Relations

The direct answer is option B, 2 to the power n squared. A relation on A is any subset of A × A. If A has n elements, then the ordered product A × A has n times n = n² ordered pairs. For each of these n² pairs, there are exactly two independent choices: include it in the relation or leave it out. Therefore the total number of subsets, and hence relations, is 2^(n²). Option B is correct. Option A, 2^n, counts choices based on only n objects and ignores that ordered pairs number n². Option C, n², counts the total possible ordered pairs, not all possible subsets of them. Option D, n!, counts a permutation-type arrangement and has no role here. For example, if there are two possible pairs, the subsets are four in number, not two. Memory cue: relations are subsets of the Cartesian product, so count pairs first, then use two choices for each pair.

The inverse relation R⁻¹ is formed by reversing the coordinates of every ordered pair in R. Thus (x, y) belongs to R exactly when (y, x) belongs to R⁻¹. Reversing the given pairs produces R⁻¹ = {(2, 1), (1, 2), (3, 3)}. The pair (2, 1) is therefore in the inverse, so option A is correct. The pair (1, 3) would require (3, 1) to be in R, which it is not; similarly, (3, 1) would require (1, 3), and (2, 3) would require (3, 2). Notice that (1, 2) and (2, 1) are both already present in R, so reversing them leaves the same two pairs in the inverse, while (3, 3) remains unchanged.

The direct answer is option A, (7,4)∈R. A relation is symmetric when, for every pair (a,b) in it, the reversed pair (b,a) is also in it. The question gives (4,7)∈R. Reverse the positions: the first number becomes 7 and the second becomes 4, so (7,4)∈R must follow. Option B, (4,4)∈R, is not required; symmetry does not change both entries to 4. Option C, (7,7)∈R, is also not required; symmetry only reverses the pair. Option D says (4,7) is not in R, directly contradicting the information given in the question. The conclusion is certain even if the relation contains other pairs, because symmetry guarantees the reverse of every included pair. Memory cue: symmetric means mirror pair—if (a,b) is present, (b,a) is present.

In a transitive relation, ( (1,4) ) and ( (4,9) ) require ( (1,9) ). In exams, the middle element (4) is the common element.

The direct answer is option C, (2,3). A pair (a,b) belongs to R when both entries are from A={1,2,3,4} and their sum is at most 5, meaning a+b≤5. Check every option: for A, (4,2) has sum 4+2=6, so it does not belong. For B, (3,3) has sum 3+3=6, so it does not belong. For C, (2,3) has sum 2+3=5, and 5≤5 is true, so it belongs. Equality is allowed because the symbol is ≤, not <. For D, (4,4) has sum 8, so it does not belong. Thus only option C satisfies the condition. The entries are ordered, but here the sum check gives the same result if their positions are exchanged. Memory cue: with ≤, the boundary value is included.

Because (4+2=6), which is greater than (5), ( (4,2)\notin R ). In exams, apply the condition directly to the ordered pair.

Direct answer: option A, {(1,2),(2,4),(3,6)}. A relation from A to B uses the first component from A and the second from B, then checks b=2a. For a=1, b=2×1=2, giving (1,2). For a=2, b=2×2=4, giving (2,4). For a=3, b=2×3=6, giving (3,6). These second entries all lie in B. A is therefore correct. B reverses every pair, so it does not follow the direction A to B. C uses b=2a+2 rather than b=2a: for example, when a=1 it gives 4 instead of 2. D incorrectly uses 4 as a first component, but 4 is not in A, and it also omits the pair for a=3. The relation is obtained by testing every element of A once. Memory cue: start with a, double it, and write (a,2a).

The pairs are ( (3,2),(4,3),(5,4) ), so the range is ( {2,3,4} ). In exams, take second components for range.

The condition gives pairs ( (3,2),(4,3),(5,4) ), so the domain is ( {3,4,5} ). In exams, take first components for domain.

Since ( (1,2)\in R ) and ( (2,3)\in R ), ( (1,3) ) is required, but it is missing. Therefore (R) is not transitive.

The direct answer is option C, 12. The set A has 4 elements. An ordered pair in A×A is made by choosing a first element in 4 ways and a second element in 4 ways, so there are 4×4=16 total ordered pairs. The relation requires a≠b, so we must remove pairs whose two entries are equal. The equal pairs are (1,1), (2,2), (3,3), and (4,4): exactly 4 pairs. Therefore the number remaining is 16-4=12. Equivalently, for each first element there are 3 choices for a different second element, giving 4×3=12. Option A, 4, counts only one possible different partner per element and misses two others. Option B, 8, is too small and has no correct counting basis here. Option C, 12, follows both methods. Option D, 16, counts all pairs, including forbidden equal pairs. Memory cue: for n elements and a≠b, count n(n-1).

