadrilateral has equal diagonals. If we used unequal sticks, the diagonals would be unequal, so the result could not be of type Q. Equal diagonals are necessary.
(b) It may matter. The property only tells us something every type-Q quadrilateral has — it doesn't say equal diagonals are enough. Q might also require more. For example, if Q were "rectangles", the sticks would also have to cross at their midpoints; if Q were "squares", they would have to cross at their midpoints and at right angles. With only this property given, we can't be sure that every way of placing two equal sticks produces a type-Q quadrilateral.
(ii) Property: "equal diagonals ⇒ type Q"
(a) Using equal sticks is enough — any quadrilateral with equal diagonals is guaranteed to be of type Q. Strictly speaking, equal sticks aren't required: this property doesn't say every type-Q quadrilateral has equal diagonals, so some type-Q quadrilaterals might have unequal ones. But choosing equal sticks is a sure-fire way to succeed.
(b) No, it won't matter how the equal sticks are put together (as long as they cross each other, so that joining the endpoints gives a quadrilateral). Equal diagonals alone are sufficient to make it type Q.
This question shows the practical difference between a property and its converse: "Q ⇒ equal diagonals" makes equal diagonals a necessary condition, while "equal diagonals ⇒ Q" makes it a sufficient one.
Extra Practice Questions
Seven extra questions in the style of the textbook's own exercise, for independent practice once you've gone through the solved questions above. For each one, write the converse, decide whether each statement is true, and justify or give a counterexample — then tap to check your answer.
1If a triangle is equilateral, then each of its angles is 60°.
Converse: If each angle of a triangle is 60°, then the triangle is equilateral.
Proposition — true. All sides are equal, so all angles are equal (angles opposite equal sides), and they add to 180°, so each is \(\dfrac{180^\circ}{3} = 60^\circ\).
Converse — true. All angles are equal, so all sides are equal (sides opposite equal angles — the TR result of this chapter).
2If n is an odd number, then \(n^2\) is odd.
Converse: If \(n^2\) is odd, then n is odd.
Proposition — true. \(n = 2k + 1\) gives \(n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1\), which is odd.
Converse — true. If n were even, \(n = 2k\), then \(n^2 = 4k^2\) would be even. So an odd \(n^2\) forces n to be odd.
3If a quadrilateral is a rectangle, then its diagonals are equal.
Converse: If a quadrilateral has equal diagonals, then it is a rectangle.
Proposition — true. In rectangle ABCD, △ABC ≅ △BAD (AB common, BC = AD, ∠B = ∠A = 90°, SAS), so AC = BD.
Converse — false. Counterexample: an isosceles trapezium (non-parallel sides equal) has equal diagonals but is not a rectangle.
4For a real number x: if \(x > 3\), then \(x^2 > 9\).
Converse: If \(x^2 > 9\), then \(x > 3\).
Proposition — true. If \(x > 3\), then \(x^2 = x \times x > 3 \times 3 = 9\) (multiplying positive numbers larger than 3).
Converse — false. Counterexample: \(x = -4\) gives \(x^2 = 16 > 9\), but \(-4 < 3\).
5If a number ends in 0, then it is divisible by 5.
Converse: If a number is divisible by 5, then it ends in 0.
Proposition — true. A number ending in 0 is a multiple of 10, and \(10k = 5 \times 2k\).
Converse — false. Counterexample: 15 is divisible by 5 but ends in 5.
6If n is a prime number, then n is odd.
Converse: If n is odd, then n is prime.
Proposition — false. Counterexample: 2 is prime but even.
Converse — false. Counterexample: 9 is odd but \(9 = 3 \times 3\) is not prime.
7If the sum of two whole numbers a and b is even, then a and b are both even.
Converse: If a and b are both even, then a + b is even.
Proposition — false. Counterexample: \(3 + 5 = 8\) is even, but 3 and 5 are both odd.
Converse — true. \(a = 2p\), \(b = 2q\) gives \(a + b = 2(p + q)\), which is even.