Relations

For every (a>1), (\gcd(a,a)>1), and (\gcd(a,b)=\gcd(b,a)). But (2R6) and (6R3) are true while (2R3) is false.

(11) is odd and (a-b) being even means the same parity. Hence ([11]) is the class of all odd integers.

Every line is parallel to itself and parallelism is symmetric. Lines parallel to the same direction also give transitivity.

If (l\perp m), then (m\perp l), so symmetry holds. No line is perpendicular to itself and perpendicularity is not transitive.

Every triangle is similar to itself. Similarity is also symmetric and transitive, so it is an equivalence relation.

Reflexive means all diagonal entries are \(1\). Here the diagonal of \(M_R\) is \(1,1,1\), so \(R\) is reflexive. For symmetry we need \(a_{ij}=a_{ji}\) for all \(i,j\). From the matrix, \(a_{23}=1\) but \(a_{32}=0\), so \(R\) is not symmetric. Hence \(R\) is reflexive but not symmetric.

All diagonal entries are (1) and no opposite off-diagonal pair occurs together. But ((1,2)) and ((2,3)) are present while ((1,3)) is absent.

Transitivity requires that whenever (x,y) and (y,z) are in the relation, (x,z) must also be present. The chain (1,2) followed by (2,4) forces the addition of (1,4). The chain (2,4) followed by (4,5) forces (2,5). After adding (1,4), the pair (1,4) followed by (4,5) creates another requirement, namely (1,5). Now there are no further composable chains that force new pairs. Thus the transitive closure adds exactly (1,4), (2,5), and (1,5), giving three pairs. Option B is correct. Adding only two would leave the chain (1,4),(4,5) without (1,5), so two cannot be sufficient; four or five are not minimal.

The direct answer is option A: (1,3), (2,4), and (1,4). A transitive relation must contain (a,c) whenever it contains (a,b) and (b,c). Start with the chain 1→2, 2→3, and 3→4. From 1→2 and 2→3, add (1,3). From 2→3 and 3→4, add (2,4). Once 1→3 and 3→4 are available, add (1,4); equivalently, the path 1→2→3→4 gives (1,4). These are the extra forward-reachable pairs. Option A lists exactly them. Option B reverses the arrows and is not required by transitivity. Option C adds self-pairs, but no cycle or rule requires (1,1), (2,2), or (3,3). Option D points backward from 4 and also does not follow from the given chain. The original pairs remain, while the three listed pairs are added. Memory cue: transitive closure connects every reachable start to every later endpoint.

The equivalence relation induced by a partition relates two elements exactly when they lie in the same block. If a block has k elements, every element in it can be paired with each of the k elements, including itself, so that block contributes k² ordered pairs. The blocks here have sizes 2, 3, and 1. Their contributions are therefore 2² = 4, 3² = 9, and 1² = 1. Pairs connecting different blocks are excluded, because elements in different blocks are not equivalent. The total is 4 + 9 + 1 = 14, so option C is correct. The value 36 counts every pair in A×A for six elements and ignores the partition restriction; 10 and 12 omit some valid within-block pairs.

In the equivalence relation induced by a partition, two elements are related exactly when they belong to the same block. The block \(\{1,4\}\) contributes \(2^2=4\) ordered pairs, and the block \(\{2,3,5\}\) contributes \(3^2=9\) ordered pairs. Therefore, the total is \(4+9=13\). Option 9 counts only the pairs from the larger block. Exam tip: square the size of each block and add the results.

The number of equivalence relations equals the number of partitions. A set of (3) elements has (5) partitions.

The class \([1]_R\) contains all \(b\in A\) such that \(b^2\equiv 1^2\equiv 1\pmod{8}\). Every odd number has square congruent to 1 modulo 8, so \(1^2,3^2,5^2,7^2\equiv 1\pmod{8}\). Hence \([1]_R=\{1,3,5,7\}\). Option B lists the even elements, whose squares are not congruent to 1 modulo 8. Exam tip: To find \([a]_R\), test each element \(b\) of the set using \(b^2\equiv a^2\pmod{8}\).

The class \([-1]\) contains all elements of \(A\) that are congruent to \(-1\) modulo 3, meaning their difference from \(-1\) is divisible by 3. Since \(2-(-1)=3\), we have \(2\equiv -1\pmod{3}\), so \([-1]=\{-1,2\}\). The elements in option B form the class of \(1\), not of \(-1\). Exam tip: To find a congruence class modulo 3, select from the given set the elements that differ from the representative by a multiple of 3.

For positive integers, (a\mid b) and (b\mid a) imply (a=b). Hence it behaves like the identity relation and is reflexive, symmetric and transitive.

(\frac{a}{b}>0) means (a) and (b) have the same sign. Having the same sign is reflexive, symmetric and transitive.

(ab>0) happens when both numbers have the same sign. Hence the two classes are positive and negative real numbers.

For every real number a, \(|a-a|=0\leq2\), so the relation is reflexive. Also, \(|a-b|=|b-a|\), so it is symmetric. However, 0R2 and 2R4 are true, whereas \(|0-4|=4>2\), making 0R4 false; hence the relation is not transitive. Therefore, option A is correct. Exam tip: To test transitivity, choose three numbers where the first two differences are within the allowed limit but the total difference exceeds it.

From (a+b=0), we also get (b+a=0). But (aRa) needs (2a=0), and transitivity does not hold in general.

Given \(a^3=b^3\). Over real numbers, cube is one-to-one, so \(a^3=b^3\Rightarrow a=b\). Therefore the relation is the same as equality: \(a=b\). Equality is reflexive because \(a=a\); symmetric because if \(a=b\) then \(b=a\); and transitive because if \(a=b\) and \(b=c\) then \(a=c\). Hence \(R\) is an equivalence relation.

The equivalence class is \([8]=\{a\in A:a\equiv 8\pmod{6}\}\). Since 8 leaves remainder 2 when divided by 6, we select all elements of \(A\) that also leave remainder 2: \(2,8,14\). Hence, \([8]=\{2,8,14\}\). Option B is incorrect because 15 leaves remainder 3 on division by 6. Exam tip: To find an equivalence class, collect every element having the same remainder modulo the given number.

If a relation is both symmetric and antisymmetric, no off-diagonal pair can occur. Only the (5) diagonal pairs are free, so the count is (2^5).

Reflexivity makes all diagonal pairs compulsory. Symmetry and antisymmetry together forbid off-diagonal pairs, so only the identity relation is possible.

By definition, \(R^{-1}\) is obtained by swapping each ordered pair:
\[(a,b)\in R^{-1} \iff (b,a)\in R.\]
Given \(R=R^{-1}\), any \((a,b)\in R\) is also in \(R^{-1}\). So \((b,a)\in R\).
That is exactly the definition of a symmetric relation. Therefore, \(R\) is symmetric.

In the inverse relation, the positions in each ordered pair are interchanged. Thus ((1,2),(2,4),(3,1)) become ((2,1),(4,2),(1,3)).