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I have this equivalence relation (this one is proven already):
$(a,b) R (c,d) ⇔ a + d = b + c$
and now I need to show that for $(a,b),(c,d),(e,f) ∈ ℕ x ℕ$:
$(a,b)R(c,d) ⇒ (a+e,b+f)R(c+e,d+f)$

What I did:

$a+d = b+c$
add f:
$⇒ a+d+f = b+c+f$
add e:
$⇒ a+d+f+e = b+c+f+e$
⇒$a+e+d+f = b+f+c+e$
⇒$(a+e)+(d+f) = (b+f)+(c+e)$

Am I allowed to just add e and f on both sides? If not, how should I get started instead?

Hanul Jeon
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  • Yes you are allowed to do that. – MichalisN May 04 '13 at 13:49
  • That means by doing what I did I have shown that this is true? There's nothing more? Anyway, thank you for your reply. – questarix May 04 '13 at 13:51
  • Yes, there's nothing more to it. $a+d=b+c$ implies $a+d+f=b+c+f$. I don't know how detailed you want the solution but you can add that when you reorder and put brackets around terms you use commutativity and associativity of addition on the natural numbers. – MichalisN May 04 '13 at 13:56

1 Answers1

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Re-write the original equation this way: $$(a,b)R(c,d)\textrm{ iff } a-b=c-d$$

Now you must verify that

$$a+e-b-f=c+e-d-f$$

After doing the obvious simplification, you get $$a-b=c-d$$ which is your assumption $(a,b)R(c,d)$

  • How is that any easier? – MichalisN May 04 '13 at 13:58
  • @Michalis: Use the definition of the integers in equivalence relation. Remember that $(a,b) = a - b$. As what Federica said, you can rewrite the equation and simplify it by simple algebra (as in elementary school math!). Finally, convert back to equivalence relations. This proof is easy because of rewriting the equation and using algebra to work out the proof. – NasuSama May 04 '13 at 15:10
  • @NasuSama: I still don't see how this is easier than what the OP suggests. Particularly if you want to define integers as pairs of natural numbers under the relation $R$ you can't use this approach to prove it is an equivalence relation.

    However it does provide a different perspective on the problem so I don't want to criticise the answer!

    – MichalisN May 04 '13 at 18:16