Types of relations

Correct answer: A. Reflexive but neither symmetric nor transitive

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: ((1,2)) is present but ((2,1)) is not, so symmetry fails. Step 3: ((2,3)) and ((3,4)) are present but ((2,4)) is missing, so transitivity also fails.

Correct answer: A. Reflexive and transitive but not symmetric

Step 1: Every number divides itself, so the relation is reflexive. Step 2: If (a\mid b) and (b\mid c), then (a\mid c), so it is transitive. Step 3: (2\mid4) is true but (4\mid2) is not, so it is not symmetric.

Correct answer: A. It is an equivalence relation

Step 1: (a-a=0) is divisible by (7), so the relation is reflexive. Step 2: If (a-b) is divisible by (7), then (b-a) is also divisible by (7), so it is symmetric. Step 3: The sum of two differences divisible by (7) is also divisible by (7), so it is transitive.

Correct answer: A. Reflexive and symmetric but not transitive

Step 1: (|a-a|=0\le2), so every element is related to itself. Step 2: (|a-b|=|b-a|), so the relation is symmetric. Step 3: ((1,3)) and ((3,5)) exist but ((1,5)) does not, so transitivity fails.

Correct answer: A. Equivalence relation

Step 1: For every (a), (a^2=a^2), so the relation is reflexive. Step 2: If (a^2=b^2), then (b^2=a^2), so it is symmetric. Step 3: Equality of squares passes through a third element, so it is transitive.

Correct answer: A. Equivalence relation

Step 1: (a-a=0) is rational, so the relation is reflexive. Step 2: If (a-b) is rational, then (b-a) is rational, so it is symmetric. Step 3: The sum of rational differences is rational, so it is transitive.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: (a-a=0) is rational, so the relation is not reflexive. Step 2: If (a-b) is irrational, then (b-a) is irrational, so it is symmetric. Step 3: (0R\sqrt{2}) and (\sqrt{2}R0) hold but (0R0) does not, so transitivity fails.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: (a+a=5) is not true for every (a), so the relation is not reflexive. Step 2: If (a+b=5), then (b+a=5), so it is symmetric. Step 3: ((1,4)) and ((4,1)) exist but ((1,1)) does not, so it is not transitive.

Correct answer: A. Equivalence relation

Step 1: The condition (a=b) gives all self-pairs, so it is reflexive. Step 2: The condition (a+b=5) is unchanged by reversing order, so it is symmetric. Step 3: The relation forms closed classes ({1,4}) and ({2,3}), so it is also transitive.

Correct answer: B. 2^9

The correct answer is B, 2^9. A relation on a five-element set uses ordered pairs, so there are 25 possible pairs. Reflexivity forces the five diagonal pairs (1,1), (2,2), ..., (5,5) to be included. For two different elements, symmetry makes (i,j) and (j,i) a single joint decision. There are C(5,2)=10 unordered pairs of distinct elements. Because (1,2) is required, the pair {1,2} is already fixed, and symmetry also forces (2,1). The other nine unordered pairs can independently be included or excluded, giving 2^9 relations. A is too small because only one choice is not removed; C incorrectly leaves all ten choices free; D counts all ordered non-diagonal possibilities without respecting symmetry. Memory cue: diagonal pairs are fixed by reflexivity, while each off-diagonal mirror pair contributes one binary choice.

Correct answer: A. (2^{\frac{n(n+1)}{2}})

Step 1: The (n) self-pairs can be chosen independently. Step 2: There are (\frac{n(n-1)}{2}) unordered pairs of distinct elements. Step 3: Total independent choices are (n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}).

Correct answer: A. (1:2^4)

Step 1: The total number of relations is (2^{16}). Step 2: Reflexive relations have (4) compulsory self-pairs, so their number is (2^{12}). Step 3: The ratio is (2^{12}:2^{16}=1:2^4).

Correct answer: A. ({1,4})

Step 1: (a\equiv b \pmod{3}) means both have the same remainder on division by (3). Step 2: (1) and (4) both leave remainder (1). Step 3: Hence the equivalence class of (1) is ({1,4}).

Correct answer: A. (8)

Step 1: In an equivalence relation, all ordered pairs within each class are included. Step 2: ({1,2}) gives (2^2=4) pairs and ({3,4}) gives (2^2=4) pairs. Step 3: Total pairs are (4+4=8).

Correct answer: A. (13)

Step 1: All ordered pairs are formed inside each equivalence class. Step 2: The first class gives (3^2=9) pairs and the second gives (2^2=4) pairs. Step 3: The total number is (9+4=13).

Correct answer: A. ({(1,2),(2,3),(1,3)})

Step 1: Transitivity requires ((1,3)) from ((1,2)) and ((2,3)). Step 2: Adding only ((1,3)) creates no further missing forced chain. Step 3: Hence this is the smallest transitive extension.

Correct answer: A. ((1,3),(2,4),(1,4))

Step 1: ((1,3)) is required from ((1,2)) and ((2,3)). Step 2: ((2,4)) is required from ((2,3)) and ((3,4)). Step 3: Then ((1,4)) is also forced by ((1,3)) and ((3,4)).

Correct answer: A. ((1,1)) and ((2,2))

Step 1: From ((1,2)) and ((2,1)), ((1,1)) is required. Step 2: From ((2,1)) and ((1,2)), ((2,2)) is required. Step 3: No chain involves (3), so these are the minimum additions.

Correct answer: A. (R\cap S) will be reflexive

Step 1: Since (R) is reflexive, every ((a,a)) belongs to (R). Step 2: Since (S) is reflexive, every ((a,a)) belongs to (S). Step 3: Therefore all self-pairs belong to (R\cap S), so it is reflexive.

Correct answer: A. (R\cup S) will be symmetric

Step 1: If a pair is in (R\cup S), it is in (R) or in (S). Step 2: In that relation, symmetry gives the reverse pair. Step 3: Therefore the reverse pair is also in (R\cup S), so the union is symmetric.

Correct answer: A. It is not always transitive

Step 1: Transitivity needs a third pair from two connected pairs. Step 2: In (R\cup S), one pair may come from (R) and the other from (S). Step 3: The required third pair may be missing, so the union need not be transitive.

Correct answer: A. Transitive but neither reflexive nor symmetric

Step 1: (R^{-1}={(2,1),(3,2),(3,1)}). Step 2: It has no self-pair and does not contain all reverse pairs. Step 3: From ((3,2)) and ((2,1)), ((3,1)) is present, so it is transitive.

Correct answer: A. (R^{-1}=R)

Step 1: In a symmetric relation, ((b,a)) is present whenever ((a,b)) is present. Step 2: The inverse relation reverses all pairs. Step 3: Since the reversed pairs are already present, (R^{-1}=R).

Correct answer: A. Symmetric and transitive but not reflexive

Step 1: (1) and (4) are not prime, so their self-pairs are absent. Step 2: The condition that both elements are prime is unchanged by reversing order, so it is symmetric. Step 3: The relation is complete inside ({2,3}), so it is transitive.

Correct answer: A. Symmetric and transitive but not reflexive

Step 1: Self-pairs of (3,4,5) are not in the relation, so it is not reflexive. Step 2: The condition that both elements are less than (3) is unchanged by reversal, so it is symmetric. Step 3: The relation is complete inside ({1,2}), so it is transitive.