Equivalence relation

Step 1: (a) always belongs to its own class ([a]). Step 2: Since ([a]\subseteq[b]), we get (a\in[b]), so (a) and (b) are related. Step 3: In an equivalence relation, related elements have equal classes, hence ([a]=[b]).

Step 1: In an equivalence relation, two classes are either equal or disjoint. Step 2: If (a) and (b) are not related, their classes cannot be equal. Step 3: Therefore the intersection of their classes is empty.

Step 1: The first class ({1,4}) contributes (2^2=4) pairs. Step 2: The second class ({2,3,5}) contributes (3^2=9) pairs. Step 3: Total ordered pairs are (4+9=13).

Step 1: Equivalence classes partition the set without leaving out or repeating elements. Step 2: The sizes of two classes must add up to (5). Step 3: Among the options, only (2+3=5) is correct.

Step 1: The product of digits of (23) is (2\cdot 3=6). Step 2: The product of digits of (16) is (1\cdot 6=6). Step 3: Since the digit product is the same, (16) lies in the same equivalence class.

Step 1: Reflexivity requires (aRa) for every (a). Step 2: (a-a=0), and (0) is rational, not irrational. Step 3: Hence no element is related to itself, so this is not an equivalence relation.

Step 1: (a-b\ge 0) means (a\ge b). Step 2: (2R1) is true, but (1R2) is false. Step 3: Since symmetry fails, it cannot be an equivalence relation.

Step 1: Reflexivity would require (|a|<|a|) for every (a). Step 2: No number's absolute value is less than itself. Step 3: Hence (aRa) is never true, so the relation is not equivalence.

Step 1: Every triangle has the same perimeter as itself. Step 2: Equality of perimeter works in both directions. Step 3: If the first and second perimeters are equal and the second and third are equal, then the first and third are equal too.

Step 1: Every circle has the same centre as itself. Step 2: Having the same centre is symmetric and transitive. Step 3: Hence each class contains all circles with one fixed centre, while radii may differ.

Step 1: If two lines intersect, the statement is also true in the reverse direction. Step 2: But if (l) intersects (m) and (m) intersects (n), it is not necessary that (l) intersects (n). Step 3: Failure of transitivity prevents it from being an equivalence relation.

Step 1: A person's height is equal to his or her own height. Step 2: If the first person's height equals the second's, the reverse is also true. Step 3: Equality of height also transfers through a third person, so this is an equivalence relation.

Step 1: Every student is in the same class as himself or herself. Step 2: Being in the same class works both ways and also through a third student. Step 3: Therefore the equivalence classes are the groups of students in each actual class.

Step 1: The multiples of (3) are ({3,6,9}), so their count is (3). Step 2: The remaining elements ({1,2,4,5,7,8}) have count (6). Step 3: Total pairs are (3^2+6^2=9+36=45).

Step 1: Reflexivity would require (\min(a,a)=a) to be odd for every (a). Step 2: For (a=2), (\min(2,2)=2), which is not odd. Step 3: Hence not all elements are related to themselves, so it is not an equivalence relation.

Step 1: (3^2-3=6). Step 2: (x^2-x=6) gives (x^2-x-6=0), that is ((x-3)(x+2)=0). Step 3: Hence (x=3) or (x=-2), so the class is ({-2,3}).

Step 1: ((-1)^2+2(-1)=1-2=-1). Step 2: (x^2+2x=-1) gives (x^2+2x+1=0), or ((x+1)^2=0). Step 3: Only (x=-1) is obtained, so the equivalence class is singleton.

Step 1: (\sin^2 0=0). Step 2: (\sin^2 \pi=0), so (\pi) is in the same class as (0). Step 3: The class contains angles whose squared sine value is (0).

Step 1: (\cos\frac{\pi}{3}=\frac{1}{2}), so the squared value is (\frac{1}{4}). Step 2: (\cos\frac{2\pi}{3}=-\frac{1}{2}), whose square is also (\frac{1}{4}). Step 3: After squaring, positive and negative values may become equal.

Step 1: A number (x) is related to (3) when (\frac{x}{3}=2^n), where (n\in Z). Step 2: This gives (x=3\cdot 2^n). Step 3: For ratio-based relations, write the class in multiplicative form.

Step 1: (12=2^2\cdot 3), so it has exactly power (2) of (2). Step 2: The relation checks only this power of (2). Step 3: Hence the class contains numbers divisible by (4) but not by (8).

Step 1: (R\cap S) keeps only pairs that are present in both relations. Step 2: Thus two elements remain together only when they are together in both (R) and (S). Step 3: Therefore its classes are common refined blocks of the two partitions.

Step 1: ((1,2)) belongs to (R\cup S) because it is in (R). Step 2: ((2,3)) belongs to (R\cup S) because it is in (S). Step 3: Transitivity would require ((1,3)), but it is in neither (R) nor (S).

Step 1: The largest relation contains all possible ordered pairs of (A). Step 2: (A\times A) is reflexive, symmetric, and transitive. Step 3: Therefore the universal relation is the largest equivalence relation.

Step 1: In an equivalence relation, every element must be related to itself. Step 2: Taking only the pairs ((a,a)) preserves symmetry and transitivity. Step 3: Hence the identity relation is the smallest equivalence relation.