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?
I'll prove that multiplication is well-defined which is more difficult and leave addition as exercise.
You want to show that
$(a, b)\cdot (c, d) \sim (a', b')\cdot(c', d')$
supposing that
$(a,b) \sim (a', b') \Longleftrightarrow a + b'= b + a', \qquad (c,d) \sim (c', d') \Longleftrightarrow c+d' = d+c' \quad (\mathbf{1.0})$
i.e. multiplication does not depend on the particular choice of representatives.
We start by showing that $(a,b)\cdot(c,d) \sim (a,b)\cdot(c',d')$ and then use commutativity of the product to show the previous follows.
By expanding the multiplication, it is equivalent to show that $(ac+bd,ad+bc) \sim (ac+bd, ad+bc)$, which by the definition of ~ is the same as showing that
$ac + bd + ad' + bc = ad + bc + ac' + bd' \quad (\mathbf{2.0})$
Multiplying $c+d' = d+c'$ by $a$ in 1.0 yields
$ac + ad' = ad + ac'$
Multiplying $d + c' = c + d'$ by $b$ in 1.0 yields
$bd + bc' = bc + bd'$
Summing up the two equations, you obtain 2.0.
By commutativity of the product (left as exercise to verify), we have also $(a, b)\cdot(c, d) \sim (a, b)\cdot(c', d') = (c', d')\cdot(a, b) \sim (c', d')\cdot(a', b') = (a', b')(c', d')$.