Question 1

Frame the converse for each of the propositions in Questions 1–12. Then, determine if each of the two statements is true or not. Justify the true statements and give a counterexample for each false statement.
If two lines are parallel, then the corresponding angles formed by a transversal are equal.

Accepted Answer

P is true. This is the corresponding angles property of parallel lines: a transversal meets both parallel lines in the same direction, so the angles it makes at the two crossings in matching positions are equal.The converse is true. Suppose the corresponding angles are equal but the lines are not parallel. Then they meet at some point R on one side of the transversal, forming a triangle with it. The exterior angle of that triangle at one crossing would equal the interior opposite angle at the other crossing, which is impossible, because an exterior angle is equal to the sum of the two interior opposite angles, so it is greater than either of them. So the lines cannot meet: they are parallel.Both directions hold, so ‘lines are parallel’ and ‘corresponding angles are equal’ imply each other.

Question 2

If a quadrilateral is a square, then all its angles are equal.

Accepted Answer

P is true. By definition every angle of a square is a right angle, so all four are $90^\circ$.The converse is false. If all four angles are equal, each is $360^\circ \div 4 = 90^\circ$, so the quadrilateral is a rectangle — but its sides need not be equal.Counterexample: a rectangle 4 cm long and 2 cm wide. All its angles are $90^\circ$, but it is not a square.

Question 3

* Given any $\triangle ABC$, 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: BE and CF are the bisectors, E on AC and F on AB). Proposition: I

Accepted Answer

P is true. Let AB = AC. Then $\angle B = \angle C$ (angles opposite equal sides), so their halves are equal too: $\angle IBF = \angle IBC = \angle ICB = \angle ICE$ In $\triangle IBC$, the base angles $\angle IBC$ and $\angle ICB$ are equal, so $IB = IC$.Now compare $\triangle IBF$ and $\triangle ICE$:So $\triangle IBF \cong \triangle ICE$ (ASA), and therefore IF = IE.The converse is false. IE = IF can also happen in a triangle that is not isosceles: it happens whenever $\angle A = 60^\circ$.Counterexample: $\triangle ABC$ with $\angle A = 60^\circ$, AB = 5, AC = 8. Then $BC^2 = 5^2 + 8^2 - 5 \times 8 = 49$ $BC = 7$ (this is the cosine rule with $\cos 60^\circ = \tfrac{1}{2}$; or check that the triangle with sides 5, 7, 8 has a $60^\circ$ angle). Here AB $\ne$ AC, but measuring (or computing) gives $IE = IF \approx 1.76$.Why it works: the angle between the two bisectors is $\angle BIC = 180^\circ - \tfrac{1}{2}(\angle B + \angle C)$ $= 90^\circ + \tfrac{1}{2}\angle A$ With $\angle A = 60^\circ$ this is $120^\circ$, so $\angle FIE = 120^\circ$ and $\angle A + \angle FIE = 180^\circ$: the quadrilateral AFIE has opposite angles adding up

Question 4

If $x = y$, 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.

Accepted Answer

P is true. $x$ and $y$ are the same number, so adding $a$ to each gives the same result.The converse is true. If $a + x = a + y$, subtract $a$ from both sides (that is P itself, with $-a$ added): $(a + x) - a = (a + y) - a$ $x = y$ This is why we may add or subtract the same number on both sides of an equation without changing its solutions: each step can be undone.

Question 5

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

Accepted Answer

P is true. Write $a = m^2$ and $b = k^2$ for whole numbers $m$, $k$. Then $ab = m^2k^2 = (mk)^2$ which is a perfect square.The converse is false. Counterexample: $a = 2$, $b = 8$. $ab = 2 \times 8 = 16 = 4^2$ 16 is a perfect square, but neither 2 nor 8 is.

Question 6

In Questions 6 and 7, $x$ and $y$ are real numbers. If $x = y$, then $x^2 = y^2$.

Accepted Answer

P is true. Squaring the same number twice gives the same result.The converse is false. Counterexample: $x = 3$, $y = -3$. $x^2 = 9 = (-3)^2 = y^2$ but $3 \ne -3$.In fact $x^2 = y^2$ means $(x - y)(x + y) = 0$, so $x = y$ or $x = -y$.

Question 7

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

Accepted Answer

P is true, for the same reason as Q6.The converse is true. Unlike squares, cubes keep the sign. Suppose $x^3 = y^3$. Then $x^3 - y^3 = 0$ $(x - y)(x^2 + xy + y^2) = 0$ The second bracket is never zero unless $x = y = 0$, because $x^2 + xy + y^2$ $= \left(x + \tfrac{y}{2}\right)^2 + \tfrac{3}{4}y^2$ is a sum of two squares, and it is zero only when $y = 0$ and then $x = 0$. So either $x - y = 0$ or $x = y = 0$; in both cases $x = y$.Compare Q6: $3^2 = (-3)^2$, but $3^3 = 27$ while $(-3)^3 = -27$.

