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The typical way of constructing $\mathbb{Z}$ from $\mathbb{N}$ is by defining an equivalence relation $\sim$ on $\mathbb{N}\times\mathbb{N}$ by $$ (a,b)\sim(c,d)\iff a+d=c+b. $$ Then, we define $\mathbb{Z}:=\mathbb{N}\times\mathbb{N}/\sim$.

Question: Why not just take the canonical representative of each equivalence class and define $$ \mathbb{Z}:=\{(m,0):m\in\mathbb{N}\}\cup\{(0,m):m\in\mathbb{N}\} ? $$ For rational numbers I would agree that every representative is important and that equivalence classes is probably the neatest way to treat fractions like $\frac{1}{2}$ and $\frac{2}{4}$ as if they were equal, but here I don't quite see what we gain by considering all representatives since once addition is defined in $\mathbb{Z}$, we can talk about subtracting two integers quite easily.

Guest
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2 Answers2

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The reason this isn't done is that then the definitions of addition and subtraction are complicated. With the first definition, what is the value of $$(a,b) + (c,d)?$$

Why, it's simply

$$(a+c, b+d).$$

But with your proposed definition, you need six cases:

$$\begin{align} (a, 0) + (c, 0) & = (a+c, 0) \\ (a, 0) + (0, d) & = \begin{cases}(a-d, 0)&\text { if $a\ge d$,}\\ (0, d-a)&\text{ otherwise}\end{cases} \\ (0, b) + (c, 0) & = \begin{cases}(c-b, 0)&\text { if $c\ge b$,}\\ (0, b-c)&\text{ otherwise}\end{cases} \\ (0, b) + (0, d) & = (0, b+d) \end{align} $$

If you use the right definition, it is easy to prove things about the properties of addition, which is the whole point of defining addition in the first place. For example, try proving that $m-n+n = m$, first with one definition, then the other, and see how it goes.

Subtraction, multiplication, and $≤$ are all similarly complex in your system, but easy with the standard definition:

$$\begin{eqnarray} (a,b) - (c,d) & = & (a+d, b+c) \\ (a,b)\cdot(c,d) &=& (ac+bd, ad+bc)\\ (a,b)\le(c,d) & \iff& a+d\le b+c \end{eqnarray} $$

The only drawback of the usual definition compared with your proposal is that equality is just a tiny bit more complicated: we have $(a,b) = (c,d)$ whenever $a+d=c+b$, whereas with your proposal we have $(a,b) = (c,d)$ whenever $a=c$ and $b=d$.

MJD
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    Oh right and one would then have to define $n-m$ for $n,m\in\mathbb{N}$ such that $n\geq m$ which is not necessary with the usual definition. – Guest Oct 21 '15 at 15:14
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Actually multiplication and $\le$ do not require several cases. Standard definitions work just fine when restricted to $$(a,b), (c,d)\in\mathbb{Z} = \{0\}\times \mathbb{N}\cup\mathbb{N}\times\{0\}.$$ For multipliction one only needs to observe that $ac+bd$ or $ad+bc$ is equal to zero. One can define addition with one formula as well: $$ (a, b)+ (c,d)= ( a+c - \min\{a+c,b+d\},b+d - \min\{a+c,b+d\}),$$ where $-$ denotes substraction of natural numbers which is well defined as $a+c \ge \min\{a+c,b+d\},b+d \ge \min\{a+c,b+d\}$.

A Dz
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