Equivalence relation

Step 1: (\gcd(6,9)=3). Step 2: In the given set, (\gcd(3,9)=3) and (\gcd(6,9)=3). Step 3: Elements with the same greatest common divisor lie in the same class.

Step 1: The coordinate sum of ((2,-1)) is (1). Step 2: All related points must satisfy (x+y=1). Step 3: In a same-sum relation, the equivalence class is a line.

Step 1: The value of (x-y) stays fixed in the relation. Step 2: The equation (x-y=c) represents a straight line. Step 3: Different values of (c) give parallel equivalence classes.

Step 1: If (aRb), then (a) and (b) lie in the same equivalence class. Step 2: Therefore ([a]=[b]). Step 3: The difference of equal sets is the empty set.

Step 1: Equivalence classes are either equal or disjoint. Step 2: If ([a]\setminus[b]\neq\varnothing), the classes are not equal. Step 3: Therefore their intersection is empty.

Step 1: The class sizes are (2,2,2,1). Step 2: The number of pairs is (2^2+2^2+2^2+1^2). Step 3: The total is (4+4+4+1=13).

Step 1: The condition (a=b) gives all diagonal pairs. Step 2: The condition (a+b=8) links (1,7), (2,6), (3,5), while (4) remains alone. Step 3: These form complete disjoint classes, so the relation is an equivalence relation.

Step 1: (10-3=7), which is divisible by (7). Step 2: Therefore (3) and (10) are in the same equivalence class. Step 3: The intersection of equal classes is that same class.

Step 1: (11-3=8), which is not divisible by (7). Step 2: Thus (3) and (11) lie in different equivalence classes. Step 3: Distinct equivalence classes have empty intersection.

Step 1: ((\sqrt{2}+5)-\sqrt{2}=5). Step 2: Since (5) is rational, (\sqrt{2}+5) lies in the same class. Step 3: Adding a rational number does not change the class in this relation.

Step 1: From (aRb), (\frac{a}{b}) is rational, and from (bRc), (\frac{b}{c}) is rational. Step 2: Their product is (\frac{a}{c}). Step 3: A product of rational numbers is rational, proving transitivity.

Step 1: There are (6) remainder classes, each with (2) elements. Step 2: Each class contributes (2^2=4) ordered pairs. Step 3: The total number of pairs is (6\cdot4=24).

Step 1: If no two distinct elements are in the same class, every class is a singleton. Step 2: This gives only the identity relation. Step 3: Hence there is exactly (1) such equivalence relation.

Step 1: If all elements are in one class, every element is related to every element. Step 2: This is only the universal relation. Step 3: Therefore there is exactly (1) such equivalence relation.

Step 1: (A\times A) has (5^2=25) ordered pairs. Step 2: (R) contains (2^2+1^2+2^2=9) pairs. Step 3: The complement has (25-9=16) pairs.

Step 1: Since (R) is an equivalence relation, every ((a,a)) is in (R). Step 2: Therefore no ((a,a)) remains in the complement. Step 3: Without reflexivity, the complement cannot be an equivalence relation.

Step 1: (1,2,3) are mutually connected and form one class. Step 2: (4) is related only to itself through ((4,4)). Step 3: Hence the class of (4) is ({4}).

Step 1: If one relation is contained in the other, the union is just the larger relation. Step 2: The larger relation is already an equivalence relation. Step 3: Therefore the union is definitely an equivalence relation in this case.

Step 1: The two class sizes must add to (7). Step 2: The sum of squares is minimized when the sizes are as balanced as possible, (3) and (4). Step 3: The number of pairs is (3^2+4^2=25).

Step 1: The two non-empty class sizes must add to (7). Step 2: The sum of squares is maximum when the sizes are most unequal, (6) and (1). Step 3: The pair count is (6^2+1^2=37).