Consider the relation ∼ on X = N×N such that (a, b) ∼ (c,d) ⇐⇒ a+d = b+c. Write down a bijection f : (X/ ∼) → Z such that f([a, b] ⊕ [c,d]) = f([a,b]) + f([c,d]).
Added by Marissa G.
Step 1
The relation states that two pairs (a, b) and (c, d) are equivalent if a + d = b + c. This means that the difference between the first and second elements of the pairs is the same. Show more…
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