# Do 'symmetric integers' have some other name?

$-1 \cdot -1 = +1$, but there seems to me to be no reason we couldn't define a number system where negative number's and positive numbers were completely symmetric. Where:

$$1 \cdot 1 = 1$$

$$-1 \cdot -1 = -1$$

I understand that in order to do this, multiplication could no longer be commutative and we'd have to decide what the result of $1 \cdot -1$ should be. I think we could choose that resulting sign of a multiplication could be the sign of the second term, so:

$$1 \cdot -1 = -1$$

$$-1 \cdot 1 = 1$$

or more generally, the sign of any multiplication is determined by the sign of the second term.

But where they otherwise behave roughly as expected, i.e. $1 - 2 = -1$.

Some other consequences I'm aware of:

$$\sqrt{-1} = -1$$

$$\sqrt{1} = 1$$

$f(x) = x^2$ would behave in a way that can only be described piecewise in the normal reals as $x^2$ when $x \geq 0$, and $-(x^2)$ when $x < 0$.

Is there already research or another name for such a number system? Or perhaps is there a ring that matches this? After looking at the properties of a ring, on http://en.wikipedia.org/wiki/Ring_%28mathematics%29 what I've described cannot be a ring since it does not have a multiplicative identity. There is no element i_m such that a * $i_m = a$ and $i_m \cdot a = a$ since multiplying by $1$ in the system I've described may change the sign of $a$ to be positive.

• There's a book called Negative Math in which the author develops such a system and explores some of its properties, although it's not very in depth. – Jack M Jan 14 '15 at 21:43
• @JackM How do I accept this comment as an answer? This seems like the closest thing to a positive answer that's been put forth so far. – Shufflepants Jan 14 '15 at 22:45
• I'm not sure it is, the book isn't a textbook or anything, the system is just developed as an example, really. – Jack M Jan 15 '15 at 0:02
• More of an answer and example than these other answers that are describing properties of systems other than the one I've described here. – Shufflepants Jan 15 '15 at 1:10
• @BCLC Yes it did thx – emcor Jul 14 '15 at 7:57

The greatest problem with your system, I think, is that the distributive law fails to hold:

$$1 = 1\cdot 1 = (-1+2)\cdot 1 \ne (-1)\cdot1+2\cdot 1 = 1+2 = 3$$

This is much more important than failure of your multiplication to have nice properties in itself.

If we have a weird "multiplication" operation that is at least associative and distributes over addition like it should, that would be a (non-commutative) rng, and there is good theory for those available. And there are various well-studied structures that dispense with associativity of multiplication (non-associative algebras), but structures with two binary operations that don't distribute are not mainstream, to say the least.

What you do have seems to be a left near-ring, though.

• The distributive law would still hold if $2=0$, that is, the ring has characteristic two. But I agree with your larger point—there are non-commutative rings, and non-associative rings/algebras, but not much need to study non-distributive rings! – Matthew Leingang Jan 15 '15 at 9:46
• @MatthewLeingang: Similar examples would require $3=0$ and so forth, so everything would need to collapse. But that doesn't seem to be what the OP is doing anyway. He's not proposing an axiomatic theory and looking for models of it. On the contrary he has a concrete structure where the elements are what they are and the operations behave in such-and-such way, and then he's looking for theory about that structure. – hmakholm left over Monica Jan 15 '15 at 11:40

$(-1)*(-1)=1$ is always going to be true in any ring. If $(-1)*(-1)=-1$ as well, then $1=-1$.

Formally speaking, this means the ring will have characteristic $2$. In practice, it tells you that every element is its own additive inverse.

• If (-1)*(-1) = 1 is true for any ring, then what I have described is not a ring. – Shufflepants Jan 14 '15 at 20:39
• @Shufflepants, it is a ring, as long as 1=-1, which is what Matthew is suggesting. Rings where 1=-1 are said to have characteristic 2. – Dylan Yott Jan 14 '15 at 20:55
• Then I guess I need to give more examples. Describing something where 1 != -1. – Shufflepants Jan 14 '15 at 20:59
• I guess this answer tells me that what I described isn't a ring. But I'm really looking for what it IS, rather than what it isn't. – Shufflepants Jan 14 '15 at 22:40

The positive integers with product $x \star y = xy$ form a commutative monoid.

