Relations

A relation on A is reflexive when every element of A is related to itself. For A={1,2}, reflexivity requires both ordered pairs (1,1) and (2,2) to be present in R. The empty relation R=∅ contains no ordered pairs at all.

Since neither (1,1) nor (2,2) belongs to the empty relation, the required self-relations are missing. Therefore R is not reflexive, and option B, “No,” is correct. It is not enough that the relation is defined on A; the diagonal pair for every element must actually be included. An empty relation can be reflexive only on an empty underlying set, not on this nonempty set.

The set (A\times A) includes both ( (1,1) ) and ( (2,2) ). A universal relation is always reflexive.

In (A\times A), every pair has its reverse pair too. Hence the universal relation is symmetric.

The set (A\times A) contains all possible pairs, so the transitive condition is satisfied. The universal relation is also transitive.

The reverse of a diagonal pair ( (a,a) ) is the same pair ( (a,a) ). Hence a relation with only diagonal pairs is symmetric.

The number (3) is one less than (4), so ( (3,4) ) is correct. Convert word conditions into equations.

Both (3) and (5) are odd, so they have the same parity. In a parity relation, even-even or odd-odd pairs are correct.

The governing concept is parity. A sum is odd exactly when one addend is odd and the other is even. In option A, 1+2=3, which is odd, so (1,2) belongs to R. In option B, 2+3=5, also odd, so (2,3) belongs to R. In option D, 2+1=3, so that pair also belongs to R. However, in option C, 1+3=4, which is even rather than odd. Therefore (1,3) is not a member of R, and option C is correct. Although reversing a pair does not change the sum or its parity, each pair must still be tested against the stated condition and the membership of its entries in A.

In a reflexive relation, every element must be related to itself, so \((a,a) \in R\) for every \(a \in A\). Option D states the condition for an irreflexive relation. Exam tip: check all diagonal ordered pairs first.

All diagonal pairs ( (1,1),(2,2),(3,3) ) are present. The extra pair ( (1,3) ) does not stop reflexivity.

The direct answer is option A: yes, R is symmetric. A relation is symmetric when every time \((a,b)\) belongs to it, the reversed pair \((b,a)\) also belongs to it. Here \((1,2)\) is present and its reverse \((2,1)\) is also present. The pair \((3,3)\) reverses to itself, so it also satisfies the condition. Therefore R is symmetric. Option A is correct. Option B is false because the required reverse pair is present. Option C says it is only reflexive, but that is not the requested classification and is also not established unless the underlying set is specified; the important fact here is symmetry. Option D is false because the listed pairs provide enough information to decide symmetry. A common mistake is to think a pair must have different entries; a self-reversing pair such as \((3,3)\) causes no problem. Check every pair and its reverse.

By the transitive rule, ( (1,2) ) and ( (2,3) ) require ( (1,3) ). This pair is present in the relation.

For an ordered pair (a, b) to belong to R, it must satisfy a + b = 6. For (2, 4), 2 + 4 = 6, so it belongs to R. Each of the other options has a sum of 5 and therefore does not belong to R. Exam tip: Check the defining condition of the relation using both entries of the ordered pair.

In ( (2,2) ), both components are equal, so (a\ne b) is not satisfied. Diagonal pairs do not appear in a not-equal relation.

Since (4=2^2), ( (4,2)\in R ). In such questions, check the first and second components in the correct order.

A relation from (A) to (B) is always a subset of (A\times B). In exams, identify the Cartesian product first.

The governing concept is the identity relation on a set. For every element x in A, the identity relation contains the ordered pair (x, x), and it contains no pair whose two components are different. Since A = {5, 7}, we write one self-pair for 5 and one self-pair for 7: R = {(5,5), (7,7)}. Therefore option B is correct. Option A contains reverse pairs connecting different elements, so it is not the identity relation. Option C is the empty relation and does not relate any element to itself. Option D is the universal relation A × A, which contains all four possible pairs, not only the diagonal self-pairs.

The universal relation is (A\times A), and (n(A\times A)=3^2=9). A universal relation contains all possible pairs.

In every ordered pair of \(R\), both the first and second components belong to set \(A\). Therefore, \(R\subseteq A\times A\), so \(R\) is a relation from \(A\) to \(A\). It is not a relation from \(R\) to \(R\), because the numbers are not elements of \(R\). Exam tip: To identify a relation on \(A\), check whether every ordered pair belongs to \(A\times A\).

A pair belongs to R only when both of its elements are in A and their sum is 7. For (3,4), 3+4=7, so (3,4)∈R. The sums for (1,4), (2,4), and (2,3) are 5, 6, and 5 respectively, so those pairs are not in R. Exam tip: directly check the sum of the two entries in each ordered pair.

From the given set, ( (3,4) ) and ( (4,3) ) are formed. In ordered pairs, changing the order changes the pair.

In every ordered pair \((x,y)\), the second component is the square of the first: \(4=2^2\), \(9=3^2\), and \(16=4^2\). Hence, the correct rule is \(y=x^2\). The rule \(y=x+2\) works only for the first pair, while \(y=2x\) and \(x=y^2\) do not satisfy all the pairs. Exam tip: Test a proposed relation by substituting the values of \(x\) and \(y\) from each ordered pair.

The governing concept is membership in a relation defined by a condition. An ordered pair (x, y) belongs to R exactly when its coordinates satisfy y = x + 3. Test option A: x = 2 and y = 5, and 5 = 2 + 3, so (2, 5) satisfies the rule and belongs to R. Option B gives y = 2 but x + 3 = 5 + 3 = 8, so it fails. In option C, x + 3 = 3 + 3 = 6, not y = 3. In option D, x + 3 = 1 + 3 = 4, not y = 3. Thus only option A is correct. The order of coordinates is essential: (2, 5) and (5, 2) are different ordered pairs, even though they contain the same two numbers. Direct substitution is sufficient to verify relation membership.

The governing concept is membership in a relation defined by an inequality. An ordered pair (a,b) belongs to R exactly when the first component is smaller than the second, that is, when a < b. Check each choice: 1 < 2 is true, so (1,2) belongs; 1 < 3 is true, so (1,3) belongs; and 2 < 3 is true, so (2,3) belongs. However, 3 < 1 is false, so (3,1) does not belong to R. Thus option D is correct. The order of the components matters: (1,3) and (3,1) are different ordered pairs, even though they contain the same two numbers.

An ordered pair \((a,b)\) belongs to \(R\) only when \(a\ge b\). For \((5,3)\), the inequality \(5\ge3\) is true, so \((5,3)\in R\). In all the other options, the first component is smaller than the second, so they do not satisfy the relation. Exam tip: Treat the first and second entries as \(a\) and \(b\), respectively, and then test the given inequality.