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Let R be the relation on Z≥ (the set of integers) defined by (x, y) ∈ R iff x2 + y2 = 2k for some integers k ≥0.R is symmetric. Which one of the following is a valid proof showing that R is symmetric?a.Let x, y ∈ Z≥ be given.Suppose (x, y) ∈ R then x2 + y2 = 2k for some k ≥0.ie y2 + x2 = 2k for some k ≥0.thus (x, y) ∈ R.b.Let x, y ∈ Z≥ be given.Suppose (x, y) ∈ R then x2 + y2 = 2k for some k ≥0.ie y2 + x2  = 2k for some k ≥0.thus (y, x) ∈ R.c.Let x, y ∈ Z≥ be given.Suppose (x, y) ∈ R then x2 + y2 = 2k for some k ≥0.thus (y, x) ∈ R.d.Let x, y ∈ Z≥ be given.Suppose (x, x) ∈ R then x2 + y2 = 2k for some k ≥0.ie y2 + x2 = 2k for some k ≥0.thus (y, y) ∈ R.

Question

Let R be the relation on Z≥ (the set of integers) defined by (x, y) ∈ R iff x2 + y2 = 2k for some integers k ≥0.R is symmetric. Which one of the following is a valid proof showing that R is symmetric?a.Let x, y ∈ Z≥ be given.Suppose (x, y) ∈ R then x2 + y2 = 2k for some k ≥0.ie y2 + x2 = 2k for some k ≥0.thus (x, y) ∈ R.b.Let x, y ∈ Z≥ be given.Suppose (x, y) ∈ R then x2 + y2 = 2k for some k ≥0.ie y2 + x2  = 2k for some k ≥0.thus (y, x) ∈ R.c.Let x, y ∈ Z≥ be given.Suppose (x, y) ∈ R then x2 + y2 = 2k for some k ≥0.thus (y, x) ∈ R.d.Let x, y ∈ Z≥ be given.Suppose (x, x) ∈ R then x2 + y2 = 2k for some k ≥0.ie y2 + x2 = 2k for some k ≥0.thus (y, y) ∈ R.

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Solution

The correct proof showing that R is symmetric is option b.

b. Let x, y ∈ Z≥ be given. Suppose (x, y) ∈ R then x² + y² = 2k for some k ≥0. This implies y² + x² = 2k for some k ≥0. Thus, (y, x) ∈ R.

This is because in a symmetric relation, if (x, y) is in R, then (y, x) must also be in R. In this case, if x² + y² = 2k (meaning (x, y) is in R), then it must also be true that y² + x² = 2k (meaning (y, x) is in

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