Relations

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

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.

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 \(\{(a,a):a\in A\}\).

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

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.

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 \((b,a)\).

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.

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.

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

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

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.

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.

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}.

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 \(b=m\).

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.

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.

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.

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

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

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

(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}).

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

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.

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

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