A relation R on A is reflexive if and only if every element is related to itself; in symbols, (a, a) must belong to R for every a ∈ A. Here the defining condition is a ≠ b. For a diagonal pair such as (1, 1), the statement 1 ≠ 1 is false, so (1, 1) is not in R. The same reasoning excludes (2, 2) and (3, 3). Since even one required self-pair is missing—and in fact all three are missing—the relation is not reflexive. Option B states this correctly. The presence of (1, 2) confirms that some unequal pairs belong to R, but it cannot establish reflexivity; options A, C, and D confuse an ordinary membership fact with the defining self-relatedness requirement.

The governing definition is that a relation R on A is symmetric when, for every a and b in A, (a,b) ∈ R implies (b,a) ∈ R. Here membership means a ≠ b. Inequality is reversible: if a and b are different, then b and a are also different. Thus, for example, (1,2) and (2,1) both belong to R, and the same holds for every distinct pair. The missing diagonal pairs such as (1,1) do not affect symmetry; they show that the relation is not reflexive. Option C actually supports symmetry rather than disproving it, while D gives a false reason because the relation is not reflexive. Therefore option A is correct.

Since (a\leq a) for every (a), (R) is reflexive. But ( (1,2)\in R ) while ( (2,1)\notin R ), so it is not symmetric.

A relation on A is reflexive when every element is related to itself; in symbols, (x,x) must belong to R for every x in A. Since A contains 1, 2, and 3, the required diagonal pairs are (1,1), (2,2), and (3,3), and all three are explicitly present. The additional pair (1,2) does not remove reflexivity. The relation is not symmetric, because (1,2) is present but (2,1) is absent. It is actually transitive as well: the only nontrivial chain using (1,2) continues through (2,2), and (1,2) is already present. Nevertheless, the question asks which listed property it satisfies, and option B states the certain required property. Option D is false because the relation has four pairs.

By definition, an equivalence relation must satisfy three properties simultaneously. Reflexivity requires (a,a) ∈ R for every a in A. Symmetry requires that (a,b) ∈ R imply (b,a) ∈ R. Transitivity requires that (a,b) and (b,c) in R imply (a,c) in R. All three conditions are necessary; omitting any one means the relation is not an equivalence relation. Antisymmetry is not part of the definition. In fact, a relation can be both symmetric and antisymmetric only in restricted situations, but that combination is not what defines equivalence. Option D lacks reflexivity, and option C lacks symmetry and transitivity. Therefore option A states exactly the required trio.

A relation from A to B is any subset of the Cartesian product A × B. Since A has 3 elements and B has 3 elements, the product contains 3 × 3 = 9 distinct ordered pairs. A relation with exactly 2 pairs is therefore a two-element subset of these 9 pairs. The number of such subsets is the combination C(9,2) = 9!/(2!7!) = 9 × 8/2 = 36. Hence option C is correct. The order inside an ordered pair is already fixed by the direction from A to B; the counting step chooses which two pairs are included. The value 9 counts all possible product pairs, not two-pair relations, while 18 and 72 result from incorrect counting.

For each fixed first coordinate a in A, the congruence a+b ≡ 0 (mod 5) determines exactly one required residue for b: b ≡ −a (mod 5). Because A contains exactly one representative of each residue modulo 5, this value of b is always in A. Listing the pairs gives (0,0), (1,4), (2,3), (3,2), and (4,1). Thus each of the five possible values of a produces one and only one valid ordered pair, for a total of 5. Option B is correct. The value 25 is the size of A×A before imposing the condition; 10 and 4 do not count the unique valid partner for every first coordinate.

The relation ‘having the same birth month’ satisfies all three defining properties of an equivalence relation. It is reflexive because every student has the same birth month as himself or herself. It is symmetric because if a and b have the same month, then b and a necessarily have the same month. It is transitive because if a shares a month with b and b shares that month with c, then a and c share it too. Therefore R is an equivalence relation, so option A is correct. It is not irreflexive, since every pair (a,a) is included, and it is transitive, making option D false. Although antisymmetry may hold in special cases, it is not the complete defining classification here.

Both a and b must belong to A = {1,2,3,4,5}, and the relation requires a = 2b. Test each possible value of b: for b = 1, a = 2, which is in A, giving (2,1); for b = 2, a = 4, giving (4,2). For b = 3, 4, or 5, the corresponding values a = 6, 8, or 10 are outside A, so they produce no valid ordered pair. Hence R = {(2,1),(4,2)} and its cardinality is 2. The relation is counted by valid ordered pairs, not merely by possible b-values or by all numerical solutions without checking membership in A. Therefore option B is correct; options A, C, and D result from omitting or adding invalid cases.

Substituting \(x=0,1,2\) into \(y=2x+1\) gives \(y=1,3,5\), respectively. Hence, the ordered pairs are \((0,1),(1,3),(2,5)\), so option A is correct. Option B results from omitting the \(+1\) term. Exam tip: in an ordered pair, the first component is \(x\) and the second component is its corresponding \(y\)-value.