# Is this structure a field?

I'm wondering whether the following addition and multiplication over the set $$(\mathbb{R}\setminus\{0\}\times \mathbb{Z}) \cup \{0\}$$ define a field:

$$(a,a')+(b,b')= \begin{cases} (a,a') \text{ if } a'>b'\\ (b,b') \text{ if } b'>a'\\ (a+b,a') \text{ if } b'=a' \text{ and } a\neq -b\\ 0 \text{ if } b'=a' \text{ and } a= -b\\ \end{cases}$$ $$(a,a')(b,b')=(ab,a'+b')$$ $$-(a,a')=(-a,a')$$ $$(a,a')^{-1}=(a^{-1},-a')$$

[$$0$$ is the additive unit, which fixes addition and multiplication with $$0$$. $$(1,0)$$ the multiplicative unit.]

If yes, does this field have a name? If no, which of the axioms fail?

I'm a bit confused because I thought there are only "relatively few" different fields, such as the rational, real, complex numbers, or finite fields.

• another field: rational expressions (fractions of polynomials in $X$) – J. W. Tanner May 15 at 16:52

No, addition is not associative. For instance, $$((1,0)+(-1,0))+(1,-1)=0+(1,-1)=(1,-1)$$ but $$(1,0)+((-1,0)+(1,-1))=(1,0)+(-1,0)=0.$$
Note that you can tell something must be wrong with just the additive axioms, since your operation $$+$$ does not allow cancellation and so cannot be a group operation. Since there clearly is an identity and inverses, associativity must fail.
By the way, there are lots and lots of different fields; there are just a few that are familiar in elementary mathematics. For instance, the following is a field: the underlying set is $$\mathbb{Q}\times\mathbb{Q}$$, addition is $$(a,b)+(c,d)=(a+c,b+d),$$ and multiplication is $$(a,b)\cdot(c,d)=(ac+2bd,ad+bc).$$ (To make this more familiar, this field is isomorphic to the subfield of real numbers consisting of numbers of the form $$a+b\sqrt{2}$$ with $$a,b\in\mathbb{Q}$$, sending $$(a,b)$$ to $$a+b\sqrt{2}$$.)
The most glaring issue I see is that your addition is not one-to-one. For instance, $$(10,10)+(b,b')$$ is $$(10,10)$$ for any $$b'<10$$. But addition has to be one-to-one: if we have $$x+y=x+z$$, then $$x+y+(-x)=x+(-x)+z \rightarrow y=z$$.
But then $$0=0*(b,b')$$. If we take $$(a,a')$$ with $$a', by the distributive property we we have $$0=0*(b,b')=((a,a')+(-a,a'))*(b,b')=(b,b')+(b,b')=(2b,b')$$.