Relations

It is reflexive and the reverses of ((1,2),(2,1)) are present. The required transitive pairs ((1,1),(2,2)) are also present.

Two numbers are congruent modulo 2 exactly when they have the same parity. Thus, A is partitioned into two equivalence classes: {1, 3} (odd numbers) and {2, 4} (even numbers). Therefore, the correct answer is 2. The answer 4 would incorrectly treat every element as a separate class, even though elements with the same parity belong to one class. Exam tip: for congruence modulo 2, group the elements into odd and even numbers.

The equivalence class of \(1\) contains all elements that leave the same remainder as \(1\) upon division by \(3\). In \(A\), only \(1\) and \(4\) leave remainder \(1\); the remainders of \(2,3,5,6\) are \(2,0,2,0\), respectively. Hence, \([1]_R=\{1,4\}\). Exam tip: For a modulo relation, group elements having the same remainder.

The direct answer is option A, a reflexive relation. A relation R on A is reflexive if every element a in A is related to itself; symbolically, (a,a) must belong to R for every a in A. This is exactly the condition stated in the question, so no calculation or example is needed beyond matching the definition. Option A is correct because its definition is precisely the required self-relationship. Option B, symmetric relation, requires that whenever (a,b) is present, (b,a) is also present; it says nothing about (a,a) for every element. Option C, empty relation, contains no pairs and generally cannot be reflexive on a nonempty set because its self-pairs are missing. Option D, inverse relation, is formed by reversing the components of pairs; it is an operation or a related construction, not the property described here. Memory cue: reflexive means each element points back to itself.

In transitivity, two connected pairs must imply a pair between the first and last elements. Therefore check the presence of ((a,c)).

The relation is defined by x + y = 5. To test symmetry, suppose (x, y) belongs to R. Then x + y = 5. Addition is commutative, so y + x = x + y = 5, which means (y, x) also belongs to R. Therefore R is symmetric, making option A correct. For example, (1, 4) and (4, 1), as well as (2, 3) and (3, 2), occur in reverse pairs. It is not reflexive because (x, x) would require 2x = 5, which is not true for every x in A. It is not universal because many pairs, such as (1, 1), do not satisfy the sum condition, and it is not empty because valid pairs do exist.

Every (x\leq x), so it is reflexive, and (x\leq y\leq z) gives (x\leq z). It is generally not symmetric.

All elements of the set are even, so the difference of any two elements is divisible by (2). Hence (R=A\times A).

In an inverse relation, every ordered pair \((a,b)\) is reversed to \((b,a)\). In \(R\), the reverse of \((1,3)\), namely \((3,1)\), is already present, while \((1,1),(2,2),(3,3)\) remain unchanged when reversed. Hence \(R^{-1}=R\), so option A is correct. Option B contains only the diagonal pairs and omits \((1,3)\) and \((3,1)\). Exam tip: If the reverse of every ordered pair in a relation is also in the relation, the relation is symmetric and \(R^{-1}=R\).

Direct answer: option D, (1,3), is not in R. Membership requires the product ab to be even. A product is even if at least one factor is even; it is odd only when both factors are odd. A: 1×2=2, which is even, so (1,2) belongs. B: 2×3=6, even, so (2,3) belongs. C: 3×4=12, even, so (3,4) belongs. D: 1×3=3, odd, so (1,3) does not belong. The numbers 1 and 3 are both odd, and odd multiplied by odd remains odd. Thus D is the only pair outside R. The order does not change the product here, because ab=ba, although ordered pairs themselves are still written in their given order. Memory cue: for an even product, look for at least one even number.

Since (|3-5|=2) and (2\leq 2), ((3,5)\in R). In absolute difference, equality with the bound is included.

A relation from (A) to (B) is always a subset of (A\times B). Order matters, so (B\times A) can be different.

The direct answer is option A: domain {1,2} and range {2,3}. The domain is the set of all first components, and the range is the set of all second components. From (1,2), we record 1 for the domain and 2 for the range. From (1,3), 1 is already present in the domain and 3 is added to the range. From (2,3), 2 is added to the domain while 3 is already present in the range. After removing repetitions, domain={1,2} and range={2,3}. Option A has both sets in the correct order. Option B reverses domain and range. Option C incorrectly includes 3 in the domain and 1 in the range, although neither occurs in the required position. Option D keeps only the first value 1 and last value 3, leaving out valid components 2 in the domain and 2 in the range. Memory cue: scan left entries for domain and right entries for range.

The ordered pairs of the relation must come from \(A\times A\) and satisfy \(a+b=6\). The valid pairs are \((2,4), (3,3), (4,2)\). The pair \((1,5)\) is not valid because \(5\notin A\). Hence, the relation contains 3 pairs. Exam tip: For each \(a\in A\), check whether \(6-a\) also belongs to \(A\).

The relation requires \(a=b^2\), with both \(a\) and \(b\) belonging to \(A\). For \(b=1\), \(a=1\), giving \((1,1)\); for \(b=2\), \(a=4\), giving \((4,2)\). For \(b=3\) and \(b=4\), the values of \(a\) are 9 and 16, which are not in \(A\). Hence option A is correct. Exam tip: Check the defining condition and membership in the given set for every ordered pair.

Here ((1,2)\in R) but ((2,1)\notin R), so it is not symmetric. Equivalence needs reflexive, symmetric, and transitive properties.

The odd elements in (A) are (1) and (3), so all ordered pairs are formed from them. Check both components in the condition.

For ((3,3)), (3^2+3^2=18), and (18\leq 10) is false. In exams, substitute each option into the condition.

Since (|A\times B|=3\times2=6), the number of relations is (2^6). In exams, first find the number of elements in (A\times B).

The domain is the set of first components of the ordered pairs, so ( {1,2,3} ) is correct. In exams, look at the first entries.

The range is the set of second components, and repeated values are written once. Hence ( {5,6,7} ) is correct.

The possible pairs are ( (1,2),(1,3),(1,4),(2,3),(2,4),(3,4) ), so the count is (6). In exams, list pairs systematically.

Each element is related only to itself, so it is the identity relation. In exams, identify pairs of the form ( (a,a) ).

A universal relation contains all pairs of (A\times A), and (|A\times A|=2\times2=4). In exams, treat universal relation as the complete Cartesian product.

To write R in roster form, consider ordered pairs from A × A and retain exactly those whose coordinates add to 4. Starting with first coordinate 1, the second coordinate must be 3, giving (1, 3). For first coordinate 2, the second must be 2, giving (2, 2). For first coordinate 3, the second must be 1, giving (3, 1). All three coordinates belong to A, so R = {(1, 3), (2, 2), (3, 1)}. Option A is therefore correct. Option B lists equal-coordinate pairs, option C lists pairs summing to 3, and option D contains a pair whose sum is 6. Because these are ordered pairs, both (1, 3) and (3, 1) must be included separately.