Mock 2 – Sets, Relations & Functions (Part 2)

Symmetric difference = (A − B) ∪ (B − A). By De Morgan, A − (B ∪ C) = (A − B) ∩ (A − C) and A − (B ∩ C) = (A − B) ∪ (A − C). Intersection distributes over difference: A ∩ (B − C) = (A ∩ B) − (A ∩ C).

(2,2) is missing, so R is not reflexive. (1,2) is in R but (2,1) is not, so not symmetric. (3,1) and (1,2) are in R but (3,2) is not, so not transitive.

(x, y) R (x, y) would need x < x, so R is not reflexive. It is therefore neither a partial order nor an equivalence relation (it is a strict, irreflexive order).

|x − x| = 0 < ε, so R_ε is reflexive, and |x − y| = |y − x| makes it symmetric. It is not transitive, so the correct answer is (A) and (B) only.

Note on the official key: the official key says option A. Official/PW key is D. R_ε is not transitive: 0 and 0.0004 are related, 0.0004 and 0.0008 are related, but 0 and 0.0008 are not.

(c, c) and (d, d) are missing, so R is not reflexive. No pair appears in both directions, so it is antisymmetric (and not symmetric).

Under divisibility, GLB = gcd and LUB = lcm. gcd(3, 9, 12) = 3, lcm = 36; gcd(1, 2, 4, 5, 10) = 1, lcm = 20.

'Brother of' is not symmetric (b may be a sister) and not reflexive, so it is neither an equivalence nor a partial order. It is transitive (taken as the intended answer).

A reflexive relation must contain the 4 diagonal pairs; each of the other 16 − 4 = 12 pairs is in or out: 2¹².

No element divides all the others (3 does not divide 5), so there is no least element, and no element is divisible by all the others (45 is not divisible by 24), so there is no greatest element.

1 → 3 → 2 gives (1, 2); 3 → 1 → 3 gives (3, 3). Row 1 becomes 111, row 2 stays 010, row 3 becomes 111.

≤ defined by x ∨ y = y is a partial order: it is transitive (i) and antisymmetric (ii). In (iii) and (iv) the conclusion is already one of the hypotheses (x < y implies x ≤ y), so both hold too.

Note: no option is fully correct; the official key (A) is shown. As printed, all four statements are true (≤ is a partial order, and (iii)/(iv) follow directly from x < y). No option is literally right; the official key is (A) (iii) only.

For convex problems a local minimum is global, the set of minima is convex (not concave), and a strictly convex function has at most one minimum.

f(x) = f(y) → x = y is injectivity; ∀y ∃x f(x) = y is surjectivity; ∀x f(x) = k is a constant function.

Only the identity relation is reflexive, symmetric and transitive. (A) misses (1,1); (C) misses (2,2); (D) is not symmetric.

Differing by a constant is reflexive (k = 0), symmetric (−k) and transitive (k₁ + k₂). (A) isn't reflexive, (B) isn't transitive and (C) isn't reflexive.

⊆ is reflexive, antisymmetric and transitive. (Z, ≤) is total; ≠ isn't reflexive; the graph (loops on a and b plus a → b) is a partial order.