Relations

For each \(a\in A\), the condition \(b=a+2\) gives the pairs \((1,3),(2,4),(3,5),(4,6)\). In every case, the second component belongs to \(B\), so all four pairs are in \(R\). Therefore, the correct answer is 4. Exam tip: To count the elements of a relation, apply its defining condition to each domain element rather than counting all possible pairs in \(A\times B\).

Reflexivity requires ((1,1),(2,2),(3,3),(4,4)). Only ((1,1)) is present, so (3) pairs must be added.

((1,2)) and ((2,1)) are already paired. For ((2,4)), add ((4,2)), and for ((3,4)), add ((4,3)).

The relation \(R\) is reflexive because \((1,1),(2,2),(3,3)\) are present. It is also symmetric because \((2,1)\) accompanies \((1,2)\), and \((3,1)\) accompanies \((1,3)\). However, \((2,1)\in R\) and \((1,3)\in R\), whereas \((2,3)\notin R\). Thus transitivity fails, so \(R\) is not an equivalence relation. Exam tip: Check all three properties—reflexivity, symmetry, and transitivity—when testing an equivalence relation.

The subset relation \(\subseteq\) is reflexive because every set is a subset of itself. It is antisymmetric: if \(A\subseteq B\) and \(B\subseteq A\), then \(A=B\). It is also transitive, so it is a partial order relation. It is not an equivalence relation because it is generally not symmetric; for example, \(\{1\}\subseteq\{1,2\}\), but \(\{1,2\}\nsubseteq\{1\}\). Exam tip: To identify a partial order, check reflexivity, antisymmetry, and transitivity.

(A\cap B=B\cap A), so the relation is symmetric. But ({1}\cap{1}\neq\varnothing), and transitivity does not hold in general.

For every A in P(X), |A| = |A|, so R is reflexive. If |A| = |B|, then |B| = |A|, so it is symmetric. If |A| = |B| and |B| = |C|, then |A| = |C|, so it is transitive. Since it has all three properties, R is an equivalence relation.

([\sqrt{2}]) contains all (x) such that (x-\sqrt{2}\in\mathbb{Q}). Hence (x=\sqrt{2}+q) where (q\in\mathbb{Q}).

In a symmetric relation, if ((a,b)\in R), then ((b,a)\in R) also. Therefore ((5,9)) forces ((9,5)).

A relation on a four-element set is a subset of A×A, which contains 4×4 = 16 ordered pairs. Reflexivity requires all four diagonal pairs (a,a) to be present. The remaining 16−4 = 12 off-diagonal pairs may each be included or excluded independently. Therefore the number of reflexive relations is 2^12, making option A correct. Option B counts arbitrary relations, whereas options C and D do not account for the independent choices among all remaining pairs.

The empty relation contains no ordered pairs. Symmetry is vacuously true because there is no pair (a,b) that could violate the requirement; transitivity is also vacuously true because there are no pairs (a,b) and (b,c) to test. However, reflexivity requires all three diagonal pairs (1,1), (2,2), and (3,3), none of which is present. Thus option A is correct, and the relation cannot be an equivalence relation because reflexivity fails.

The relation consists of pairs whose coordinates add to 5: R = {(1,4),(2,3),(3,2),(4,1)}. By the definition of composition, (x,z) belongs to R∘R if there is some intermediate y such that (x,y) belongs to R and (y,z) also belongs to R. For (1,1), choose y = 4. Since 1+4=5, (1,4) is in R; since 4+1=5, (4,1) is also in R. Therefore (1,1) belongs to R∘R. In fact, applying the relation twice returns each element to itself, so all diagonal pairs occur. For (1,2), the intermediate value would have to be both 4 and 3, which is impossible. The other off-diagonal options fail similarly, so A is correct.

For A with three elements, A×A has nine ordered pairs. In a symmetric relation, the three diagonal pairs (1,1), (2,2), and (3,3) are independently chosen. The six off-diagonal pairs form three linked blocks: {(1,2),(2,1)}, {(1,3),(3,1)}, and {(2,3),(3,2)}. Each block is either wholly included or wholly excluded, giving 3+3=6 binary choices and hence 2^6 relations. Option A is correct.