Chapter 2: Relations and Functions (Sets and Functions)

Equal ordered pairs have equal components: $\frac{x}{3}+1=\frac{5}{3}$ ⇒ $\frac{x}{3}=\frac{2}{3}$ ⇒ x = 2; $y-\frac{2}{3}=\frac{1}{3}$ ⇒ y = 1.

x = 2; y = 1

n(A × B) = n(A) × n(B) = 3 × 3 = 9.

9

G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}; H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}.

{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}; {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

(i) False: P × Q = {(m, n), (m, m), (n, n), (n, m)} (four pairs). (ii) True. (iii) B ∩ φ = φ, and A × φ = φ: True.

(i) False; (ii) True; (iii) True

{(−1, −1, −1), (−1, −1, 1), (−1, 1, −1), (−1, 1, 1), (1, −1, −1), (1, −1, 1), (1, 1, −1), (1, 1, 1)} (2³ = 8 triplets).

{(−1, −1, −1), (−1, −1, 1), (−1, 1, −1), (−1, 1, 1), (1, −1, −1), (1, −1, 1), (1, 1, −1), (1, 1, 1)}

First components: A = {a, b}; second components: B = {x, y}.

A = {a, b}; B = {x, y}

(i) B ∩ C = φ, so A × (B ∩ C) = φ. A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}; A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}: no common pair, so (A × B) ∩ (A × C) = φ: equal. (ii) A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}; since 1, 2 ∈ B and 5, 6 ∈ D, each of these pairs is in B × D: A × C ⊂ B × D.

equal; A × C ⊂ B × D

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}: 4 elements, so 2⁴ = 16 subsets: φ; {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}; {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}; {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}; and A × B itself.

2⁴ = 16 subsets

x, y, z are distinct first elements and A has 3 elements: A = {x, y, z}. 1 and 2 are second elements and B has 2: B = {1, 2}.

A = {x, y, z}; B = {1, 2}

n(A)² = 9 ⇒ n(A) = 3; −1, 0, 1 appear as components, so A = {−1, 0, 1}. The other seven elements: (−1, −1), (−1, 1), (0, −1), (0, 0), (1, −1), (1, 0), (1, 1).

A = {−1, 0, 1}; (−1, −1), (−1, 1), (0, −1), (0, 0), (1, −1), (1, 0), (1, 1)

y = 3x ≤ 14 ⇒ x ≤ 4: R = {(1, 3), (2, 6), (3, 9), (4, 12)}. Domain {1, 2, 3, 4}; codomain A; range {3, 6, 9, 12}.

{1, 2, 3, 4}; A; {3, 6, 9, 12}

x = 1, 2, 3: R = {(1, 6), (2, 7), (3, 8)}; domain {1, 2, 3}, range {6, 7, 8}.

R = {(1, 6), (2, 7), (3, 8)}; {1, 2, 3}; {6, 7, 8}

Check each pair: |x − y| odd when one is odd and the other even. R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

(i) R = {(x, y) : y = x − 2, x ∈ P} (or x ∈ P, y ∈ Q). (ii) {(5, 3), (6, 4), (7, 5)}. Domain {5, 6, 7}, range {3, 4, 5}.

R = {(x, y) : y = x − 2, x ∈ P}; {(5, 3), (6, 4), (7, 5)}; {5, 6, 7}; {3, 4, 5}

(i) {(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) {1, 2, 3, 4, 6}. (iii) {1, 2, 3, 4, 6}.

(i) {(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) {1, 2, 3, 4, 6}; (iii) {1, 2, 3, 4, 6}

R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}: domain {0, 1, 2, 3, 4, 5}, range {5, 6, 7, 8, 9, 10}.

{0, 1, 2, 3, 4, 5}; {5, 6, 7, 8, 9, 10}

Primes 2, 3, 5, 7: {(2, 8), (3, 27), (5, 125), (7, 343)}.

{(2, 8), (3, 27), (5, 125), (7, 343)}

n(A × B) = 3 × 2 = 6, so the number of relations = 2⁶ = 64.

