3

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?

Olba12
  • 2,579
  • Yes, I made a typo. – Olba12 Feb 07 '16 at 22:50
  • 2
    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. – hmakholm left over Monica Feb 07 '16 at 22:52
  • 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. – Andrew D. Hwang Feb 07 '16 at 22:55
  • 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? – Olba12 Feb 07 '16 at 23:02
  • By "well-defined", do you mean it satisfies certain axioms, such as associativity and commutativity, and distributivity over addition? By the way, you're working with split-complex numbers. – J.G. Feb 05 '21 at 21:17

1 Answers1

1

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 $(a, b)\cdot(c, d) \sim (a, b)\cdot(c', d') = (c', d')\cdot(a, b) \sim (c', d')\cdot(a', b') = (a', b')(c', d')$.

gioretikto
  • 306
  • You write this as though the product were defined two pairs of naturel numbers; it's not, it's defined on two equivalence classes of such pairs. And you seem to have lost either equal or equivalence signs at various stages. – ancient mathematician Dec 23 '20 at 12:20
  • there were two missing relations, I hope it is ok now – gioretikto Dec 23 '20 at 13:08