Chapter 9: Propositions and Their Converses (0904CH09)
Ganita Manjari Part II · 23 questions · 2 exercises
Chapter code: 0904ch09 · Book: Ganita Manjari Part II · Class 9 Maths
Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part II Chapter 9. Every exercise has a clear, step-by-step answer for chapter code 0904CH09.
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Exercise Set 9.1(17 questions)
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.
1
If two lines are parallel, then the corresponding angles formed by a transversal are equal.
School Level Solution
Proposition: If two lines are parallel, their corresponding angles are equal.Converse: If a pair of corresponding angles formed by a transversal are equal, then the two lines are parallel.The proposition is the corresponding-angles property of parallel lines. The converse is the converse of that theorem: equality of a pair of corresponding angles implies that the two lines are parallel. Hence, both statements are true.
2
If a quadrilateral is a square, then all its angles are equal.
School Level Solution
Proposition: A square has four right angles, so all its angles are equal. This is true.Converse: If all four angles of a quadrilateral are equal, then it is a square. Since a quadrilateral has angle sum 360°, each angle is 90°. It is therefore a rectangle, but its adjacent sides need not be equal. For example, a 4 cm by 2 cm rectangle has four equal angles but is not a square. Thus the converse is false.
★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. Proposition: If AB = AC, then IE = IF.Fig. 9.1
School Level Solution
Proposition: If AB = AC, then △ABC is isosceles at A, so ∠B = ∠C. The figure is symmetric about the bisector of ∠A. The B- and C-angle-bisectors and their intersection I are interchanged by this symmetry, as are E and F. Hence IE = IF.Converse: If IE = IF, then AB = AC. This is false. For example, take ∠A = 60°, ∠B = 40° and ∠C = 80°. The angles between the extended bisectors and the opposite sides are supplementary in pairs, so their sines are equal; since the inradius is the same, IE = IF, while ∠B ≠ ∠C and therefore AB ≠ AC. Hence this is a counterexample.
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.
School Level Solution
Proposition: If x = y, adding the same number a to both sides gives a + x = a + y.Converse: If a + x = a + y, subtract a from both sides to obtain x = y.Both statements are true by the addition and cancellation properties of equality.
5
If a and b are perfect squares, then ab is a perfect square.
School Level Solution
Write a = r² and b = s². Then ab = r²s² = (rs)², a perfect square; so the proposition is true.Converse: If ab is a perfect square, then a and b are both perfect squares. This is false: a = 2 and b = 8 are not perfect squares, but ab = 16 = 4².
In Questions 6 and 7, x and y are real numbers.
6
If x = y, then x² = y².
School Level Solution
If x = y, squaring both sides gives x² = y².Converse: If x² = y², then x = y. From x² − y² = 0, we get (x − y)(x + y) = 0, so x = y or x = −y. For x = 2 and y = −2, x² = y² but x ≠ y. Thus the converse is false.
7
If x = y, then x³ = y³.
School Level Solution
Cubing x = y proves x³ = y³.Converse: If x³ = y³, then x³ − y³ = (x − y)(x² + xy + y²) = 0. Alternatively, the cube function is one-to-one on the real numbers. Since x² + xy + y² is non-negative and can be zero only when x = y = 0, the equality forces x = y. Thus the converse is true.
In Questions 8–12, n is a positive integer.
8
If n is divisible by 24, then it is divisible by both 4 and 6.
School Level Solution
Since 24 = 4 × 6, every multiple of 24 is divisible by both 4 and 6.Converse: If n is divisible by 4 and 6, then n is divisible by 24. A number divisible by both 4 and 6 need only be divisible by lcm(4,6) = 12. Counterexample: n = 12 is divisible by 4 and 6 but not by 24.
9
If n is divisible by 60, then it is divisible by both 5 and 12.
School Level Solution
Because 60 is divisible by both 5 and 12, any multiple of 60 is divisible by each of them.Converse: If n is divisible by 5 and 12, then n is divisible by 60. Since gcd(5,12)=1, lcm(5,12)=5×12=60. Every common multiple is therefore a multiple of 60. The converse is true.
10
If n is the square of a prime number, then it has exactly 3 factors.
School Level Solution
If n = p² for a prime p, its positive factors are exactly 1, p and p².Converse: If n has exactly three positive factors, then n is the square of a prime. A number with exactly three factors must have factor pairs 1·n and one repeated pair p·p, so n = p². Any non-prime p would introduce more factors. Hence the converse is true.
11
If n is a product of two unequal prime numbers, then it has exactly 4 divisors.
School Level Solution
If n = pq for distinct primes p and q, its positive divisors are 1, p, q and pq, so it has exactly four.Converse: If n has exactly four divisors, then it is a product of two unequal primes. False: 8 = 2³ has exactly four positive divisors, 1, 2, 4 and 8, but it is not a product of two unequal primes.
12
If n and n + 3 have no factors in common, then n is not a multiple of 3.
School Level Solution
If 3 divides n, it also divides n + 3, so the two numbers have a common factor 3. The contrapositive proves that if they have no common factor, n is not a multiple of 3.Converse: If n is not a multiple of 3, then n and n + 3 have no factors in common. Any common divisor divides their difference (n + 3) − n = 3, so it can only be 1 or 3. Since 3 does not divide n, their only common positive divisor is 1. The converse is true.
