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Consider the following two relations on Z8 = {0, 1, 2, 3, 4, 5, 6, 7): (i) aRba- bezt (ii) aSbab € 2Z For each relation, determine whether it is an equivalence relation, or a poset, or neither

User Corydoras
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Answer:

Let's first define what it means for a relation to be an equivalence relation or a partial order (poset):

Equivalence relation: A relation on a set is an equivalence relation if it is reflexive, symmetric, and transitive. That is, for all a, b, and c in the set:

Reflexivity: aRa (a is related to itself)

Symmetry: If aRb then bRa (if a is related to b, then b is related to a)

Transitivity: If aRb and bRc, then aRc (if a is related to b and b is related to c, then a is related to c)

Partial order (poset): A relation on a set is a partial order if it is reflexive, antisymmetric, and transitive. That is, for all a, b, and c in the set:

Reflexivity: aRa

Antisymmetry: If aRb and bRa, then a = b (if a is related to b and b is related to a, then a and b are equal)

Transitivity: If aRb and bRc, then aRc

Now let's apply these definitions to the two relations given:

(i) aRb if and only if a = b or a - b is even

Reflexivity: aRa since a = a or a - a = 0 (which is even)

Symmetry: If aRb, then either a = b or a - b is even. If a = b, then bRa since b = a or b - a = 0 (which is even). If a - b is even, then b - a is also even, so bRa. Therefore, the relation is symmetric.

Transitivity: If aRb and bRc, we have two cases to consider:

If a = b and b = c, then a = c and aRc.

If a - b and b - c are both even, then a - c is even (the sum of two even numbers is even), so aRc.

If a - b and b - c are both odd, then a - c is even (the sum of two odd numbers is even), so aRc. Therefore, the relation is transitive.

Thus, we can conclude that relation (i) is an equivalence relation

Explanation:

User Jrushing
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