Relations

Both (1) and (3) are odd, so they have the same parity. In a parity relation, even-even or odd-odd pairs are correct.

In \((a,b)\in R\), \(a\) must divide \(b\). Check each pair: \(2\mid 6\) so \((2,6)\in R\); \(3\mid 6\) so \((3,6)\in R\); \(5\mid 5\) so \((5,5)\in R\). But \(6\nmid 3\), so \((6,3)\notin R\). Hence the correct answer is \((6,3)\).

The number (6) is a multiple of (3), so ( (6,3)\in R ). Remember the order in multiple and factor relations.

The governing concept is forming a relation by listing every ordered pair that satisfies its defining equation. Both a and b must come from A = {0,1,2}, and their sum must equal 2. If a = 0, then b must be 2, giving (0,2). If a = 1, then b must be 1, giving (1,1). If a = 2, then b must be 0, giving (2,0). These are all possibilities because every value of a in A has been checked, and no other value is available. Thus R = {(0,2),(1,1),(2,0)}, so option A is correct. Option B gives sums of 1, option C gives a sum of 4, and option D is impossible because valid pairs exist.

Check the condition \(a+b>5\) for every ordered pair in \(A\times A\). Only \((3,3)\) satisfies it because \(3+3=6>5\). The pairs \((2,3)\) and \((3,2)\) have sum exactly 5, so they are not included. Hence, \(R=\{(3,3)\}\). Exam tip: the symbol \(>\) excludes equality.

Check the condition a+b<3 for the ordered pairs in A×A. Only (1,1) satisfies it because 1+1=2<3. The pairs (1,2) and (2,1) have sum 3, which is not less than 3, so they are excluded. Therefore, R={(1,1)}. Exam tip: List the ordered pairs and test the given inequality one pair at a time.

The relation is understood as a relation on A, so both a and b are selected from A. The condition a=1 fixes the first coordinate. The second coordinate is unrestricted by the condition, so it may be any one of the four elements of A. We therefore form one pair for each possible second coordinate.

The pairs are (1,1), (1,2), (1,3), and (1,4). There are four distinct pairs, so option C is correct. The answer is not 1 because fixing a does not fix b. It is not 8 or another larger number because only one first-coordinate value is allowed and A has four choices for the second coordinate. This uses the standard convention that the displayed relation is on A×A.

Here the relation is on A, so both coordinates must come from A. The condition b=5 fixes the second coordinate, while the first coordinate may be any element of A. Since A has five elements, exactly one pair is obtained for each possible first coordinate.

The complete list is (1,5), (2,5), (3,5), (4,5), and (5,5). Thus the relation contains 5 pairs, making option B correct. The value 25 would count all possible pairs in A×A, but most of those do not have second coordinate 5. The value 1 would overlook the four other choices for the first coordinate. Counting the free coordinate gives the answer directly.

The direct answer is option A: \((2,4)\). Transitivity says that if \((a,b)\in R\) and \((b,c)\in R\), then \((a,c)\) must also be in R. Here the first pair is \((2,3)\), so the middle value is 3. The second pair is \((3,4)\), whose first value is the same 3. Joining them gives first value 2 and final value 4, so the required pair is \((2,4)\). Option A correctly follows the transitive rule. Option B, \((4,2)\), reverses the required order. Option C, \((3,2)\), is unrelated to the forward chain and reverses the first pair. Option D, \((1,4)\), uses a starting value 1 not supplied by these two pairs. The existing relation contains \((1,3)\), but that does not change the pair required from \((2,3) o(3,4)\). Exam cue: connect the repeated middle number and retain the outside numbers.

The domain contains the first components of the ordered pairs. Hence the domain of (R) is ( {1,2,3} ).

This is the identity relation on A. It is reflexive because every diagonal pair (1,1), (2,2), and (3,3) is present. It is symmetric because reversing any diagonal pair gives the same pair, and it is transitive because a chain such as (a,a) followed by (a,a) again gives (a,a). Thus all three properties hold, making A correct.

