Relations

Correct answer: A. ((7,3))

Explanation: For distinct elements (3) and (7), the reverse pair cannot occur together. This is the key test of antisymmetry.

Correct answer: A. \((1,6)\)

Explanation: By transitivity, if \((a,b)\in R\) and \((b,c)\in R\), then \((a,c)\in R\) must hold. Here, \(a=1\), \(b=4\), and \(c=6\), so \((1,6)\in R\). The other options reverse the order of the elements, and transitivity does not guarantee such reverse pairs. Exam tip: in a transitive relation, match the second element of the first pair with the first element of the second pair, then connect the remaining elements.

Correct answer: A. \(\{(1,1),(2,2),(3,3)\}\)

Explanation: A relation is reflexive if \((a,a)\) belongs to it for every \(a\in A\). Thus, \((1,1),(2,2),(3,3)\) are the only required pairs, giving the smallest reflexive relation. Option D is also reflexive, but it contains all 9 ordered pairs, so it is not minimal. Exam tip: the smallest reflexive relation on a set \(A\) is \(\{(a,a):a\in A\}\).

Correct answer: A. It is the identity relation

Explanation: Squares of distinct positive elements are not equal, so only ((a,a)) pairs occur. Hence it is the identity relation.

Correct answer: A. It is transitive but not reflexive

Explanation: The condition \(a-b>0\) is equivalent to \(a>b\). If \(a>b\) and \(b>c\), then \(a>c\), so the relation is transitive. However, for every \(a\in A\), \(a-a=0>0\) is false; hence \((a,a)\notin R\), and the relation is not reflexive. It is also not symmetric, since \((2,1)\in R\) but \((1,2)\notin R\). Exam tip: A strict greater-than relation is transitive and irreflexive, so it cannot be an equivalence relation.

Correct answer: A. It is symmetric

Explanation: If \((a,b)\in R\), then \(a+b\le 6\). Since addition is commutative, \(b+a=a+b\le 6\), so \((b,a)\in R\) as well. Hence, \(R\) is symmetric. It is not reflexive because \((4,4)\notin R\); it is not antisymmetric because both \((1,2)\) and \((2,1)\) belong to \(R\). It is also not transitive since \((2,1),(1,5)\in R\), but \((2,5)\notin R\). Exam tip: To test symmetry, check whether every \((a,b)\) is accompanied by \((b,a)\).

Correct answer: A. Because \((1,1)\notin R\)

Explanation: A relation is reflexive only if \((a,a)\in R\) for every \(a\in A\). For \((1,1)\) to belong to \(R\), we need \(1+1\ge 5\), but \(2\ge 5\) is false. Therefore, \((1,1)\notin R\), so the relation is not reflexive. Although \((4,4)\in R\), one self-pair does not establish reflexivity for the entire set. Exam tip: To test reflexivity, check every pair of the form \((a,a)\); one failure is enough.

Correct answer: B. 2

Explanation: The relation connects 1 with 2 in both directions and 3 with 4 in both directions, while every element is related to itself. Hence, the equivalence classes are [1]=[2]={1,2} and [3]=[4]={3,4}. Therefore, there are 2 equivalence classes. Exam tip: Count the distinct groups in the partition, not the number of ordered pairs or elements.

Correct answer: A. ([a]=[b])

Explanation: In an equivalence relation, related elements lie in the same class. Hence their equivalence classes are equal.

Correct answer: A. Reflexive, antisymmetric and transitive

Explanation: A partial order relation is reflexive, antisymmetric, and transitive. (\le) and (\subseteq) are good examples.

Correct answer: A. Partial order relation

Explanation: The inclusion relation \(\subseteq\) is reflexive because \(A\subseteq A\) for every \(A\in P\). It is antisymmetric: if \(A\subseteq B\) and \(B\subseteq A\), then \(A=B\). It is also transitive, since \(A\subseteq B\) and \(B\subseteq C\) imply \(A\subseteq C\). Therefore, \(R\) is a partial order relation. It is not an equivalence relation because it is not symmetric; for example, \(\{1\}\subseteq\{1,2\}\), but \(\{1,2\}\not\subseteq\{1\}\). Exam tip: To identify a partial order, check reflexivity, antisymmetry, and transitivity.