13
Find counterexamples:
(i) all numbers of the form 4n² + 1 are prime;
(ii) all numbers of the form n² + n + 11 are prime;
(iii) all numbers of the form 4ⁿ + 3 are prime.
School Level Solution
(i) For n = 6, 4n² + 1 = 4(36) + 1 = 145 = 5 × 29, which is composite.(ii) For n = 10, n² + n + 11 = 100 + 10 + 11 = 121 = 11², which is composite.(iii) For n = 4, 4ⁿ + 3 = 256 + 3 = 259 = 7 × 37, which is composite. Each is a counterexample to the corresponding claim.
14
Find counterexamples:
(i) if n is prime, then 2ⁿ − 1 is prime;
(ii) if n is even, then 2ⁿ + 1 is prime.
School Level Solution
(i) Let n = 11, which is prime. Then 2¹¹ − 1 = 2048 − 1 = 2047 = 23 × 89, so it is not prime.(ii) Let n = 6, which is even. Then 2⁶ + 1 = 64 + 1 = 65 = 5 × 13, so it is not prime.
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 divisible by 2 and 4? Why or why not?
School Level Solution
(i) Let n be a number divisible by 8, so n = 8k for some integer k. Then n = 2 × 4k, so n is divisible by 2, and n = 4 × 2k, so n is divisible by 4. Hence the statement is true.(ii) No. A number is divisible by 4 when the number formed by its last two digits is a multiple of 4, and every such number is automatically divisible by 2 as well. For example, 12 is divisible by both 2 and 4, but 12 is not divisible by 8. So checking divisibility by 2 and 4 is not enough to decide whether a number is divisible by 8.
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.
School Level Solution
1. If a number is divisible by 3, then the sum of its digits is a multiple of 3.
2. If the sum of the digits of a number is a multiple of 3, then the number is divisible by 3.Reason: Powers of 10 leave remainder 1 on division by 3 (10 = 3 × 3 + 1, 100 = 3 × 33 + 1, and so on), so a number and the sum of its digits always leave the same remainder when divided by 3. Hence each of the two if-then sentences is true.
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 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?Fig. 9.2 — Two thin sticks placed as the diagonals; joining their endpoints gives the quadrilateral.
School Level Solution
(i) (a) Yes, the two sticks must be of equal length. The given property says every quadrilateral of type Q has equal-length diagonals, and the sticks are used as those diagonals. If the sticks were unequal, the quadrilateral formed could never have equal diagonals, so it could not be of type Q.
(b) Yes, it can matter. The property is only necessary: equal diagonals alone do not guarantee that the quadrilateral obtained belongs to category Q, so how the sticks are put together can decide whether the result is of type Q.(ii) (a) Yes, the sticks must still be of equal length, because this property also requires the diagonals to be equal.
(b) No, it will not matter. Any quadrilateral formed by joining the endpoints of two equal sticks has equal diagonals, and the property says every such quadrilateral is of type Q — so any arrangement works.
In-text and Think and Reflect(6 questions)
1
In a triangle, if two angles are equal, prove that the sides opposite them are equal. (Use the hint in the book: draw an altitude from the vertex containing the third angle.)
School Level Solution
Let △ABC have ∠B = ∠C. Draw AD ⟂ BC. Since ∠B = ∠C, the right triangles △ABD and △ACD have ∠ABD = ∠ACD, ∠ADB = ∠ADC = 90°, and the common side AD. They are congruent by AAS. Hence AB = AC, the sides opposite the equal angles.
2
Give an example in which both a proposition and its converse are false.
School Level Solution
The proposition is false for n = 2: 2 is divisible by 2 but not by 3. Its converse, “If an integer is divisible by 3, then it is divisible by 2,” is false for n = 3. One counterexample to each universal statement is enough to show it is false.
3
The Pythagoras theorem says that if a triangle is right-angled with legs a,b and hypotenuse c, then a²+b²=c². Prove the converse.
School Level Solution
Construct a right triangle with perpendicular legs a and b; by Pythagoras its hypotenuse has length √(a²+b²)=c. The constructed triangle therefore has side lengths a,b,c, exactly as the given triangle does. The triangles are congruent by SSS. Corresponding to the constructed right angle, the given triangle’s angle opposite side c is also 90°. Thus the converse is true.
4
Give one real-life example of a true proposition whose converse is false, and state a counterexample to the converse.
School Level Solution
Proposition: If a person lives in Delhi, then the person lives in India. This is true because Delhi is in India.Converse: If a person lives in India, then the person lives in Delhi. This is false; for example, a person who lives in Mumbai lives in India but not in Delhi.
5
Give one example each from everyday life, geometry and number theory of a true proposition whose converse is false.
School Level Solution
Everyday life: If a person lives in Delhi, then they live in India; the converse is false because a person may live in Mumbai.Geometry: Every square is a rectangle, but a non-square rectangle is a counterexample to the converse.Number theory: Every multiple of 4 is even, but 6 is even and is not a multiple of 4.
6
The factor-partner argument says that a positive integer with an odd number of positive factors is a perfect square. Which implication does this prove? Why is the converse also true?
School Level Solution
Pair each positive divisor d of n with its partner n/d. If n has an odd number of divisors, at least one pair must consist of the same number twice, so d=n/d and n=d². This proves the first implication. Conversely, if n=k², the divisor pair (k,k) is the one repeated pair; every other divisor pairs with a distinct partner. Hence n has an odd number of positive divisors.
