Prove that the relation R in the set A=1,2,3,4,5A = {1, 2, 3, 4, 5} given by R=(a,b):∣a−b∣isevenR = {(a, b) : |a - b| is even} is an equivalence relation.

Answer: The relation R is an equivalence relation.

Step-by-step solution

Step 1: Understand Equivalence Relation

An equivalence relation partitions a set into disjoint subsets where all elements within a subset are related to each other. To prove that a given relation is an equivalence relation, we must demonstrate that it satisfies three fundamental properties: reflexivity, symmetry, and transitivity.

Step 2: Check for Reflexivity

For a relation to be reflexive, every element must be related to itself. In this case, we need to check if ∣a−a∣|a - a| is even for any element aa in the set AA. The absolute difference between an element and itself is always 0.

Step 3: Evaluate Reflexivity Condition

Since 0 is an even number (as it can be written as 2×02 \times 0), the condition ∣a−a∣|a - a| is even is satisfied for all a∈Aa \in A. Therefore, the relation R is reflexive.

Step 4: Check for Symmetry

For symmetry, if an element aa is related to an element bb, then bb must also be related to aa. We assume that (a,b)∈R(a, b) \in R, which means that the absolute difference between aa and bb is an even number.

Step 5: Evaluate Symmetry Condition

The absolute value of a number is the same as the absolute value of its negative. Therefore, ∣a−b∣|a - b| is equal to ∣b−a∣|b - a|. Since we assumed ∣a−b∣|a - b| is even, it follows that ∣b−a∣|b - a| must also be even. This implies that (b,a)∈R(b, a) \in R, so the relation R is symmetric.

Step 6: Check for Transitivity

For transitivity, if aa is related to bb, and bb is related to cc, then aa must be related to cc. We assume that (a,b)∈R(a, b) \in R and (b,c)∈R(b, c) \in R. This means that a−ba - b and b−cb - c are both even integers (positive or negative, as their absolute values are even).

Step 7: Express differences as even integers

If the absolute value of a difference is even, then the difference itself must be an even integer (it could be positive or negative). So, a−ba-b can be written as an even integer 2m2m and b−cb-c can be written as an even integer 2n2n, where mm and nn are integers.

Step 8: Find the difference between a and c

To check if (a,c)∈R(a, c) \in R, we need to evaluate the difference a−ca - c. We can express a−ca - c as a sum of the differences a−ba - b and b−cb - c, which we already know are even integers.

Step 9: Substitute and sum the even integers

Substituting the representations for a−ba - b and b−cb - c as 2m2m and 2n2n respectively, we find that their sum 2m+2n2m + 2n simplifies to 2(m+n)2(m + n). Since mm and nn are integers, their sum (m+n)(m + n) is also an integer. This means that a−ca - c is twice an integer.

Step 10: Evaluate Transitivity Condition

Since a−ca - c can be expressed as 2×an integer2 \times \text{an integer}, it is an even integer. Therefore, its absolute value ∣a−c∣|a - c| is also even. This proves that (a,c)∈R(a, c) \in R, and thus the relation R is transitive.

Step 11: Conclusion

Since the relation R satisfies all three properties required for an equivalence relation (reflexivity, symmetry, and transitivity), we can conclude that R is an equivalence relation on the set A.

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