I have a good understanding of why we can’t divide by zero given our understanding of the real numbers. I’m not looking for any explanation tide to the real numbers. Rather what I’m trying to understand is why it’s not possible to construct a set (or is it?) that satisfies all the field axioms but without the exception to the rule that all elements have a multiplicative inverse excluding the additive identity.
Also, of all the potential pairs of identity and inverse elements is this the bad one? Presumably it has something to do with the directionality of the distributive axiom, but I can’t piece it together.
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Well you might have seen the following proof of why 0 has no multiplicative inverse in the reals:
Suppose 0x = 1
Then (0 + 0)x = 1 (additive identity)
But (0 + 0)x = 0x + 0x = 1 + 1 (distributivity)
But 1 is not the additive identity, so 1 + 1 =/= 1, contradiction
You can substitute whatever the identity elements are for your field for 0 and 1 and get the same result. As another commenter said, there are lots of ways to define division by zero (e.g. the projective reals), but (edit) none that form a field.
There exist algebraic structures that support zero division. They are called wheels. They are not fields by definition, but they can be consistently defined. The problem is, that as u/stumblewiggins pointed out, their usefulness is limited.
As long as you require an additive identity and distributivity of multiplication, you must have this exception for 0. Notice that in any field F, for all x in F we have 0x=(0+0)x = 0x+0x implying 0x=0. Thus if F has more than 1 element, multiplication by 0 on the left is not injective, i.e. 0x=0y does not imply x=y. This implies that 0 has no inverse in F.
>why it’s not possible to construct a set (or is it?) that satisfies all the field axioms but without the exception to the rule that all elements have a multiplicative inverse excluding the additive identity.
Okay so we want a commutative ring with identity (R, +, *, 0, 1) such that every element of R has a multiplicative inverse?
Since every element of R has a multiplicative inverse, write z for the multiplicative inverse of zero. Then for all r in R we have
r = r * 1 = r * (0 * z) = (r * 0) * z = 0 * z = 1
So therefore every element of R is equal to 1, which is in turn equal to every other element of R. In other words, R consists of a single element, specifically 0. (And also 0 = 1.)
So you can have this thing you describe, but there’s only one of them and it’s not very interesting.
(Technically the field axioms also include that 0 is NOT equal to 1, so I’m assuming we’re discarding that as well here.)
(People bring up other things in the comments that have some sort of “division by zero” that are slightly better-behaved, but this is answering your actual question, about whether we can simply remove the exception from the field axioms that 0 doesn’t need an inverse.)
For a field we want that multiplicative inverses are unique. Suppose we define division by 0 such that 1/0 = w which we can informally think of as “infinity”; but then we also have that 2/0 = w, now we no longer have uniqueness of inverses since then w*0 = 1 and w*0 = 2 which implies that (w/2)*0 = 1. That is, the multiplicative inverse of 0 is both w and w/2 (and also w/3 and so on). Worse, now what is 0/0? Is it 0 or w? Either way we lose some other useful fact. What is w+w? If w+w = w then the additive identity is no longer unique and 2w = w is a similarly strange result in multiplication.
We might say “okay, multiplicative inverses are unique except for 0” but that leads us down the same path of having an exception somewhere in our axiom. Since in general division by 0 causes problems it’s easiest to makes the exception that division by 0 is not defined.
The uniqueness of multiplicative inverses makes division well defined: that is a/b = a*b^-1 for b=/= 0 and we know this works because in a field inverses are uniquely defined. From there it becomes easy to prove additional algebraic laws that we expect to hold; for example the cancellation law.
As to why this is important; the field structure (or any other algebraic structure) exists to make it easy to prove facts about the structure which allows us to solve problems. If we make the structure too complicated or over-generalized it becomes more difficult to prove things or impossible to prove things at all.
I don’t know if you can construct a structure where all the axioms hold and division by zero is allowed (ie, is that set of axioms still consistent), but I do know that if you constructed any such structure that you’d lose a lot of theorems that are useful and the structure no longer resembles the usual arithmetic we expect on real numbers, which is the entire point of a field (that is, a field is a generalization of the rational and real numbers with addition and multiplication).
Edit: I believe it creates a contradiction in that the additive and multiplicative identities must be distinct and there would be elements for which this is no longer true.
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In a field with a set and two operations, analogous to addition and multiplication, with the latter distributive over the former, we define a zero element to be the identity element for the addition operation and a unity element to be the identity element of the multiplication operation.
We define an additive inverse or negative of an element to be an element which when added to it produces the zero element.
We define a multiplicative inverse or reciprocal of an element to be an element which when multiplied by it produces the unity element.
We assume the axioms of a field apply to it.
Let 1 be the unity element and 0 be the zero element.
1 + (-1) = 0 by definition of negative.
For any element N in the field, let’s see what happens when multiplied by the zero element.
N 0 = N (1 + -1) by substitution
= (N 1) + (N -1) by distribution
= N + -N = 0
Therefore any element multiplied by the zero element is the zero element.
This means that the zero element has no reciprocal, since nothing to multiply it produce the unity element.
In a field, we define division to be multiplying by the reciprocal of the other.
But since the zero element has no reciprocal, it is not possible to divide by the zero element.
QED
Think of division as separating into piles. If you have 10 apples and you put them in 5 equal piles, there will be 2 apples in each pile. Right?
Dividing by zero is putting those 10 apples in zero piles. It can’t be done.
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By definition, when a / b = c,
it must also be true that b * c = a
So if b = 0 we end up having to solve the equation 0 * c = a, which is unsolvable for any a other than zero, because whatever you multiply by 0 still is 0.
For a = 0 on the other hand, any number for c will satisfy the equation, so we must conclude that division by 0 is just not defined in any case.
There is a reason some calculators output +/- infinity when dividing by zero, and a reason the IEEE floating point standards have positive and negative infinity bit patterns. I don’t know the reasons, but figured I’d mention.