Equivalence relation

Step 1: (18=2\cdot 3^2), so its greatest prime factor is (3). Step 2: The relation checks only the greatest prime factor. Step 3: Therefore the class contains all natural numbers whose greatest prime factor is (3).

Step 1: For ((4,1)), (4-1=3). Step 2: Every pair related to it must have first component minus second component equal to (3). Step 3: Hence the equivalence class is ({(x,y):x-y=3}).

Step 1: The (x)-coordinate of ((3,-2)) is (3). Step 2: All points in the same class must have (x)-coordinate (3). Step 3: The (y)-coordinate can be any real number, so the class is a vertical line.

Step 1: The distance of ((6,8)) from the origin is (\sqrt{6^2+8^2}=10). Step 2: All points related to it must also have distance (10). Step 3: All points at the same distance from the origin form a circle centered at the origin.

Step 1: (1) is related to (2), and (2) is related to (3). Step 2: Transitivity requires (1) to be related to (3). Step 3: To preserve symmetry, both ((1,3)) and ((3,1)) must be added.

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.