Introduction to Relations

Step 1: The (3) diagonal pairs are independently optional. Step 2: For (\frac{3\cdot2}{2}=3) pairs of distinct elements, each pair gives three choices. Step 3: Hence the number of relations is (2^3\cdot3^3).

Step 1: Reflexive symmetric relations have (6) independent off-diagonal reverse-pair groups, so there are (2^6) such relations. Step 2: Among these, only the identity relation is also antisymmetric. Step 3: Therefore the required number is (2^6-1).

Step 1: A reflexive relation must contain all (4) diagonal pairs. Step 2: The remaining (16-4=12) pairs are optional. Step 3: On a non-empty set, a reflexive relation cannot be irreflexive, so the count is (2^{12}).

Step 1: Total relations are (2^9), and symmetric relations are (2^6). Step 2: Antisymmetric relations are (2^3\cdot3^3), and relations having both properties are (2^3). Step 3: By inclusion-exclusion, the answer is (2^9-2^6-2^3\cdot3^3+2^3).

Step 1: To split (5) elements into (4) classes, one class must have (2) elements and the others must be singletons. Step 2: The two-element class can be chosen in (\binom{5}{2}=10) ways. Step 3: The remaining elements form singleton classes.

Step 1: Splitting five elements into three classes can have sizes (3,1,1) or (2,2,1). Step 2: For (3,1,1), there are (\binom{5}{3}=10) ways; for (2,2,1), there are (\frac{\binom{5}{2}\binom{3}{2}}{2}=15) ways. Step 3: Total equivalence relations are (10+15=25).

Step 1: Treat (1) and (2) together as one combined block. Step 2: Now there are three units: ({1,2},3,4). Step 3: The number of partitions of three units is (5), so there are (5) relations.

Step 1: Total equivalence relations on (4) elements are (15). Step 2: Those in which (1) and (2) are together are (5). Step 3: Hence those in which they are in different classes are (15-5=10).

Step 1: (2a+3a=5a) is always divisible by (5), so the relation is reflexive. Step 2: (2a+3b\equiv0\pmod5) also gives (2b+3a\equiv0\pmod5). Step 3: The condition is equivalent to (a\equiv b\pmod5), so it is also transitive.

Step 1: For reflexivity, (a+2a=3a) must be divisible by (4) for every integer (a). Step 2: For (a=1), this gives (3), which is not divisible by (4). Step 3: One counterexample is enough to show that reflexivity fails.

Step 1: If (aRb), then (b-a) and (b+a) are also integers, so it is symmetric. Step 2: Reflexivity would require (2a\in\mathbb{Z}) for every real (a), which is false. Step 3: The condition is not reflexive on all real numbers, so it cannot be an equivalence relation.

Step 1: A number (x) related to (\sqrt{2}) must satisfy (x-\sqrt{2}\in\mathbb{Q}). Step 2: Hence (x=\sqrt{2}+q), where (q\in\mathbb{Q}). Step 3: Convert the defining condition directly into set-builder form.

Step 1: ([0]) contains all rational numbers. Step 2: (\frac{3}{5}) is rational, so (\frac{3}{5}-0) is rational and both are in the same class. Step 3: An irrational representative usually gives a different class unless the difference is rational.

Step 1: For (xR3), we need (x^2-3^2=0). Step 2: This gives (x^2=9), so (x=3) or (x=-3). Step 3: When solving square-based class questions, remember both signs.

Step 1: For (xR2), we need (x^3=2^3=8). Step 2: Over real numbers, (x^3=8) has the single solution (x=2). Step 3: The cube function is one-one on real numbers, so the class is a singleton.

Step 1: (|a-a|=0\le1), so it is reflexive. Step 2: (|a-b|=|b-a|), so it is symmetric. Step 3: (0R1) and (1R2) hold, but (0R2) fails, so it is not transitive.

Step 1: (|a-a|=0<1), so it is reflexive. Step 2: Distance is the same in both directions, so it is symmetric. Step 3: (0R0.6) and (0.6R1.2) hold, but (0R1.2) fails; hence it is not transitive.

Step 1: (a-a=0), which is not greater than (1), so it is not reflexive. Step 2: If (3R1) holds, (1R3) does not, so it is not symmetric. Step 3: If (a-b>1) and (b-c>1), then (a-c>2), so it is transitive.

Step 1: (ab>0) means both numbers have the same non-zero sign. Step 2: The condition is unchanged when order is reversed, so it is symmetric. Step 3: Same sign passes through a middle element, but (0R0) is false; hence it is not reflexive.

Step 1: (a^2\ge0), so it is reflexive. Step 2: (ab=ba), so it is symmetric. Step 3: (-1R0) and (0R1) hold, but (-1R1) fails, so transitivity fails.