Relations

Since ((1,2)\in R) and ((2,3)\in R), transitivity requires ((1,3)). It is missing, so (R) is not transitive.

The relation requires \(a+b=6\) with \(a,b\in A\). The valid ordered pairs are \((2,4),(3,3),(4,2)\). Therefore, the set of their first components is \(\{2,3,4\}\), so option B is correct. Option C incorrectly omits 4, while 1 cannot be included because it would require \(b=5\), which is not in \(A\). Exam tip: The domain of a relation is the set of all first components of its ordered pairs.

Check the ordered pairs formed from elements of \(A\) against \(a^2+b^2=25\). Since \(3^2+4^2=9+16=25\) and \(4^2+3^2=16+9=25\), the pairs \((3,4)\) and \((4,3)\) belong to \(R\). No other pair from \(A\times A\) satisfies the equation. Hence, \(R\) contains \(2\) ordered pairs. Exam tip: In a relation, \((a,b)\) and \((b,a)\) are counted separately because order matters, unless the two entries are equal.

If (a\ne b), then (b\ne a), so it is symmetric. But ((a,a)) never occurs, so it is not reflexive.

The equivalence class of \(a\) is \([a]_R=\{x\in A:(a,x)\in R\}\). Here, the pairs \((1,1)\) and \((1,2)\) belong to \(R\), whereas \((1,3)\) does not. Therefore, \([1]_R=\{1,2\}\). Option D is incorrect because \(3\) is not related to \(1\). Exam tip: To find \([a]_R\), identify the ordered pairs whose first component is \(a\), then collect their second components.

The class \([2]\) contains all elements of \(A\) that are congruent to \(2\) modulo \(3\). Both \(2\) and \(5\) leave remainder \(2\) on division by \(3\), so \([2]=\{2,5\}\). Option B, \(\{1,4\}\), is actually the class \([1]\), since both elements leave remainder \(1\). Exam tip: To find an equivalence class, collect all elements having the same remainder modulo the given number.

An even sum means both numbers have the same parity. Thus the classes are odd ({1,3}) and even ({2,4}).

By the definition of a symmetric relation, if \((a,b)\in R\), then \((b,a)\in R\) must also hold. Here, \(a=2\) and \(b=5\), so \((5,2)\in R\) is necessary. The presence of \((2,2)\) or \((5,5)\) is related to reflexivity, not symmetry. Exam tip: To test symmetry, interchange the two elements of every ordered pair and check whether the reversed pair is also present.

Transitivity states that if \((a,b)\in R\) and \((b,c)\in R\), then \((a,c)\in R\). Substituting \(a=1\), \(b=4\), and \(c=6\) gives \((1,6)\in R\), so option C is correct. Options A and B reverse the direction of the given ordered pairs, while option D is not required by transitivity. Exam tip: in a transitive chain, match the second element of the first pair with the first element of the second pair, then join the remaining outer elements.

The only ordered pairs from A satisfying a+b=7 are (3,4) and (4,3). Therefore, the range—the set of second components of the ordered pairs—is {3,4}. Option A incorrectly includes every element of A; for instance, pairing 2 would require 5, but 5∉A. Exam tip: the range of a relation is formed from the second coordinates of its ordered pairs.

The required ((1,3)) from ((1,2)) and ((2,3)) is present, so the key transitivity condition holds. Diagonal pairs are absent, so it is not reflexive.

Symmetry lets us infer ((4,3)\notin R) from ((3,4)\notin R). But the presence of diagonal pairs is not guaranteed by symmetry.

For ((1,1)), (1+1=2) is prime, so it is in (R). However, all listed pairs are actually in (R), so this is a trap question.

For an ordered pair \((a,b)\) to belong to \(R\), the sum \(a+b\) must be prime. In option D, \(3+4=7\), and 7 is prime; hence \((3,4)\in R\). The sums for the other options are 4, 8, and 6, none of which is prime. Exam tip: calculate the sum of the two entries and then test whether the sum is prime.

If (a+b=6), then (b+a=6), so the relation is symmetric. But all ((a,a)) are not present, so it is not reflexive.

This relation sends each (a) to (5-a), and applying it twice returns (a) to itself. Therefore (R\circ R=I_A).

Reflexivity fixes the (5) diagonal pairs, and symmetry lets the remaining pairs be chosen in unordered pairs. Therefore the number is (2^{\frac{5(5-1)}{2}}=2^{10}).

For any \(a\in A\), we have \(a\mid a\), so \(R\) is reflexive. If \(a\mid b\) and \(b\mid c\), then \(a\mid c\), so \(R\) is transitive. However, symmetry fails: for example, \(2\mid 4\) is true, so \((2,4)\in R\), but \(4\mid 2\) is false, so \((4,2)\notin R\). Hence \(R\) is reflexive and transitive but not symmetric.

(a+a) is even, (a+b) even implies (b+a) even, and same parity gives transitivity. In exams, identify the even and odd equivalence classes.

(A\times A) has (n^2) ordered pairs, and a relation can be any subset of it. Hence the number of relations is (2^{n^2}).

(A\times B) has (mn) ordered pairs. Each pair may be included or excluded, so there are (2^{mn}) possible relations.

A reflexive relation must contain ((1,1),(2,2),(3,3)). The remaining (9-3=6) pairs are optional, so the answer is (2^6).

There are (4^2=16) ordered pairs, and (4) diagonal pairs are compulsory. Thus (12) pairs are optional, giving (2^{12}).

For a symmetric relation, the number of independent choices is (\frac{4(4+1)}{2}=10). Hence the number of relations is (2^{10}).

Reflexivity fixes the (5) diagonal pairs. Symmetry leaves only (\frac{5(5-1)}{2}=10) off-diagonal blocks free, so the answer is (2^{10}).