Relations

The number (2) divides (4). In a divisibility relation, ( (2,4) ) and ( (4,2) ) are not the same.

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

A relation \(R\) on \(A\) is reflexive only if \((a,a)\in R\) for every \(a\in A\). Here \((1,1),(2,2)\), and \((3,3)\) are present, but \((4,4)\) is missing; therefore, \(R\) is not reflexive. Exam tip: To test reflexivity, check the diagonal pair \((a,a)\) for every element of the set.

The direct answer is option A, Yes. A relation on a set A is reflexive when every element of A is related to itself. Here A has exactly two elements, x and y. Therefore we must check the two self-pairs: (x,x) and (y,y). Both of them are present in R. The pair (x,y) is an additional pair; it does not cause any problem, because reflexivity requires certain pairs to be present but does not forbid other pairs. Option A is correct because both required self-pairs occur. Option B, No, is wrong because neither required self-pair is missing. Option C, only empty, is wrong because R is not empty; it contains three pairs, and being empty is not the definition of reflexivity. Option D, only universal, is wrong because a universal relation must contain every pair in A × A, including (y,x), but that pair is absent. A reflexive relation need not be universal. Memory cue: for reflexive, check the diagonal pairs (a,a) for every element.

Here ( (3,4)\in R ), but ( (4,3)\notin R ). One missing reverse pair makes the relation non-symmetric.

The direct answer is option A, (1,5) is missing. Transitivity means: whenever (a,b) and (b,c) are in a relation, (a,c) must also be in it. In this relation, (1,3) is present and (3,5) is present. The middle entries match, so these pairs form a chain from 1 to 5. Transitivity therefore requires (1,5). That pair is absent, so the relation is not transitive. Option A is correct. Option B, (3,1), is not required; that would reverse the first pair and concerns neither transitivity nor the given chain. Option C, (5,3), is also not required; it reverses the second pair. Option D, (1,1), is not required because transitivity does not demand that an element relate to itself. It demands only the shortcut from the first starting element to the final ending element. Thus the exact failure is the missing pair (1,5). Exam cue: match the ending of the first pair with the beginning of the second pair, then connect the two outside entries.

A relation R on A is reflexive if every element of A is related to itself. For A={1,2,3}, reflexivity requires all three diagonal ordered pairs (1,1), (2,2), and (3,3) to be members of R. The empty relation contains no ordered pairs at all.

Since none of the required self-pairs is present in R=∅, the relation is not reflexive. Therefore option B, No, is correct. It is not enough that the relation is defined on A; it must actually contain each required pair. The empty relation can satisfy some other properties in suitable contexts, but it cannot be reflexive on a nonempty set because it lacks every diagonal pair.

Taking all pairs of (A\times A) gives the universal relation. You can think of it as a complete relation.

The governing concept is the equality relation on A. A pair (a,b) belongs to R exactly when its two components are equal. Thus the relation on A contains the diagonal pairs (1,1), (2,2), and (3,3). Option A satisfies 1 = 1, option B satisfies 2 = 2, and option C satisfies 3 = 3, so all three belong to R. In option D, the components are 1 and 3, and 1 is not equal to 3; therefore (1,3) does not belong to the relation. Option D is correct. The fact that both numbers are elements of A is not enough: they must also satisfy the defining condition a = b.

An ordered pair \((x,y)\) belongs to \(R\) when \(x+y\) is even. For option C, \(3+7=10\), which is even; hence \((3,7)\in R\). The sums in the other options are 3, 7 and 11, all odd. Exam tip: add the two components and check whether the sum is even or odd.

Since (4+8=12) is even, it will not belong to the odd-sum relation. For an odd sum, one number should be even and the other odd.

The governing concept is testing membership by substituting the ordered pair into the rule b = a + 2. For option A, a = 1 and b = 2, but 1 + 2 = 3, so it fails. For option B, a = 2 and b = 4, and 2 + 2 = 4, so the condition is satisfied; therefore (2,4) belongs to R. Option C reverses the suitable values: for (4,2), the rule would require b = 6, not 2. For option D, when a = 5, the required b is 7, not 3. Hence option B is the only correct answer. The order of an ordered pair must be preserved when applying the formula.