The negative integers with product $x \star y = -xy$ form a commutative monoid.

In fact, both of these monoids are isomorphic, with the isomorphism given by (in either direction!) $x \mapsto -x$.

Rather than thinking of all nonzero integers as a single entity, you might do better to instead think of having two copies of this monoid and ponder how to combine them into a larger monoid (or other algebraic structure).

Short answer: No, it doesn't have a name.

As Jack M points out, Alberto A. Martinez's book Negative Math: How Mathematical Rules Can Be Positively Bent spends some time exploring the consequences of your altered sign rule. (Actually, Martinez takes the sign of the product to be the sign of the first term, not the second, but this is only a cosmetic difference.) Briefly, some points of discussion are:

• The symmetry between positive and negative is attractive.
• Imaginary numbers are no longer introduced by square roots. This is sometimes convenient and sometimes not. If one wishes to express rotations in two dimensions, one can fall back on an explicit rotation operator, which is no longer tied to a particular number.
• Multiplication is still associative, but not commutative, similarly to the multiplication of quaternions.
• Multiplication is distributive over addition on one side, but not the other.
• Expressions such as $\sqrt{2+\sqrt{-121}}$ and $\sqrt{-1}^{\sqrt{-1}}$ have simpler values than before.
• One needs extra absolute-value signs to state the Pythagorean Theorem, which is arguably appropriate anyway.
• We must take care in expanding and interpreting expressions like $(a+b)^2$. The result has a different geometric meaning depending on the signs of $a$, $b$, and $a+b$.
• Extraction of roots now has a unique solution, $\sqrt[x]a$.
• Roots and powers are now exact inverses: $\sqrt[x]{a^x}={\sqrt[x]a}^x=a$.
• Powers and roots now commute with one another: $\sqrt[x]{a^y}={\sqrt[x]a}^y$.
• Powers and roots now commute with changes of sign: $-(a^x)=(-a)^x$ and $-\sqrt{a}=\sqrt{-a}$.
• Square roots now distribute over multiplication: $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$.
• Similar simplifications are obtained for logarithms.
• Division, fractional exponents, and trigonometry are not considered. Coordinate geometry is sketched out to illustrate how polynomials change.

Martinez briefly remarks that the new operation might be called by a different name and symbol, perhaps "replication" and $\otimes$, to distinguish it from the traditional multiplication operation… but he doesn't really recommend that.

• This should be accessed as a answer. – mikea Jan 15 '15 at 16:14

Note that elements of a ring which satisfy $x^2=x$ are called idempotents, and these become important in some contexts - for example a square matrix with $1$ as the top left entry and zero everywhere else is a non-trivial idempotent. Such things become significant, for example, in representation theory.

[note $0,1$ are trivial idempotents and in characteristic $2$ you have $-1=1$ being a trivial idempotent - I'm addressing other situations]

One issue with idempotents is that if they are invertible, then you trivially have $x=1$ simply by multiplying through by the inverse. Your $-1$ is ambiguous, because it is unclear whether you imagine it to be invertible.

Algebraic structures where $ij=-ji$ for significant elements are also important. The Quaternions are a significant example. Note that (provided your system is associative) you have $i^2j=ji^2$ so some products don't have the minus sign. You can have $i,j$ invertible in this kind of system, but then they can't be idempotent.

So there are mathematical structures of significance which capture the ideas in your post, but there are come challenges to having everything you want at once, and also a big question of what use it might be if you did.

• What do you mean by -1 being invertible? – Shufflepants Jan 14 '15 at 20:47
• @Shufflepants I mean there being an element $a$ with $a\cdot (-1)=1$ – Mark Bennet Jan 14 '15 at 20:49
• The system I described has no such element a such that a * (-1) = 1 (at least not without introducing a complex variant of this system). There is an a such that (-1) * a = 1, namely a = 1. – Shufflepants Jan 14 '15 at 20:57