Relations

Every element of \\(R\\) is an ordered pair. In both \\( (1,2) \\) and \\( (2,3) \\), the first and second components belong to \\(A\\), so both pairs belong to \\(A\\times A\\). Therefore, \\(R\\subseteq A\\times A\\). Option A is incorrect because the elements of \\(R\\) are ordered pairs, not individual numbers from \\(A\\). Exam tip: Any relation defined on a set \\(A\\) is a subset of \\(A\\times A\\).

There are (3^2=9) pairs in (A\times A), so it has (2^9) subsets. The number of relations is (2^{n(A)^2}).

The direct answer is option B: \(2^6\) relations. A relation from A to B is any subset of \(A\times B\). Since A has 2 elements and B has 3 elements, \(A\times B\) has \(2\times3=6\) ordered pairs. For each of these six pairs, a relation has two choices: include it or do not include it. Therefore the total number is \(2\times2\times2\times2\times2\times2=2^6\). Option A, \(2^5\), uses only five pairs and misses one. Option B is correct. Option C, 6, counts the pairs in the Cartesian product, not all possible relations. Option D, \(3^2\), reverses the relevant counting idea and is not the number of subsets of six pairs. A useful rule is: if \(|A|=m\) and \(|B|=n\), the number of relations from A to B is \(2^{mn}\).

In every ordered pair, the second component is twice the first: \(2=2(1)\), \(4=2(2)\), and \(6=2(3)\). Hence the condition is \(y=2x\), with \(x\) restricted to \(\{1,2,3\}\). Option A uses \(y=x\), while option D does not reproduce all the given ordered pairs. Exam tip: in a restricted set-builder form, check both the rule and the stated domain or Cartesian product.

The relation is defined by \(a-b=0\). Adding \(b\) to both sides gives \(a=b\), so every ordered pair in \(R\) has equal components and is of the form \((a,a)\). Options A and B describe inequalities, while option D would require \(a+b=0\). In the exam, simplify the defining condition first to identify the form of the ordered pairs.

Relation \(R\) contains an ordered pair \((a,b)\) only when \(a<b\). For \((2,4)\), the condition becomes \(2<4\), which is true; therefore, \((2,4)\in R\). Option B is incorrect because the pair does satisfy the condition, and option C is incorrect because reversing the pair would require \(4<2\), which is false. Exam tip: In an ordered pair, changing the order of the elements can change whether the pair belongs to the relation.

The relation is defined by the condition a+b=4, with both entries taken from A={1,2,3}. Test each offered ordered pair using the sum of its first and second components. For (1,1), the sum is 2, so it fails. For (1,3), the sum is 1+3=4, so it satisfies the defining condition and belongs to R. For (2,3), the sum is 5, and for (3,3), it is 6; neither belongs to R. Therefore option B is correct. In a relation described by a rule, the order of the components matters because the first component is a and the second is b.

A relation \(R\) on \(A\) is reflexive if \((a,a)\in R\) for every \(a\in A\). Here, all three required diagonal pairs—\((1,1)\), \((2,2)\), and \((3,3)\)—are present in \(R\). Therefore, \(R\) is reflexive on \(A\). Option B is incorrect because no required pair is missing. Exam tip: To test reflexivity, check whether every element of the set is related to itself.

A relation R on A is reflexive when every element of A is related to itself; equivalently, (a,a) must belong to R for every a ∈ A. Although (1,1) and (2,2) are present, (3,3) is missing. One missing required diagonal pair is enough to disprove reflexivity, so option C is correct. The absence of (1,2) is irrelevant to this property.

In symmetric relation, if ( (a,b)\in R ), then ( (b,a)\in R ) must also be present. Here both ( (1,2) ) and ( (2,1) ) are present.

The definition of transitivity says that if (a,b) ∈ R and (b,c) ∈ R, then (a,c) must also be in R. Since (1,2) and (2,3) are both in R, the required consequence is (1,3). That pair is absent, so the relation fails the condition and is not transitive. Reverse pairs and diagonal pairs are not required for this specific test; therefore A is correct.

There are three ordered pairs listed in (R). While counting, count pairs and not repeated elements.

The relation \(R\) contains ordered pairs whose two components have an even sum. Here, \(1+3=4\), and 4 is even; therefore, \((1,3)\in R\). Option B is incorrect because the sum is not odd. In an exam, substitute the given values into the defining condition first.

For \((x,y)\) to belong to \(R\), \(x\) must divide \(y\) exactly. Here, \(2\mid 6\) because \(6\div 2=3\) with remainder 0, so \((2,6)\) is correct. In \((2,5)\) and \((3,8)\), a remainder is left, while \((5,2)\) reverses the required order. Exam tip: \(x\mid y\) means that \(y/x\) is an integer.

