Relations

By definition, \([1]=\{b\in A:(1,b)\in R\}\). Thus, \(1+b\) must be even, which happens exactly when \(b\) is odd. The odd elements of \(A\) are \(1\) and \(3\), so \([1]=\{1,3\}\). Option D is incorrect because it includes only 1 and omits the related element 3. Exam tip: To find \([a]\), list every \(b\) for which \((a,b)\in R\).

For a relation \(R\), the condition \(R^{-1}=R\) means that whenever \((a,b)\in R\), the pair \((b,a)\) must also belong to \(R\). This is exactly the definition of a symmetric relation. Reflexivity concerns only pairs of the form \((a,a)\), while transitivity and antisymmetry do not guarantee equality with the inverse relation. Exam tip: associate \(R^{-1}=R\) directly with symmetry.

The inverse relation \(R^{-1}\) is formed by reversing every ordered pair \((a,b)\) in \(R\) to \((b,a)\). Reversing \((1,2),(2,3),(1,3)\) gives \((2,1),(3,2),(3,1)\), which are exactly the elements of \(S\). Therefore, \(S=R^{-1}\). Option B would contain pairs from both the original and inverse relations, whereas S contains only the reversed pairs. Exam tip: to find an inverse relation, interchange the two entries in every ordered pair.

For each element a of A, we need b = 6 − a, with b also belonging to A. This gives (2, 4), (3, 3), and (4, 2); (1, 5) is not valid because 5 ∉ A. Hence, R contains 3 ordered pairs. Exam tip: (2, 4) and (4, 2) are distinct ordered pairs because their order is different.

The valid pairs are ((3,5),(4,4),(4,5),(5,3),(5,4),(5,5)). For conditional counting, count possible (b) for each (a).

(1) and (2) are related to each other, while (3) is related only to itself. Hence the classes are ({1,2}) and ({3}).

In a partition, subsets are non-empty, disjoint, and cover the whole set. Only ({{1,3},{2}}) satisfies these conditions.

In an equivalence relation, every element of a class is related to every element of the same class. The class \(\{1,4\}\) contributes \(2^2=4\) ordered pairs, and \(\{2,3\}\) contributes another \(2^2=4\). Thus, the total is \(4+4=8\). No pair is formed between elements belonging to different equivalence classes. Exam tip: if the class sizes are \(n_1,n_2,\ldots\), the number of ordered pairs is \(n_1^2+n_2^2+\cdots\).

The relation connects exactly those pairs whose elements have the same parity. Its equivalence classes are therefore {1,3} and {2,4}. Since a-a=0 for every a∈A, the relation is reflexive. If a-b∈{0,2,-2}, then b-a∈{0,2,-2} as well, so it is symmetric. Any two elements related through an intermediate element have the same parity, making the relation transitive. Hence R is an equivalence relation. Exam tip: When a relation is defined through possible differences, check whether those differences characterize a familiar grouping such as parity.

Every ((a,a)) is present in both, so every ((a,a)) remains in the intersection. Reflexivity is preserved under intersection.

The reverse of every pair is present in the same relation, so it is also present in the union. Symmetry is preserved under union.

For example, (R={(1,2)}) and (S={(2,3)}) are transitive, but their union lacks ((1,3)). Hence the union is not always transitive.

For transitivity, whenever \((a,b)\in R\) and \((b,c)\in R\), the pair \((a,c)\) must also belong to \(R\). Since \((1,2)\) and \((2,3)\) are in \(R\), the pair \((1,3)\) is required. All three diagonal pairs are already present, so no other additional pair is needed. For instance, adding \((3,1)\) does not satisfy the requirement generated by the chain \(1\to2\to3\). Exam tip: check every two-step chain \(a\to b\to c\) and verify that \(a\to c\) is present.

