Introduction to Relations

Step 1: The (5) diagonal pairs are independently optional. Step 2: For (\frac{5\cdot4}{2}=10) unordered pairs of distinct elements, each gives three choices. Step 3: Hence the total number is (2^5\cdot3^{10}).

Step 1: Irreflexivity forbids all diagonal pairs. Step 2: Symmetry leaves (6) independent off-diagonal reverse-pair groups. Step 3: Removing the empty relation from (2^6) gives (2^6-1).

Step 1: Reflexivity forces all diagonal pairs to be present. Step 2: Symmetry together with antisymmetry allows no pair between distinct elements. Step 3: Only the identity relation is possible, so the number is (1).

Step 1: An equivalence relation partitions a set into disjoint classes. Step 2: The number of partitions of a (5)-element set is (52). Step 3: Therefore the number of equivalence relations is (52).

Step 1: Exactly two classes means splitting (5) elements into two non-empty parts. Step 2: There are (2^5-2=30) non-empty proper subsets, but each partition is counted twice. Step 3: Hence the number is (\frac{30}{2}=15).

Step 1: A three-element class can be chosen in (\binom{6}{3}=20) ways. Step 2: The remaining three elements automatically form the other class. Step 3: Since interchanging the two classes does not create a new partition, the count is (\frac{20}{2}=10).

Step 1: We must split six elements into three two-element classes. Step 2: The count is (\frac{6!}{(2!)^3\cdot3!}). Step 3: This equals (15), so there are (15) such equivalence relations.

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

Step 1: First treat (1) and (2) as one block. Step 2: Element (3) must not be in that block, while (4) and (5) may be placed freely. Step 3: Partitions with (1,2) together are (15), and those with (1,2,3) together are (5), so the answer is (15-5=10).

Step 1: (-3) has the same remainder as (5) modulo (8). Step 2: Its class contains all integers of the form (8k+5). Step 3: Converting a negative representative to a positive remainder helps avoid mistakes.

Step 1: (3a+4a=7a), so the relation is reflexive. Step 2: (3a+4b\equiv0\pmod7) gives (a\equiv b\pmod7), so reversing the order also works. Step 3: Same-remainder relations are transitive, so this is an equivalence relation.

Step 1: (2a+2a=4a), so the relation is reflexive. Step 2: (2a+2b) treats (a) and (b) equally, so it is symmetric. Step 3: The condition means (a+b) is even, so (a) and (b) have the same parity; hence transitivity holds.

Step 1: (1+b) is even exactly when (b) is odd. Step 2: Therefore the class of (1) is the set of all odd integers. Step 3: In parity-based relations, identify even and odd groups first.

Step 1: A number (x) must satisfy (x-\frac{1}{2}\in\mathbb{Z}). Step 2: Hence (x=\frac{1}{2}+n), where (n\in\mathbb{Z}). Step 3: To write the class, add all allowed integer differences to the representative.

Step 1: Two classes are equal when the difference of their representatives is rational. Step 2: ((\sqrt{2}+3)-\sqrt{2}=3), which is rational. Step 3: Therefore ([\sqrt{2}+3]=[\sqrt{2}]).

Step 1: For (xR0), we need (x^2=0^2). Step 2: This gives (x^2=0), so (x=0) only. Step 3: Negative zero and positive zero are not distinct elements.

Step 1: For every (a), (\lfloor a\rfloor=\lfloor a\rfloor), so it is reflexive. Step 2: Equality remains true when reversed, so it is symmetric. Step 3: Having the same greatest-integer value passes through a middle element, so it is transitive.

Step 1: (\frac{5}{2}=2.5) has greatest-integer value (2). Step 2: Real numbers with greatest-integer value (2) lie in ([2,3)). Step 3: Do not include the right endpoint (3), because its greatest-integer value is (3).

Step 1: (|a-b|=0) holds exactly when (a=b). Step 2: Thus the relation contains only pairs of the form ((a,a)). Step 3: It is the identity relation and is reflexive, symmetric and transitive.

Step 1: (|a|<|a|) is never true, so it is not reflexive. Step 2: (|1|<|2|) is true but (|2|<|1|) is false, so it is not symmetric. Step 3: If (|a|<|b|) and (|b|<|c|), then (|a|<|c|), so it is transitive.

Step 1: The condition (a^2+b^2=1) is unchanged when (a) and (b) are interchanged, so it is symmetric. Step 2: (2a^2=1) is not true for every (a), so it is not reflexive. Step 3: ((1,0)) and ((0,1)) belong, but ((1,1)) does not, so it is not transitive.

Step 1: If (a+b=6), then (b+a=6), so the relation is symmetric. Step 2: Not all diagonal pairs are present, for example ((1,1)) is absent, so it is not reflexive. Step 3: From ((1,5)) and ((5,1)), transitivity would need ((1,1)), which is absent.

Step 1: If (a+b\le6), then (b+a\le6), so the relation is symmetric. Step 2: ((5,5)) is absent because (10\le6) is false, so it is not reflexive. Step 3: Since ((5,1)) and ((1,5)) are present but ((5,5)) is absent, it is not transitive.

Step 1: Modulo (4), the counts are remainder (0:1), remainder (1:2), remainder (2:1), and remainder (3:1). Step 2: Sum divisible by (4) needs remainder pairs ((0,0),(1,3),(3,1),(2,2)). Step 3: The count is (1\cdot1+2\cdot1+1\cdot2+1\cdot1=6).

Step 1: Each of the three remainder classes modulo (3) has two elements. Step 2: Only ordered pairs with the same remainder belong to the relation. Step 3: The total number is (2^2+2^2+2^2=12).