Relations

Since (3) does not divide (4), ( (3,4) \notin R ). In exams, check the direction of divisibility carefully.

The direct answer is option A, {(3,1),(4,2),(5,3)}. The inverse relation is obtained by reversing the order of every ordered pair: (a,b) becomes (b,a). Start with (1,3), which becomes (3,1). Next, (2,4) becomes (4,2). Finally, (3,5) becomes (5,3). Collecting these gives exactly option A. Option B is the original relation, so no components have been reversed. Option C contains incorrect changes: it does not reverse the given pairs in a consistent way. Option D contains the original pairs in reverse listing order, but merely changing the order in which pairs are written is not the same as taking the inverse; each pair itself must be reversed. Therefore A is correct. Exam cue: inverse means flip every arrow, not merely reorder the list.

For a reflexive relation, ( (a,a)\in R ) is required for every (a\in A). Here ( (3,3)\notin R ), so it is not reflexive.

For every ( (a,b)\in R ), ( (b,a)\in R ) is also present, so (R) is symmetric. In exams, match each pair with its reverse.

Since ( (1,2)\in R ) and ( (2,3)\in R ) imply ( (1,3)\in R ). For transitivity, connect the middle element carefully.

If (a-b) is even, then (a) and (b) have the same parity. In exams, make odd and even groups separately.

A relation from (A) to (B) must be a subset of (A\times B). So first components must come from (A), and second components must come from (B).

Total relations are (2^{|A\times B|}=2^{12}), and removing the empty relation gives (2^{12}-1). In exams, remember to subtract (1) for non-empty relations.

In every ordered pair, the second component is (2) times the first, so (b=2a). In exams, test the rule on all pairs.

An empty relation has no ordered pair, so it is ( \varnothing ). In exams, keep empty relation and identity relation separate.

Direct answer: option A, (2,1), belongs to R. The defining condition is a=b+1, so substitute the first component for a and the second for b. A: for (2,1), 2=1+1, which is true; hence it belongs. B: for (1,2), the condition would require 1=2+1=3, false. C: for (3,5), it would require 3=5+1=6, false. D: for (4,4), it would require 4=4+1=5, false. All numbers shown are natural numbers, so the membership test is simply the equation and the order of the components. Do not use b=a+1, because that reverses the given relation and changes the answer. The relation contains pairs in which the first entry is exactly one greater than the second. Memory cue: “first = second + 1.”

For reflexivity, all pairs ( (a,a) ) are required, and they are present here. The extra pair ( (1,2) ) does not break reflexivity.

For symmetry, the reverse of ( (1,2) ) is ( (2,1) ), and the reverse of ( (2,2) ) is itself. ( (1,1) ) is not necessary.

For (a\leq b), the number of pairs is (4+3+2+1=10). In exams, include diagonal pairs ( (a,a) ) also.

The first component must be greater than some smaller (b), so (2,3,4) appear. (1) is not greater than any (b\in A).

The second component must have some greater (a\in A), so (1,2,3) appear. There is no element in (A) greater than (4).

A relation from (A) to (B) is any subset of (A\times B). In exams, treat a relation as a set of ordered pairs.

It is reflexive because ( (1,1),(2,2),(3,3) ) are present, and symmetric because ( (1,2),(2,1) ) are paired. It is not universal because all pairs are not present.

The direct answer is option A, (1,3). Transitivity requires that if (a,b) and (b,c) belong to R, then (a,c) must also belong to R. Here (1,2) is present and (2,3) is present. The middle entries match at 2, so the required shortcut is from 1 directly to 3: (1,3). Adding it makes the displayed chain satisfy the transitive condition. Option A is correct. Option B, (3,1), goes backward and is not demanded by the rule. Option C, (2,1), reverses the first relation and is irrelevant. Option D, (3,2), reverses the second relation and is also irrelevant. Notice that transitivity does not mean adding every possible pair; it means adding the pair forced by a two-step chain. Since the set has only these three elements and the relation has the chain 1 to 2 to 3, (1,3) is the needed addition. Remember: aRb and bRc imply aRc.

The pairs are ( (1,2),(2,1),(2,3),(3,2),(3,4),(4,3) ), so the count is (6). In exams, take both directions because of absolute difference.

(4) is a multiple of (2), so ( (4,2)\in R ). In exams, understand the direction of multiple of and divides separately.

The domain is formed from first components and the range from second components. Therefore domain is ( {1,2,3} ) and range is ( {4,5,6} ).

(1+3=4) is even, so ( (1,3)\notin R ). In exams, an odd sum needs one odd and one even element.

The smallest reflexive relation contains only all pairs ( (a,a) ). It is also called the identity relation.

The largest relation is (A\times A) because it contains all possible ordered pairs. It is called the universal relation.