The governing concept is the direction of a relation defined by divisibility. The statement aRb means that the first entry a must be a multiple of the second entry b; equivalently, a=kb for some integer k. In option B, 6=2×3, so 6 is a multiple of 3, and both numbers belong to A. Thus (6,3) is in R. In option A, 2 is not a multiple of 4. In option C, 3 is not a multiple of 6, and in option D, 1 is not a multiple of 4. Therefore option B is the only valid pair. The reverse pair (3,6) would represent the opposite divisibility direction and cannot be accepted under the given definition.

The governing concept is membership in a relation defined by a condition. An ordered pair (a, b) belongs to R exactly when the sum of its coordinates satisfies the inclusive inequality a + b ≤ 5. For (1, 4), the sum is 5, so it is included. For (2, 3), the sum is also 5, and equality is allowed by the symbol ≤. For (3, 1), the sum is 4, so it is included as well. For (4, 4), the sum is 8, and 8 ≤ 5 is false. Thus (4, 4) is the only listed pair outside R, so option D is correct. The equality case explains why options A and B are not excluded.

A relation R on a set A is symmetric when every ordered pair in R has its reversed ordered pair in R as well. Symbolically, for all a, b ∈ A, (a, b) ∈ R implies (b, a) ∈ R. Therefore, if (2, 5) belongs to R, then (5, 2) must also belong to R. Option A states exactly this defining condition. Option B describes the requirement for reflexivity, not symmetry. Option C is too restrictive: the empty relation is symmetric, but it is not the only symmetric relation. Option D merely describes that R is a subset of A, which is not the usual correct form for a relation and says nothing about reversing pairs.

The governing condition is that the sum of the two entries must be a prime number. Test each option: for (1, 4), 1 + 4 = 5, and 5 is prime, so (1, 4) belongs to R. For (2, 4), the sum is 6, which is composite. For (3, 5), the sum is 8, also composite. For (4, 4), the sum is 8, again not prime. Since only option A produces a prime sum, it is the unambiguous correct answer. Notice that order matters for a relation, although in this particular sum condition reversing a pair would give the same sum. The decisive test is simply whether the resulting integer has exactly two positive divisors.

An ordered pair belongs to R precisely when the sum of its two coordinates is 6. Check the options individually: (2, 5) has sum 7, so it does not satisfy the condition. (3, 3) has sum 3 + 3 = 6, so it belongs to R. (4, 4) has sum 8 and is excluded. (5, 2) has sum 7 and is also excluded. Therefore option B is the correct answer. The set A supplies the allowable entries, while the equation a + b = 6 decides membership. Although (2, 5) and (5, 2) are reversals of one another, neither qualifies because their sum is 7; this illustrates that the condition must be checked rather than inferred from the entries alone.

Transitivity requires that whenever (a,b) and (b,c) are in R, the pair (a,c) must also be in R. In this relation, (1,2) and (2,3) provide a valid two-step chain, so transitivity demands (1,3). Because (1,3) is not in R, the condition fails. The missing pairs (2,1) or (1,1) concern other properties, and R is certainly not all of A × A. Thus A is correct.

Start with the defining equation a + b = 2a. Subtract a from both sides: b = a. Rewriting the same equality in the usual order gives a = b. Therefore the relation consists exactly of the diagonal pairs (1,1), (2,2), (3,3), (4,4), and (5,5), so it is the identity relation on A. No division or restriction introduces any additional solution, because the simplification is simply subtraction of a from both sides. Option B would correspond to a different condition, a = 2b; option C reverses the coefficient incorrectly; and option D does not follow from the given equation. Hence option A is the correct simplified identity.

In an ordered pair \((x,y)\), \(x\) is the input and \(y\) is its image. Here, the pair \((3,3)\) has 3 as its second component, so the input with image 3 is 3. The other inputs 1 and 2 have image 2, while 4 has image 4. Exam tip: Match the given image with the second component of each ordered pair.