2⁶ = 64

For any integers a, b, a − b is an integer, so every pair is in R (R = Z × Z): domain Z, range Z.

Z

(i) Each first element appears once: function; domain {2, 5, 8, 11, 14, 17}, range {1}. (ii) Function; domain {2, 4, 6, 8, 10, 12, 14}, range {1, 2, 3, 4, 5, 6, 7}. (iii) 1 has two images (3 and 5): not a function.

(i) function, {1}; (ii) Function; (iii) not a function

(i) Defined for all x: domain R; |x| ≥ 0 so −|x| ≤ 0: range (−∞, 0]. (ii) Need 9 − x² ≥ 0 ⇔ −3 ≤ x ≤ 3: domain [−3, 3]; 9 − x² takes all values from 0 to 9, so the root takes 0 to 3: range [0, 3].

R; (−∞, 0]; [−3, 3]; [0, 3]

(i) 0 − 5 = −5 (ii) 14 − 5 = 9 (iii) −6 − 5 = −11.

(i) −5; (ii) 9; (iii) −11

(i) 32 (ii) 9 × 28/5 + 32 = 50.4 + 32 = 82.4 (iii) −18 + 32 = 14 (iv) 9C/5 + 32 = 212 ⇒ 9C/5 = 180 ⇒ C = 100.

(i) 32; (ii) 82.4; (iii) 14; (iv) C = 100

(i) x > 0 ⇒ 3x > 0 ⇒ 2 − 3x < 2, and it takes every value below 2: range (−∞, 2). (ii) x² ≥ 0 ⇒ x² + 2 ≥ 2: range [2, ∞). (iii) range R.

(i) (−∞, 2); (ii) [2, ∞); (iii) R

The only point covered by both rules of f is x = 3, where x² = 9 and 3x = 9: one image. So every x in [0, 10] has a single image: f is a function. For g at x = 2: x² = 4 but 3x = 6: two images, so g is not a function.

f is a function; g is not a function

$\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=$ 2.1.

2.1

x² − 8x + 12 = (x − 2)(x − 6) ≠ 0: domain R − {2, 6}.

R − {2, 6}

x − 1 ≥ 0 ⇒ domain [1, ∞); √(x − 1) takes every value ≥ 0: range [0, ∞).

[1, ∞); [0, ∞)

Defined for all x: domain R; |x − 1| ≥ 0 and takes every non-negative value: range [0, ∞).

R; [0, ∞)

y = x²/(1 + x²) = 1 − 1/(1 + x²). As 1 + x² ≥ 1, 0 < 1/(1 + x²) ≤ 1, so 0 ≤ y < 1 (y = 0 at x = 0; y → 1 but never reaches it). Range [0, 1).

[0, 1)

(f + g)(x) = 3x − 2; (f − g)(x) = x + 1 − 2x + 3 = 4 − x; (f/g)(x) = (x + 1)/(2x − 3), x ≠ 3/2.

(x) 3x − 2; (x) 4 − x; (x) (x + 1)/(2x − 3)

f(0) = b = −1; f(1) = a + b = 1 ⇒ a = 2. Check: f(2) = 3, f(−1) = −3. a = 2, b = −1.

a = 2, b = −1

(i) False: (2, 2) ∉ R since 2 ≠ 2² (true only for a = 1). (ii) False: (4, 2) ∈ R (4 = 2²) but (2, 4) ∉ R (2 ≠ 16). (iii) False: (16, 4) ∈ R and (4, 2) ∈ R, but (16, 2) ∉ R (16 ≠ 4).

(i) False; (ii) False; (iii) False

(i) Every pair has its first element in A and second in B, so f ⊂ A × B: yes, a relation. (ii) 2 has two images, 9 and 11: not a function.

yes, a relation; not a function

Take a = 0, b = 1: (0, 1) ∈ f; a = 0, b = 2: (0, 2) ∈ f. The element 0 has two images: not a function.

not a function

f(9) = 3, f(10) = 5, f(11) = 11, f(12) = 3, f(13) = 13: range {3, 5, 11, 13}.

{3, 5, 11, 13}