Question 8

In Questions 8–12, $n$ is a positive integer. If $n$ is divisible by 24, then it is divisible by both 4 and 6.

Accepted Answer

P is true. If $n = 24k$, then $n = 4 \times (6k)$ $n = 6 \times (4k)$ so 4 and 6 both divide $n$.The converse is false. Counterexample: $n = 12$. It is divisible by 4 and by 6, 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 $4 \times 6 = 24$.

Question 9

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

Accepted Answer

P is true. If $n = 60k$, then $n = 5 \times (12k)$ $n = 12 \times (5k)$ The converse is true. If 12 divides $n$, write $n = 12m$. Now 5 divides $12m$, and 5 is a prime that does not divide 12, so 5 must divide $m$: $m = 5j$. Then $n = 12 \times 5j = 60j$ Compare Q8: here 5 and 12 have no common factor, so ‘divisible by both’ does give ‘divisible by their product’.

Question 10

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

Accepted Answer

P is true. Let $n = p^2$ with $p$ prime. A factor of $p^2$ can only be built from the prime $p$, so the factors are $1, p, p^2$ — exactly 3 (e.g. 25: 1, 5, 25).The converse is true. Suppose $n$ has exactly 3 factors. 1 and $n$ are two of them, so there is exactly one more, say $d$.(a) An odd number of factors means $n$ is a perfect square (Example 3 of the chapter), and the extra factor must be its own partner: $n = d \times d$.(b) $d$ must be prime: if $d$ had a factor $q$ with $1 < q < d$, then $q$ would be a fourth factor of $n$.So $n = d^2$ with $d$ prime.

Question 11

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

Accepted Answer

P is true. If $n = pq$ with primes $p \ne q$, its divisors are $1, p, q, pq$ — exactly 4 (e.g. 15: 1, 3, 5, 15).The converse is false. Counterexample: $n = 8$. Its divisors are 1, 2, 4, 8 — exactly 4 — but $8 = 2 \times 2 \times 2$ is the cube of a prime, not a product of two unequal primes. (Any $p^3$ works: 27 has divisors 1, 3, 9, 27.)

Question 12

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

Accepted Answer

P is true. If $n$ were a multiple of 3, then $n + 3$ would be a multiple of 3 too, and 3 would be a common factor. So if there is no common factor, $n$ is not a multiple of 3.The converse is true. Let $d$ be any common factor of $n$ and $n + 3$. Then $d$ divides their difference: $(n + 3) - n = 3$ so $d = 1$ or $d = 3$. But $d = 3$ would mean 3 divides $n$, which is not the case. So the only common factor is 1.Example: $n = 7$, $n + 3 = 10$: no common factor. Compare $n = 6$, $n + 3 = 9$: common factor 3.

Question 13

There 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 $4^n + 3$ are prime.

Accepted Answer

Try $n = 1, 2, 3, \ldots$ until a value is not prime.(i) $4n^2 + 1$ gives 5, 17, 37 for $n = 1, 2, 3$ (all prime), then for $n = 4$: $4 \times 4^2 + 1 = 65 = 5 \times 13$ So $n = 4$ is a counterexample.(ii) $n^2 + n + 11$ gives 11, 13, 17, 23, 31, 41, 53, 67, 83, 101 for $n = 0$ to 9 — all prime. But for $n = 10$: $10^2 + 10 + 11 = 121 = 11 \times 11$ (It had to fail by $n = 10$ at the latest: then $n^2 + n + 11 = 10 \times 11 + 11 = 11 \times 11$.)(iii) $4^n + 3$ gives 7, 19, 67 for $n = 1, 2, 3$, then for $n = 4$: $4^4 + 3 = 259 = 7 \times 37$ So $n = 4$ is a counterexample.

Question 14

Find 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.

Accepted Answer

(i) For $n = 2, 3, 5, 7$ we get 3, 7, 31, 127 — all prime. The next prime is $n = 11$: $2^{11} - 1 = 2047 = 23 \times 89$ So $n = 11$ is a counterexample: 11 is prime but 2047 is not.(ii) For $n = 2, 4$ we get 5 and 17 (prime). For $n = 6$: $2^6 + 1 = 65 = 5 \times 13$ So $n = 6$ is a counterexample.

