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Practice Exercise 1.1

Relations & Functions: Detailed step-by-step solutions for 16 similar practice questions.

Chapter Index Start Solving
Q1 Types Q2 Real Q3 Finite Q4 Inequality Q5 Cubes Q6 Symm Q7 Books Q8 Modulus Q9 Integer Q10 Ex Q11 Points Q12 Tri Q13 Poly Q14 Lines Q15 MCQ Q16 MCQ

Q1. Determine Properties

Similar to Q1(i)
Determine whether the relation R in the set A = {1, 2, ..., 10} defined as R = {(x, y) : 2x - y = 0} is reflexive, symmetric or transitive.
View Step-by-Step Solution
Step 1: Understand the Relation
The condition is 2x - y = 0, which means y = 2x.
We need to find pairs (x, y) where both x and y are in set A = {1, 2, ..., 10}.
Step 2: List the Elements
Let's find the pairs:
• If x = 1, y = 2(1) = 2. Pair: (1, 2)
• If x = 2, y = 2(2) = 4. Pair: (2, 4)
• If x = 3, y = 2(3) = 6. Pair: (3, 6)
• If x = 4, y = 2(4) = 8. Pair: (4, 8)
• If x = 5, y = 2(5) = 10. Pair: (5, 10)
• If x = 6, y = 12. But 12 is not in A. Stop here.
R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)}
Step 3: Check Reflexivity
A relation is Reflexive if (a, a) ∈ R for every a ∈ A.
Does (1, 1) belong to R? No, because 2(1) - 1 ≠ 0.
Conclusion: Not Reflexive.
Step 4: Check Symmetry
A relation is Symmetric if (a, b) ∈ R implies (b, a) ∈ R.
We have (1, 2) ∈ R. Is (2, 1) in R? No, because 2(2) - 1 = 3 ≠ 0.
Conclusion: Not Symmetric.
Step 5: Check Transitivity
A relation is Transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R.
We have (1, 2) ∈ R and (2, 4) ∈ R. For transitivity, (1, 4) must be in R.
Is (1, 4) in R? Check: 2(1) - 4 = -2 ≠ 0.
Conclusion: Not Transitive.
Answer: Neither Reflexive, nor Symmetric, nor Transitive.
Similar to Q1(ii)
Determine properties of Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 3 and x < 5}.
View Step-by-Step Solution
Step 1: List Elements
Condition: x is a natural number less than 5. So x ∈ {1, 2, 3, 4}.
• x=1, y=1+3=4 → (1, 4)
• x=2, y=2+3=5 → (2, 5)
• x=3, y=3+3=6 → (3, 6)
• x=4, y=4+3=7 → (4, 7)
R = {(1, 4), (2, 5), (3, 6), (4, 7)}
Step 2: Check Reflexive
(1, 1) is not in R because 1 ≠ 1 + 3.
Not Reflexive.
Step 3: Check Symmetric
(1, 4) ∈ R. But (4, 1) ∉ R because 1 ≠ 4 + 3.
Not Symmetric.
Step 4: Check Transitive
We need pairs like (a, b) and (b, c).
In our set R = {(1, 4), (2, 5), (3, 6), (4, 7)}, we do have one chain:
(1, 4) is in R. Does (4, something) exist? Yes, (4, 7).
For transitivity, (1, 7) must be in R. Is it? No.
Not Transitive.
Answer: Neither Reflexive, nor Symmetric, nor Transitive.
Similar to Q1(iii)
Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is a multiple of x}.
View Step-by-Step Solution
Step 1: Check Reflexivity
Is every number a multiple of itself? Yes, x = 1 * x.
So (1,1), (2,2)... are in R.
Reflexive: Yes.
Step 2: Check Symmetry
(2, 4) ∈ R because 4 is a multiple of 2.
Is (4, 2) ∈ R? Is 2 a multiple of 4? No.
Symmetric: No.
Step 3: Check Transitivity
If y is a multiple of x (y = mx) and z is a multiple of y (z = ny), then z = n(mx) = (nm)x.
So z is a multiple of x.
Transitive: Yes.
Answer: Reflexive and Transitive, but not Symmetric.
Similar to Q1(iv)
Relation R in the set Z of all integers defined as R = {(x, y) : x - y is an even integer}.
View Step-by-Step Solution
Step 1: Reflexivity
Check x - x. Result is 0. 0 is an even integer.
So (x, x) ∈ R for all x.
Reflexive: Yes.
Step 2: Symmetry
If x - y is even, then -(x - y) = y - x is also even.
So if (x, y) ∈ R, then (y, x) ∈ R.
Symmetric: Yes.
Step 3: Transitivity
Let x - y = 2m (even) and y - z = 2n (even).
Add them: (x - y) + (y - z) = 2m + 2n.
x - z = 2(m + n), which is even.
So (x, z) ∈ R.
Transitive: Yes.
Answer: Equivalence Relation (Reflexive, Symmetric, and Transitive).
Similar to Q1(v)
Relation R in the set of human beings defined as R = {(x, y) : x is exactly 10 cm taller than y}.
View Step-by-Step Solution
Analysis
Reflexive: A person cannot be 10cm taller than themselves. (No)
Symmetric: If A is taller than B, B cannot be taller than A. (No)
Transitive: If A is 10cm taller than B, and B is 10cm taller than C, then A is 20cm taller than C, not 10cm. (No)
Answer: Neither Reflexive, Symmetric, nor Transitive.