When (a=b), the pairs are ( (1,1),(2,2),(3,3),(4,4) ). In identity relation, the number of pairs is (n(A)).

The direct answer is option D,
(1,3). A relation defined by a>b contains an ordered pair (a,b) only when the first number is greater than the second number. Check each pair carefully in its given order: for (2,1), 2>1 is true, so option A belongs to R. For (3,1), 3>1 is true, so option B belongs to R. For (3,2), 3>2 is true, so option C belongs to R. For (1,3), 1>3 is false; in fact, 1<3, so option D does not belong to R. The order cannot be reversed: (1,3) is not the same as (3,1). Remember: in an inequality relation, always compare the first component with the second component exactly as written. Memory cue: first number must be bigger.

The direct answer is option A: the reverse of \((2,3)\) is \((3,2)\), and it is present. An ordered pair’s reverse is formed by interchanging its coordinates. Thus the reverse of \((a,b)\) is \((b,a)\). Applying this to \((2,3)\) gives \((3,2)\), which appears explicitly in R. Option A is correct. Option B, \((2,1)\), is the reverse of \((1,2)\), not of \((2,3)\). Option C, \((1,2)\), is also unrelated as the reverse of the given pair; it is the reverse of \((2,1)\). Option D, \((3,3)\), keeps equal coordinates and cannot be obtained by reversing \((2,3)\). The presence of both \((2,3)\) and \((3,2)\) shows this pair is matched in reverse, although the question only asks which reverse pair is present. Memory cue: reverse an ordered pair by swapping first and second positions, never by changing their values.

A reflexive relation must contain the diagonal pair of every element. For (A={1,2}), ( (1,1) ) and ( (2,2) ) are compulsory.

For every element of A, the required ordered pair (a,a) is present in R: (1,1), (2,2), and (3,3). Hence, R is reflexive. The additional pairs (1,2) and (2,1) do not affect reflexivity. Option C is incorrect because R is not only symmetric; it is both symmetric and reflexive. Exam tip: To test reflexivity, check whether every pair of the form (a,a) belongs to the relation.

The governing concept is membership in a relation defined by a condition on an ordered pair. A pair (a,b) belongs to R only when its second component is exactly one more than its first, so b=a+1. For option A, (1,3), the required value would be 1+1=2, not 3. For option B, (2,3), we get 2+1=3, so the condition is satisfied and both entries belong to A. For option C, 2 is not equal to 3+1, and for option D, 1 is not equal to 4+1. Therefore (2,3) is the only valid pair, making option B correct. The order of the components must not be reversed.

The pairs from the given set are ( (2,3) ) and ( (3,2) ). In ordered pairs, ( (2,3) ) and ( (3,2) ) are different.

The relation R is defined by the rule a ≤ b, with both a and b taken from A. Therefore a pair belongs to R exactly when its first number is less than or equal to its second number. Equality is allowed, so pairs such as (1,1) and (3,3) satisfy the rule. The order of the two components must be preserved.

For (1,1), 1 ≤ 1 is true. For (2,4), 2 ≤ 4 is true. For (3,3), 3 ≤ 3 is true. For (4,2), the required comparison is 4 ≤ 2, which is false. Thus (4,2) is the only listed pair not in R, and option D is correct. Reversing it to (2,4) would make the inequality true.

The governing rule is x=y+1. Thus, in every valid ordered pair, the first coordinate must be exactly one greater than the second coordinate. For option A, x=2 and y=1, and 2=1+1, so (2,1) satisfies the definition. Option B fails because its first coordinate is 1, whereas 2+1=3. Option C fails because 3 is not equal to 5+1, and option D fails because 1 is not equal to 1+1. Hence option A is the unique correct answer. This question also tests the importance of order: even if two numbers are related in one direction, reversing their positions generally produces a different ordered pair and need not satisfy the same relation.

The identity relation on a set A is defined as I_A={(a,a):a∈A}; it pairs every element with itself and no other element. Since A={1,2,3}, write one self-pair for each member: (1,1), (2,2), and (3,3). Therefore option B is correct. Option A contains pairs connecting different elements, so it is not the identity relation. Option C is the empty relation and does not include the required self-pairs. Option D is not written as a set of ordered pairs and, even as a set, it is not the identity relation under the standard definition. The diagonal pairs are also exactly what make the identity relation reflexive on A.

In an empty relation, there is no pair that can break symmetry. Hence the empty relation is considered symmetric.