The direct answer is option C, (5,3). The condition says that a is two more than b. In simple equation form, a = b + 2, or a − b = 2. Test each option carefully. For (1,3), the first number is not two more than the second; 1 is two less than 3, so option A is wrong. For (2,4), 2 is also two less than 4, so option B is wrong. For (5,3), 5 = 3 + 2, so the condition is exactly satisfied and option C is correct. For (3,5), 3 is two less than 5, not two more, so option D is wrong. All numbers belong to A, but belonging to A alone is not enough; the stated numerical condition must also hold. Helpful method: translate words into a subtraction and check whether first number minus second number equals 2.

The governing concept is relation membership determined by the equation b = 2a. Evaluate the rule for each proposed ordered pair. For (4,2), the required second component is 2 × 4 = 8, so it fails. For (8,4), the required second component is 16, so it also fails. For (2,4), the required second component is 2 × 2 = 4, exactly matching the pair; hence (2,4) belongs to R. For (1,4), the required second component is 2, not 4. Therefore option C is correct. The first coordinate is multiplied by 2 to obtain the second coordinate; multiplying the second coordinate or reversing the pair would apply the rule incorrectly.

Since (2\cdot3=6), ( (2,3) ) belongs to the relation. In a product condition, check the product of both components.

The direct answer is option B: 2 ordered pairs. The relation is understood on A, so both a and b belong to \(A=\{1,2,3\}\), and they must satisfy \(ab=6\). Check the possible first values. If \(a=1\), then \(b=6\), which is not in A. If \(a=2\), then \(b=3\), giving \((2,3)\). If \(a=3\), then \(b=2\), giving \((3,2)\). Thus R contains exactly two ordered pairs. Option A, 1, forgets that reversing the order gives another ordered pair. Option B is correct. Option C, 3, includes an invalid possibility such as \((1,6)\), because 6 is outside A. Option D, 4, overcounts the solutions. The pairs \((2,3)\) and \((3,2)\) are different because order matters. Memory cue: for ordered pairs, swapping coordinates generally creates a separate pair.

An ordered pair (a, b) belongs to R only when a + b < 5. The sums for (1, 2), (2, 2), and (3, 1) are 3, 4, and 4, respectively, so these pairs belong to R. For (4, 2), a + b = 6, and 6 < 5 is false; therefore, it does not belong to R. Exam tip: For each option, add the two components and test the given inequality.

The direct answer is option A: \((5,3)\). A pair \((a,b)\) belongs to R exactly when \(a-b=2\). Test each option carefully. For A, \(5-3=2\), so \((5,3)\in R\). For B, \(3-5=-2\), not 2, so it does not belong. For C, \(4-4=0\), so it fails the condition. For D, \(2-5=-3\), so it also fails. Thus option A is the only correct pair. The order is essential because subtraction is not unchanged when the entries are swapped: \(a-b\) and \(b-a\) usually have opposite signs. A weak student can solve such questions by simply subtracting the second coordinate from the first coordinate in every option. Memory cue: read \((a,b)\) as first minus second, never second minus first.

Since (|5-4|=1), ( (5,4) ) belongs to the relation. In absolute value, the sign of the difference does not matter.

For transitivity, whenever \((a,b)\) and \((b,c)\) belong to the relation, \((a,c)\) must also belong to it. Here, \((1,2)\) and \((2,3)\) are present, so \((1,3)\) must be present, but it is missing from \(R\). Therefore, \(R\) is not transitive. Exam tip: Check the pair formed by the first element of the first ordered pair and the second element of the next pair.

A relation is reflexive if \((a,a)\in R\) for every \(a\in A\). Here, all three diagonal pairs \((1,1),(2,2)\), and \((3,3)\) are present, so \(R\) is reflexive. It is not irreflexive because an irreflexive relation contains no pair of the form \((a,a)\), and it is not universal because pairs such as \((1,3)\) are missing. Exam tip: check the diagonal pairs \((a,a)\) first when testing reflexivity.

The identity relation contains all diagonal pairs, and each reverse is the same pair. It is also transitive.

A universal relation contains all possible pairs. Hence reflexive, symmetric, and transitive conditions are all satisfied.

For a relation to be reflexive, \((a,a)\) must belong to the relation for every element \(a\) of the base set. Since \(A=\{1,2\}\), both \((1,1)\) and \((2,2)\) are required, but neither is in \(R\). The presence of \((1,2)\) and \((2,1)\) does not make the relation reflexive. Exam tip: to test reflexivity, check all diagonal pairs \((a,a)\).

The universal relation on A contains every ordered pair in A × A. For A = {1, 2}, these are exactly the four pairs listed in R, so C is correct. If A included 3, pairs involving 3 would also be needed.