Q2 - Q6. Proving Properties

Similar to Q2
Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b³} is neither reflexive nor symmetric nor transitive.
View Step-by-Step Solution
Step 1: Check Reflexivity
Take a = 1/2. Is 1/2 ≤ (1/2)³?
0.5 ≤ 0.125 is False.
Not Reflexive.
Step 2: Check Symmetry
Take a=1, b=2. 1 ≤ 2³ (1≤8) is True.
Swap: Is 2 ≤ 1³? 2 ≤ 1 is False.
Not Symmetric.
Step 3: Check Transitivity
Let a=100, b=5, c=2.
(100, 5): 100 ≤ 5³ (100 ≤ 125) → True.
(5, 2): 5 ≤ 2³ (5 ≤ 8) → True.
Check (100, 2): Is 100 ≤ 2³? 100 ≤ 8 is False.
Not Transitive.
Answer: Proven neither R, S, nor T.
Similar to Q3
Check whether the relation R defined in the set {1, 2, 3, 4, 5} as R = {(a, b) : b = a + 2} is reflexive, symmetric or transitive.
View Step-by-Step Solution
Step 1: List Pairs
R = {(1, 3), (2, 4), (3, 5)}
Step 2: Verify Properties
Reflexive: (1,1) is missing. No.
Symmetric: (1,3) is in R, but (3,1) is not. No.
Transitive: We have (1,3) and (3,5). For transitivity, (1,5) must be in R. But 5 ≠ 1 + 2. So (1,5) is not in R. No.
Answer: Neither R, S, nor T.
Similar to Q4
Show that the relation R in R defined as R = {(a, b) : a ≥ b}, is reflexive and transitive but not symmetric.
View Step-by-Step Solution
Explanation
Reflexive: a ≥ a is always true. (Yes)
Symmetric: 5 ≥ 2 is true, but 2 ≥ 5 is false. (No)
Transitive: If a ≥ b and b ≥ c, mathematically it implies a ≥ c. (Yes)
Answer: Reflexive and Transitive only.
Similar to Q5
Check whether the relation R in R defined by R = {(a, b) : a ≤ b²} is reflexive, symmetric or transitive.
View Step-by-Step Solution
Analysis
Reflexive: Counter-example a=0.5. 0.5 ≤ 0.25 is False. (No)
Symmetric: 1 ≤ 2² (True). But 2 ≤ 1² is False. (No)
Transitive: 2 ≤ (-2)², -2 ≤ (-1)². But 2 ≤ (-1)² is 2 ≤ 1 (False). (No)
Answer: Neither R, S, nor T.
Similar to Q6
Show that the relation R in the set {a, b, c} given by R = {(a, b), (b, a)} is symmetric but neither reflexive nor transitive.
View Step-by-Step Solution
Verification
Reflexive: (a,a) is missing. No.
Symmetric: (a,b) is there, (b,a) is there. Yes.
Transitive: (a,b) ∈ R and (b,a) ∈ R. We need (a,a) to be in R for transitivity. It is missing. No.
Answer: Symmetric only.

