Mock 1 – Sets, Relations & Functions (Part 1)

Relation A = {(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)} contains both (2,3) and (3,2) with 2≠3, so it is not antisymmetric, and it contains (2,4) without (4,2), so it is not symmetric; but checking every composition (e.g. (2,3)&(3,2)→(2,2), (2,3)&(3,4)→(2,4), (3,2)&(2,4)→(3,4)) shows the required pair is always present, so A is transitive — 'only transitive' (II). Relation B = {(2,4),(4,2)} is symmetric since both (2,4) and (4,2) are present, but it is not transitive because (2,4)&(4,2) would require (2,2), which is absent — 'only symmetric' (III). Relation C = {(1,2),(2,3),(3,4)} has no pair whose reverse is also present, so it is (vacuously) antisymmetric, but (1,2)&(2,3) would require (1,3), which is missing, so it is not transitive — 'antisymmetric' (I). Relation D = {(1,3),(1,4),(2,3),(2,4),(3,1),(3,4)} contains (1,3) and (3,1) with 1≠3 (so not antisymmetric, and also not symmetric since (1,4) lacks its reverse), and (1,3)&(3,1) would require (1,1), which is absent, so it is not transitive either — 'neither symmetric nor transitive' (IV). The match is A-II, B-III, C-I, D-IV.

Count elements of each inner set first: {a, b, {a,b}} has 3 elements, so |P(A)| = 2³ = 8. {∅, a, {a}, {{a}}} has 4 distinct elements, so |P(B)| = 2⁴ = 16. P(∅) = {∅} has 1 element, so P(P(∅)) has 2¹ = 2 elements. P({a}) = {∅, {a}} has 2 elements, so P(P({a})) has 2² = 4 elements. {{a,b}, c} has 2 elements, so |P(E)| = 2² = 4. So the cardinalities are C=2, D=4, E=4, A=8, B=16, giving the non-decreasing order C, D, E, A, B.

Using R∘S = {(a, c) : there exists b with (a, b) ∈ R and (b, c) ∈ S}, i.e. applying R first and then S: from (1, 2) ∈ R and (2, 5) ∈ S we get (1, 5); from (3, 4) ∈ R and (4, 2) ∈ S we get (3, 2); from (2, 2) ∈ R and (2, 5) ∈ S we get (2, 5). Combining these pairs gives R∘S = {(1, 5), (3, 2), (2, 5)}.

∅ has no elements (A false). {0} contains only 0 (B false). {∅} contains ∅ (C true). The only element of {∅} is ∅, not {∅} (D false). Every element of {∅}, namely ∅, is in {∅, {∅}}, and the sets differ, so {∅} is a proper subset (E true).

A reflexive relation must contain (1,1), (2,2), (3,3) and (4,4). R₁ misses (3,3), R₂ misses (3,3) and (4,4), and R₄ has no loops. Only R₃ and R₅ contain all four pairs.

xv = yu means x/y = u/v (equality of fractions). xy = yx (reflexive), the condition is symmetric, and equal ratios chain (transitive). It is an equivalence relation, not asymmetric.

Dividing by abcd: (b + c)/(bc) = (a + d)/(ad), i.e. 1/c + 1/b = 1/d + 1/a, i.e. 1/a − 1/b = 1/c − 1/d. Equality of the value 1/a − 1/b is reflexive, symmetric and transitive.

With aₖ = f(2ᵏ): aₖ = 5aₖ₋₁ + 3, a₀ = 7. The solution is aₖ = 5ᵏ·7 + 3(5ᵏ − 1)/4 = 5ᵏ(31/4) − 3/4.

The characteristic equation r² − 8r + 16 = 0 has the double root 4, so the general solution is (c₁ + c₂n)4ⁿ. 4ⁿ and n·4ⁿ satisfy it; 1, 2ⁿ and (−4)ⁿ do not.

Cardinalities (number of elements, which are sets): A3 = 1, A1 = 2, A4 = 3, A2 = 4, A5 = 5.

(2, 2) is missing (not reflexive). (1, 2) is present without (2, 1) (not symmetric). (3, 1) and (1, 2) are present without (3, 2) (not transitive).

Because the inequalities are strict, (x, y) R (x, y) never holds, so R is not reflexive. A partial order and an equivalence relation both need reflexivity.

|x − x| = 0 < ε and |x − y| = |y − x|, so it is reflexive and symmetric. It is not transitive: 0 ~ 0.0003 and 0.0003 ~ 0.0006, but |0 − 0.0006| > ε.