For every \(a\in A\), \(a\ge a\), so \(R\) is reflexive. If \(a\ge b\) and \(b\ge c\), then \(a\ge c\), so it is also transitive. However, \((2,1)\in R\), whereas \((1,2)\notin R\); hence it is not symmetric. Therefore, option B is incorrect because an equivalence relation must also be symmetric. Exam tip: A relation defined by \(\ge\) is generally reflexive, transitive, and antisymmetric; test symmetry using two unequal elements.

If (a,b)∈R, then ab is even. Since ab=ba, (b,a) also belongs to R; hence R is symmetric. It is not reflexive because (1,1)∉R, as 1×1=1 is odd. It is also not transitive: (1,2) and (2,1) are in R, but (1,1) is not. Therefore, option A is correct. Exam tip: To test reflexivity, check every pair (a,a); to disprove transitivity, look for two related pairs whose required composite pair is absent.

Since (\operatorname{lcm}(a,b)=\operatorname{lcm}(b,a)), the relation is symmetric. Not all diagonal pairs occur, for example (\operatorname{lcm}(2,2)=2).

Direct answer: Option A, symmetric but not reflexive. The condition is that a+b must be prime. Since a+b=b+a, whenever (a,b) belongs to the relation, (b,a) also belongs; symmetry is therefore certain. For reflexivity, every (a,a) would need 2a to be prime. Taking a=1 gives 2, which is prime, but a=2 gives 4, which is not prime; also (4,4) gives 8, not prime. Thus reflexivity fails. Option A is correct. Option B is wrong because it claims reflexivity, contradicted by (2,2). Option C is wrong because an equivalence relation must be reflexive, and this is not. Option D is wrong: transitivity does not hold; for example (1,1) and (1,2) are valid because their sums are 2 and 3, but (1,2) is valid here, while a clearer failure is (1,2) and (2,1) valid but (1,1) valid, so use (2,3) and (3,2): sums 5 and 5 are prime, but (2,2) has sum 4 and is not valid. Memory cue: commutative sum gives symmetry, not automatically reflexivity or transitivity.

For transitivity, whenever (a,b) and (b,c) belong to R, (a,c) must also belong to R. Here, (1,2) and (2,4) are in R, but (1,4) is not; therefore, R is not transitive. All diagonal pairs (a,a) are present, so R is reflexive, and the reverse pairs of (1,2) and (2,4) are also present, so R is symmetric. Exam tip: To test transitivity, combine two related ordered pairs and check whether the resulting pair is included in the relation.

The condition (a=b) gives reflexivity and (a+b=5) gives symmetry. But ((1,4)) and ((4,2)) would require ((1,2)), which is absent.

When 2 is divided by 2, the remainder is 0. In the set \(A\), the numbers 2, 4 and 6 also leave remainder 0 upon division by 2, so they are related to 2. Therefore, \([2]=\{2,4,6\}\). Option B contains the odd numbers, which leave remainder 1. Exam tip: To find an equivalence class, select all elements of the set that give the same remainder on division by the specified divisor.

All diagonal pairs are present and ((2,1)) accompanies ((1,2)). All pairs needed for transitivity are also present.

For an even sum, both numbers must have the same parity. With (2) odd and (2) even elements, the count is (2^2+2^2=8).

Under R, an element a belongs to [1] if a − 1 is divisible by 4. In A, this is true for 1 and 5 because 1 − 1 = 0 and 5 − 1 = 4. Therefore, [1] = {1, 5}. The option {1} is incomplete because it omits 5; in such questions, test every element of the given set for congruence with 1.

For (b=1,2,3), we get (a=5,3,1), so there are (3) pairs. In such counting, substitute each possible (b) and check whether (a) belongs to the set.

The equivalence class is \([1]=\{x\in A:x-1\equiv 0\pmod{5}\}\). Within \(A\), this condition holds only for \(1\) and \(6\), since \(1-1=0\) and \(6-1=5\) are both divisible by 5. Hence, \([1]=\{1,6\}\). The set \(\{1,5\}\) is incorrect because \(5-1=4\), which is not divisible by 5. Exam tip: To find an equivalence class, select all elements whose difference from the given element is divisible by the modulus.