Q7 - Q14. Equivalence Relations

Similar to Q7
Show that the relation R in the set A of all the books in a library, given by R = {(x, y) : x and y have the same price} is an equivalence relation.
View Step-by-Step Solution
Proof
Reflexive: Every book has the same price as itself. (Yes)
Symmetric: If book x has same price as y, y has same price as x. (Yes)
Transitive: If Price(x) = Price(y) and Price(y) = Price(z), then Price(x) = Price(z). (Yes)
Answer: Equivalence Relation.
Similar to Q8
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a - b| is even}, is an equivalence relation.
View Step-by-Step Solution
Part 1: Equivalence
Reflexive: |a-a|=0 (even). Yes.
Symmetric: |a-b| is same as |b-a|. Yes.
Transitive: Sum of two even numbers is even. Yes.
Part 2: Subsets
Elements {1, 3, 5} are all odd. Difference of any two odd numbers is even. So they are related.
Elements {2, 4} are all even. Difference is even. Related.
Odd - Even is Odd (not even). So {1,3,5} is not related to {2,4}.
Answer: Validated as Equivalence Relation.
Similar to Q9
Show that R in the set A = {x ∈ Z : 0 ≤ x ≤ 10}, given by R = {(a, b) : |a - b| is a multiple of 5}, is an equivalence relation. Find elements related to 2.
View Step-by-Step Solution
Step 1: Check Equivalence
Standard proof for divisibility relations applies (0 is divisible, symmetry holds, sum of divisible numbers is divisible).
Step 2: Find Related Elements
We need x in A such that |x - 2| is a multiple of 5.
|x - 2| = 0 ⇒ x = 2
|x - 2| = 5 ⇒ x = 7 (since 7-2=5)
Next multiple is 10, implies x=12, but 12 ∉ A.
Set of elements related to 2 = {2, 7}
Similar to Q10
Give an example of a relation that is Symmetric but neither Reflexive nor Transitive.
View Solution
Example
Let A = {5, 6, 7}.
Define R = {(5, 6), (6, 5)}.
• Symmetric: Yes.
• Reflexive: No (5,5 missing).
• Transitive: (5,6) & (6,5) are there, but (5,5) is missing. No.
R = {(5, 6), (6, 5)}
Similar to Q11
Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of P from origin is same as distance of Q from origin} is an equivalence relation.
View Step-by-Step Solution
Proof
Reflexive: OP = OP. Yes.
Symmetric: If OP = OQ, then OQ = OP. Yes.
Transitive: If OP = OQ and OQ = OS, then OP = OS. Yes.
Geometric Interpretation
The set of all points P related to a fixed point k (≠ origin) are all points at the same distance from the origin. This forms a circle.
Answer: Equivalence Relation; Locus is a Circle.
Similar to Q12
Show that relation R defined in set A of all triangles as R = {(T1, T2) : T1 is congruent to T2} is an equivalence relation.
View Solution
Congruence is always Reflexive (same as self), Symmetric (A≅B then B≅A), and Transitive (A≅B, B≅C then A≅C).
Similar to Q13
Show that R defined in set of all polygons as R = {(P1, P2) : P1 and P2 have same number of vertices} is equivalence. What is the set of elements related to a triangle?
View Solution
Part 1: Proof
"Same number of vertices" is an equality relation, which is always an equivalence relation.
Part 2: Related Set
A triangle has 3 vertices. Any polygon related to it must also have 3 vertices.
Set of all triangles.
Similar to Q14
Let L be set of lines in XY plane. R = {(L1, L2) : L1 is parallel to L2}. Show equivalence. Find set of lines related to y = 3x + 5.
View Step-by-Step Solution
Step 1: Equivalence
Line is parallel to itself (Reflexive). If L1||L2 then L2||L1 (Symmetric). If L1||L2 and L2||L3 then L1||L3 (Transitive).
Step 2: Find Related Lines
Given line: y = 3x + 5.
Slope m = 3.
Any line related to it must have the same slope.
Set of lines: y = 3x + c (where c is any constant).

Q15 - Q16. MCQs

Similar to Q15
Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
View Solution
Step 1: Check Reflexive
Set is {1, 2, 3, 4}. R contains (1,1), (2,2), (3,3), (4,4).
Yes, Reflexive. (Eliminates C)
Step 2: Check Symmetric
We have (1, 2). Do we have (2, 1)? No.
Not Symmetric. (Eliminates A and D)
Step 3: Check Transitive
(1, 3) and (3, 2) are in R. Is (1, 2) in R? Yes.
It holds for other pairs too.
Yes, Transitive.
Correct Option: (B) Reflexive and transitive but not symmetric.
Similar to Q16
Let R be the relation in the set N given by R = {(a, b) : a = b - 2, b > 6}. Choose the correct answer.

(A) (2, 4) ∈ R
(B) (3, 8) ∈ R
(C) (6, 8) ∈ R
(D) (8, 7) ∈ R
View Solution
Step 1: Apply Condition
Condition 1: b must be > 6.
Condition 2: a = b - 2.
Step 2: Check Options
(A) b=4 (Not > 6). Incorrect.
(B) b=8 (>6). a = 8-2 = 6. But option says a=3. Incorrect.
(C) b=8 (>6). a = 8-2 = 6. Option matches (6, 8). Correct.
(D) b=7 (>6). a = 7-2 = 5. Option says a=8. Incorrect.
Correct Option: (C) (6, 8) ∈ R