Question 15

Consider 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 divisibl

Accepted Answer

(i) If $n = 8k$, then $n = 2 \times (4k)$ $n = 4 \times (2k)$ so $n$ is divisible by 2 and by 4.(ii) No. That would be using the converse (‘divisible by 2 and 4 ⇒ divisible by 8’), and the converse is false. Counterexample: 12 is divisible by 2 and by 4, but not by 8. (In fact every number divisible by 4 is already divisible by 2, so checking 2 adds nothing.)The correct shortcut for 8 is: a number is divisible by 8 if and only if the number formed by its last three digits is divisible by 8 (e.g. 5136: $136 = 8 \times 17$, so 5136 is divisible by 8).

Question 16

Recall 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.

Accepted Answer

Two sentences are needed, a proposition and its converse, and both are true:

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

Why both hold: take a 3-digit number with digits $a, b, c$ (the idea is the same for any length):

$100a + 10b + c$

$= 99a + 9b + (a + b + c)$

$99a + 9b$ is always a multiple of 3. So the number and its digit sum differ by a multiple of 3: one is a multiple of 3 exactly when the other is.

Because both directions are true, the shortcut is safe: ‘a number is divisible by 3 if and only if the sum of its digits is a multiple of 3’. Example: 456 has digit sum 15, a multiple of 3, and $456 = 3 \times 152$.

Question 17

We 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 call

Accepted Answer

(i) Property: type Q ⇒ equal diagonals.(a) Yes, the sticks must be of equal length. Every type-Q quadrilateral has equal diagonals, so if the sticks were unequal the result could not be of type Q.(b) Yes, it may matter. The property only says equal diagonals are needed; it does not say they are enough (that would be the converse). Q may demand more. For example, if Q is ‘rectangle’, the equal sticks must also bisect each other (cross at their midpoints); crossed anywhere else, the equal sticks give another shape — e.g. if they cut each other in the same ratio but not at their midpoints, the result is an isosceles trapezium, which has equal diagonals but is not a rectangle. If Q is ‘square’, they must also cross at right angles.(ii) Property: equal diagonals ⇒ type Q.(a) Equal sticks are enough, but not necessarily required. With equal sticks the quadrilateral is certainly of type Q. The property does not say that type-Q quadrilaterals must have equal diagonals, so some type-Q quadrilaterals might have unequal diagonals; but to be sure of getting type Q, use equal sticks.(b) No, it does not matter how the equal sticks are put together, as long as they cross each other so that joinin

Part	Reason
\(\angle IBF = \angle ICE\)	halves of the equal angles B and C
\(IB = IC\)	shown above
\(\angle BIF = \angle CIE\)	vertically opposite angles at I

Propositions and their Converses

Exercise Set 9.1

Statement	True?
P: If two lines are parallel, then the corresponding angles formed by a transversal are equal.	True
Converse: If the corresponding angles formed by a transversal are equal, then the two lines are parallel.	True

Statement	True?
P: If a quadrilateral is a square, then all its angles are equal.	True
Converse: If all the angles of a quadrilateral are equal, then it is a square.	False

Statement	True?
P: If AB = AC, then IE = IF.	True
Converse: If IE = IF, then AB = AC.	False

Statement	True?
P: If \(x = y\), then \(a + x = a + y\).	True
Converse: If \(a + x = a + y\), then \(x = y\).	True

Statement	True?
P: If \(a\) and \(b\) are perfect squares, then \(ab\) is a perfect square.	True
Converse: If \(ab\) is a perfect square, then \(a\) and \(b\) are perfect squares.	False

Statement	True?
P: If \(x = y\), then \(x^2 = y^2\).	True
Converse: If \(x^2 = y^2\), then \(x = y\).	False

Statement	True?
P: If \(x = y\), then \(x^3 = y^3\).	True
Converse: If \(x^3 = y^3\), then \(x = y\).	True

Statement	True?
P: If \(n\) is divisible by 24, then it is divisible by both 4 and 6.	True
Converse: If \(n\) is divisible by both 4 and 6, then it is divisible by 24.	False

Statement	True?
P: If \(n\) is divisible by 60, then it is divisible by both 5 and 12.	True
Converse: If \(n\) is divisible by both 5 and 12, then it is divisible by 60.	True

Statement	True?
P: If \(n\) is the square of a prime number, then it has exactly 3 factors.	True
Converse: If \(n\) has exactly 3 factors, then it is the square of a prime number.	True

Statement	True?
P: If \(n\) is a product of two unequal primes, then it has exactly 4 divisors.	True
Converse: If \(n\) has exactly 4 divisors, then it is a product of two unequal primes.	False

Statement	True?
P: If \(n\) and \(n + 3\) have no common factor (other than 1), then \(n\) is not a multiple of 3.	True
Converse: If \(n\) is not a multiple of 3, then \(n\) and \(n + 3\) have no common factor (other than 1).	True

Propositions and their Converses

Exercise Set 9.1

Book a Free Orientation