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To prove R is reflexive, we need to show that $(a, b) ~R~ (a, b)$ for all $(a, b) \in N \times N$. Since $a - a = 0$ and $b - b = 0$, we have $a - a = b - b$, which means $(a, b) ~R~ (a, b)$. Therefore, R is reflexive.
To prove R is symmetric, we need to show that if $(a, b) ~R~ (c, d)$, then $(c, d) ~R~ (a, b)$. Given $(a, b) ~R~ (c, d)$, we have $a - c = b - d$. Multiplying both sides by -1, we get $c - a = d - b$. This implies $(c, d) ~R~ (a, b)$. Therefore, R is symmetric.
To prove R is transitive, we need to show that if $(a, b) ~R~ (c, d)$ and $(c, d) ~R~ (e, f)$, then $(a, b) ~R~ (e, f)$. Given $(a, b) ~R~ (c, d)$, we have $a - c = b - d$. Given $(c, d) ~R~ (e, f)$, we have $c - e = d - f$. Adding these two equations, we get $(a - c) + (c - e) = (b - d) + (d - f)$, which simplifies to $a - e = b - f$. This implies $(a, b) ~R~ (e, f)$. Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
Final Answer: R is an equivalence relation.
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