Correct answer: A. Symmetric but not transitive

Explanation: If \((a,b)\in R\), then \(a\ne b\), which also implies \(b\ne a\). Hence \((b,a)\in R\), so the relation is symmetric. However, \((1,2)\in R\) and \((2,1)\in R\), whereas \((1,1)\notin R\); therefore, it is not transitive. It is also not antisymmetric because both \((1,2)\) and \((2,1)\) belong to the relation. Exam tip: To test transitivity, combine two related ordered pairs with a common middle element and check whether the resulting pair is also present.

Correct answer: A. It is an equivalence relation

Explanation: Parity means whether a number is even or odd. Two numbers have the same parity when both are even or both are odd. A relation based on same parity has the three important properties required for an equivalence relation: reflexivity, symmetry, and transitivity. Therefore, option A is correct.

Every number in the set has the same parity as itself, so the relation is reflexive. If a has the same parity as b, then b also has the same parity as a, so it is symmetric. If a has the same parity as b and b has the same parity as c, then a and c also have the same parity, so it is transitive. The equivalence classes are {1,3} and {2,4}.

Correct answer: A. \(\{4\}\)

Explanation: Substituting the first element \(a=4\) gives the condition \(4\leq b\). Since \(b\) must belong to \(A\) and \(4\) is the maximum element of the set, this condition is true only for \(b=4\). Therefore, the set of related second elements is \(\{4\}\). Option B incorrectly includes \(1,2,3\), for which \(4\leq b\) is false, while option D is wrong because \((4,4)\in R\). Exam tip: for a maximum element \(m\), the condition \(m\leq b\) usually permits only \(b=m\).

Correct answer: A. No pair

Explanation: R is already reflexive because it contains (1,1), (2,2), and (3,3). It is symmetric because (1,2) is accompanied by (2,1). Transitivity also holds: 1 and 2 form one equivalence class, while 3 forms a separate class. Thus, the equivalence classes are {1,2} and {3}, and no new pair is needed. Exam tip: To test an equivalence relation, check reflexivity, symmetry, and transitivity in that order.

Correct answer: B. (5) pairs

Explanation: By checking values, we get ((1,2),(2,1),(2,4),(3,3),(4,2)). In such questions, check possible (b) values for each (a) systematically.

Correct answer: A. It is an equivalence relation

Explanation: Elements with the same parity are related, so the relation is reflexive, symmetric, and transitive. In exams, check it by splitting into even and odd classes.

Correct answer: C. Reflexive, symmetric and transitive

Explanation: (a+a) is always even, so the relation is reflexive. The parity condition also gives symmetry and transitivity.

Correct answer: C. (3)

Explanation: There are three classes according to remainders (0,1,2). In exam, counting residues is the fastest way for modulo relations.

Correct answer: B. (2^6)

Explanation: (A\times A) has (9) pairs and (3) diagonal pairs are compulsory. The remaining (6) pairs are optional, so the number is (2^6).

Correct answer: C. (2^{16})

Explanation: (A\times A) has (4^2=16) ordered pairs. Every relation is a subset of (A\times A), so the total number is (2^{16}).

Correct answer: A. Reflexive, antisymmetric and transitive

Explanation: (a \leq a) gives reflexivity. If (a \leq b) and (b \leq a), then (a=b), so it is also antisymmetric and transitive.

Correct answer: C. It is irreflexive and transitive

Explanation: The relation a<b compares two elements of A and includes a pair only when the first element is smaller than the second. It is irreflexive because no element is smaller than itself: a<a is always false. It is transitive because if a<b and b<c, then a is smaller than c, so a<c.

It is not reflexive, since no (a,a) belongs to the relation. It is not symmetric: for example, (1,2) belongs, but (2,1) does not. Therefore it cannot be an equivalence relation, which requires reflexivity and symmetry. The correct description is “irreflexive and transitive,” so option C is correct. The finite set does not change these basic properties of the less-than relation.

Correct answer: A. It is an equivalence relation

Explanation: All diagonal pairs are present and ((2,1)) is also present with ((1,2)). The classes ({1,2}) and ({3}) show transitivity too.

Correct answer: A. ((1,3))

Explanation: From ((1,2)) and ((2,3)), transitivity requires ((1,3)). In such questions, match the middle element.