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Relations and Functions - Important Questions

⏱️ Est. Revision Time: 15 Mins
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1
Exercise 2.1
If A = {1, 2, 3} and B = {4, 5}, find A × B and B × A. Show that A × B is not equal to B × A.
▶ Show Detailed Solution
A × B = {(1,4), (1,5), (2,4), (2,5), (3,4), (3,5)}
B × A = {(4,1), (4,2), (4,3), (5,1), (5,2), (5,3)}

n(A × B) = 3 × 2 = 6 and n(B × A) = 2 × 3 = 6 (equal cardinalities),
but the ordered pairs differ. The order in each pair matters,
so A × B ≠ B × A.
2
Exercise 2.1
Let R be a relation on N defined by R = {(x, y) : y = 2x, x, y ∈ N}. Write R in roster form for x ≤ 5, and find its domain and range.
▶ Show Detailed Solution
For x = 1, 2, 3, 4, 5:
y = 2, 4, 6, 8, 10

R = {(1,2), (2,4), (3,6), (4,8), (5,10)}
Domain = {1, 2, 3, 4, 5}
Range = {2, 4, 6, 8, 10}
3
Exercise 2.1
Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by R = {(a, b) : a, b ∈ A, b is exactly divisible by a}.<br>(i) Write R in roster form.<br>(ii) Find the domain and range of R.<br>(iii) Is R an equivalence relation? Give reason.
▶ Show Detailed Solution
(i) b is divisible by a (a divides b):
1 divides all: (1,1),(1,2),(1,3),(1,4),(1,6)
2 divides: 2,4,6 → (2,2),(2,4),(2,6)
3 divides: 3,6 → (3,3),(3,6)
4 divides: 4 → (4,4)
6 divides: 6 → (6,6)

R = {(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)}

(ii) Domain = {1, 2, 3, 4, 6}, Range = {1, 2, 3, 4, 6}

(iii) Not symmetric: (1,2) ∈ R but (2,1) ∉ R. Hence R is NOT an equivalence relation (equivalence requires reflexivity, symmetry, transitivity).
4
Exercise 2.1
If f(x) = x² - 5x + 4, find f(2), f(-1), and the values of x for which f(x) = 0.
▶ Show Detailed Solution
f(2) = 4 - 10 + 4 = -2
f(-1) = 1 + 5 + 4 = 10

f(x) = 0: x² - 5x + 4 = 0
(x - 1)(x - 4) = 0
x = 1 or x = 4
5
Exercise 2.1
Find the domain and range of the function f(x) = √(x - 2) + 1.
▶ Show Detailed Solution
The expression under the square root must be non-negative:
x - 2 ≥ 0 ⇒ x ≥ 2

Domain = [2, ∞)

As x ranges over [2, ∞), √(x-2) ≥ 0, so f(x) ≥ 1.
Range = [1, ∞)
6
Exercise 2.1
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1,4), (2,5), (3,6)}. Is f a function from A to B? Is it onto? Give reason.
▶ Show Detailed Solution
Each element of A appears exactly once as first component, so f is a function from A to B.

Codomain = {4,5,6,7}; Range = {4,5,6}.
Since 7 is never mapped to, the function is NOT onto (surjective).