Let $M = \mathbb{N} \ \mathbb{x} \ \mathbb{N}$. We define the following relation on $M$. Let $(a,b)R(a',b')$ iff $a + b'=a'+b$ We define the set of intergers $\mathbb{Z}$, to be the set of equivalence classes of $M$ under the equivalence relation above. We try to define the arithmetic operations of addition and multiplication in the following way $$[(a,b)] + [(a',b')] := [(a+a',b+b')]$$ and $$[(a,b)] \cdot [(a',b')] := [(aa'+ bb',a'b+ab')] $$

I should prove that multiplication is well defined. If I understand this correctly i should given that $$(a,b) \sim (c,d) \ i.e \ a+d =c+b$$ and $$(a',b') \sim (c',d') \ i.e \ a'+d'=c'+b'$$ I should prove that $$(ac+bd,cb+ad) \sim (a'c' +b'd',c'b'+a'd')$$ which means that $$ ac+bd+c'b'+b'd' = a'c' +b'd'+cb+ad$$ I dont know how to do this?

  • $\begingroup$ Yes, I made a typo. $\endgroup$ – Olba12 Feb 7 '16 at 22:50
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    $\begingroup$ It is probably easiest to prove it in two steps: First, if $(a,b)\sim(a',b')$ then $(a,b)(c,d)\sim(a',b')(c,d)$. Then the same thing to the other side: if $(c,d)\sim(c',d')$ then $(a',b')(c,d)\sim(a',b')(c',d')$. Splitting it up like this will allow you to attack each part with the distributive law. $\endgroup$ – Henning Makholm Feb 7 '16 at 22:52
  • $\begingroup$ In this situation, it probably helps to "cheat": Multiply out $(a - b)(c - d)$ as if you know about integer arithmetic, then interpret the result as an ordered pair of natural numbers. Then use the fact that $(a - b) = (a' - b')$, etc. As long as the end result refers only to pairs of natural numbers, you haven't technically done anything illogical. $\endgroup$ – Andrew D. Hwang Feb 7 '16 at 22:55
  • $\begingroup$ Could you Henning please explain more careful, I am really confused. If $(a,b) \sim (a',b')$ then $(a,b)(a,d) \sim (a',b')(c,d)$ is this multiplicative cancellation law? $\endgroup$ – Olba12 Feb 7 '16 at 23:02

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