# Is there an algebraic formula that gives this weird multiplication?: $(-x)\circ(-x)=(-x)\circ(x)$

I'd like to know whether it's possible to give an equivalent algebraic formula, in terms of normal algebraic operations (i.e. $$+, -, ×, ÷, x^y$$), if possible avoiding $$|x|$$, for an operator $$\circ$$, in the domain ℤ such that:

$$\begin{array}{|r | r r r r | r r r | r r r r} \hline \circ & ... & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & ... \\ \hline \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\ -4 & ... & -16 & -12 & -8 & -1 & 0 & -4 & -8 & -12 & -16 & ... \\ -3 & ... & -12 & -9 & -6 & -1 & 0 & -3 & -6 & -9 & -12 & ... \\ -2 & ... & -8 & -6 & -4 & -1 & 0 & -2 & -4 & -6 & -8 & ... \\ \hline -1 & ... & -4 & -3 & -2 & -1 & 0 & -1 & -2 & -3 & -4 & ... \\ 0 & ... & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & ... \\ 1 & ... & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & ... \\ \hline 2 & ... & -8 & -6 & -4 & -2 & 0 & 2 & 4 & 6 & 8 & ... \\ 3 & ... & -12 & -9 & -6 & -3 & 0 & 3 & 6 & 9 & 12 & ... \\ 4 & ... & -16 & -12 & -8 & -4 & 0 & 4 & 8 & 12 & 16 & ... \\ \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{array}$$

As you see, this is some sort of weird multiplication where $$(-)\cdot(-)=(-)\cdot(+)=(+)\cdot(-)=(-)$$ but $$(+)\cdot(+)=(+)$$. I am mostly interested in the subcase of the square in the middle and, if possible, all the $$\circ$$ operations where at least one of $$-1$$, $$0$$ or $$1$$ is an argument. If that operation can satisfy at least the square at the middle, that will be enough for me. As a last resort I'm disposed to accept division by zero defined in such a way that for all $$x$$, $$x/0=0$$.

But it is important to note that, even if it doesn't matter too much what happens outside the domain $$\{-1, 0, 1\}$$, the operation must be defined for all integers: no modules, no restricted domains.

If that isn't possible, what strategy shall I use to prove it?

ps: Since I'm not a mathematician, I'd like to apologise for any formal or conceptual error I've made. Corrections, though, are more than encouraged.

• "If that operation can satisfy at least the square at the middle, that will be enough for me." $$x\circ y=xy\cdot\cfrac{1+x+y-xy}2$$
– user856
Commented Sep 18, 2017 at 22:39
• Just want to say that this is a really cool question, and well formatted as well. I hope to see it answered well! Commented Sep 18, 2017 at 22:48
• your multiplication can be written as $\text{sign}[\min(a,b)] |ab|$. Now note that $\min(a,b) = \frac{a+b-|a-b|}{2}$ and that $|a-b| = \sqrt{(a-b)^2}$. I just don't know how to generally represent $\text{sign}(a,b)$ in a form like you desire, and already I am pushing the line a bit by utilizing $|x| = \sqrt{x^2}$ Commented Sep 18, 2017 at 22:56
• @LuisBartolo Is there a particular reason you want to avoid divisions and absolute values? I would still argue that $\text{sign}[\min(a,b)] |ab|$ is a much simpler definition, and that in many cases a piecewise definition would be even better Commented Sep 18, 2017 at 23:17
• It is an important thing to note, though. 'How to define this operation?' - you just did. It's really, really important not to confuse formulas for definitions. It may be worth asking 'is there an algebraic formula that gives this operation?' Or even 'how close can this operation be to multiplication?' (For starters: it looks like it defines a monoid, but clearly it can't be invertible). But the way you've phrased the question is IMHO actually a bad habit that it's worth trying to shed. Commented Sep 18, 2017 at 23:34

You can avoid using the absolute value and sign functions if you change the representation of the domain! Let's use a sign-and-magnitude construction. An integer is a pair $(\sigma, n)$ where $\sigma$ is either the symbol $+$ or the symbol $-$, $n$ is an natural number, and we identify $(+,0)=(-,0)$. For a natural number $n$, we may use the shorthand $n=(+,n)$ and $-n=(-,n)$. But whenever we want to define any operation on integers, we may choose to define the operation on the sign and the magnitude independently; and the operation will be well-defined as long as we respect $(+,0)=(-,0)$.
With that setup, the definition is just: $$(\sigma,x)\circ(\tau,y)=(\min(\sigma,\tau),xy)$$ Then observe that if either $x=0$ or $y=0$, then $xy=0$, so the sign doesn't matter, and we're done.
If you're unhappy with the $\min$ function, we can do away with that too, by representing the sign symbol as a sign bit. The usual computer representation would be $0=+$ and $1=-$. But it's slightly better for us to choose the opposite, $0=-$ and $1=+$. Then we have: $$(\sigma,x)\circ(\tau,y)=(\sigma\tau,xy)$$ which is as simple as you could hope for!