# Defining division by zero

I have looked through some of the previous questions posted on this topic, and I think mine is different.

Is there a flaw in defining division by zero? For example, define

$\frac{a}{0} = \infty_a$

it would seem like things work now, for example,

$\frac{a/0}{b/0}=\frac{\infty_a}{\infty_b}=a/b$.

What could go wrong with this idea, or more specifically, is it defined in some branch of mathematics?

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What is $\infty_a/\infty_0$ when $a\neq 0$? (I assume that $\infty_0/\infty_0$ will be $1$?) – Arturo Magidin Dec 2 '11 at 17:01
I think it is $a/0=\infty_a$, yes $0/0=1$ – picakhu Dec 2 '11 at 17:02
Mind you: you can extend the number system; but you are going to end up "losing" some of the properties you had before. The question is whether what you gain makes up for what you lose. In the case of going from $\mathbb{R}$ to $\mathbb{C}$, for example, you lose the ordering, but you gain the Fundamental Theorem of Algebra (among many other things). Is what you gain from your extension enough to make up for what we're going to lose by way of traditional algebraic rules? – Arturo Magidin Dec 2 '11 at 17:08
Since $0+0=0$, we have $a=0\cdot \infty_a=(0+0)\cdot \infty_a=0\cdot \infty_a+0\cdot\infty_a=a+a$. So $a=a+a$, and therefore $a=0$. Makes life simpler, to have everything equal to $0$. – André Nicolas Dec 2 '11 at 17:10
@picakhu: If $\infty_0=1$, then $a\times\infty_0 = a$. But $a\times \infty_0 = a\times(0/0) = (a\times 0)/0 = 0/0 = \infty_0$. So $a = a\times\infty_0 = \infty_0 = 1$ and everything is equal to $1$. – Arturo Magidin Dec 2 '11 at 17:13

Since $0+0=0$, we have $a=0\cdot \infty_a=(0+0)\cdot\infty_a=0\cdot\infty_a+0\cdot\infty_a=a+a$. So $a=a+a$, and therefore $a=0$.

It certainly makes life simpler to have everything equal to $0$.

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This is soon going to happen with all the bank accounts anyway... – Lohoris May 13 '12 at 15:10

I'm not going to consider $a=0$ (i.e. the quotient $\frac{0}{0}$) as we can construct functions in calculus limit initially tends to $\frac{0}{0}$ after some work turns out to be any real number we'd like. It may be the case that allowing $\infty_0$ could be useful, for example we have $0 = 0 \cdot \infty_0$, but I don't consider it here.

I noticed in André Nicolas' post, he showed that allowing division by zero ends up being trivial when you assume that the new elements you add obey the distributive law. So I conclude that if these new numbers are well-defined, then we can't use the distributive property with them.

What follows are a few thoughts in that regard.

Assume that $a \neq 0$ and define $\frac{a}{0} = \infty_a$ for some element $\infty_a$. Certainly $\infty_a$ is not a real number, so let us extend $\mathbb{R}$ to a new set $\mathbb{R}^{\dagger}$ which includes every $\infty_x$ for all $x \in \mathbb{R} \setminus \{0\}$.

What properties will this expanded set $\mathbb{R}^{\dagger}$ have? We decided above that it should not have the distributive property.

Well, we know that $\frac{a}{0} = \infty_a$ so perhaps $a = 0 \cdot \infty_a$. This strikes us as odd, because we know for any real number, multiplication by zero always yields zero. So we are at a crossroads. We can do one of two things:

(1): We can say "the new set $\mathbb{R}^{\dagger}$ must follows the rules of multiplying by zero" in which case we would derive $a=0$, which would be a contradiction (remember, we assumed $a \neq 0$ in the beginning). If we enforced this restriction, we would find our new set of numbers paradoxical and then throw them out.

(2): Allow this strange property of zero in this new set and accept all the consequences for its use.

Here is one consequence of (2):

Proposition: If $\mathbb{R}^{\dagger}$ is associative and commutative, then it contains only three elements.

Proof: Let $a, b \in \mathbb{R} \setminus \{0\}$. Now since we assumed (2), we know $a=0 \cdot \infty_a$ and $b = 0 \cdot \infty_b$, so we consider the product $ab = (0 \cdot \infty_a) (0 \cdot \infty_b)$.

We can write this product in two ways:

$$(i): (0 \cdot \infty_a) (0 \cdot \infty_b) = (a \cdot 0) \cdot \infty_b = 0 \cdot \infty_b = b,$$

but on the other hand

$$(ii): (0 \cdot \infty_a) (0 \cdot \infty_b) = \infty_a (b \cdot 0) = \infty_a \cdot 0 = a.$$

We conclude that $a=b$. So we have $\mathbb{R}^{\dagger} = \{0, a, \infty_a \}$.

Perhaps we should not assume $\mathbb{R}^{\dagger}$ is not commutative or not associative or not both, then...

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Presumably for non-zero $a$ you have $\frac{1}{\infty_a} = \frac{0}{a} = 0$, so this suggests for non-zero $a$ and $b$ that $\frac{1}{\infty_a} = \frac{1}{\infty_b}$ which in turn suggests ${\infty_a} = {\infty_b}$ and so perhaps $a=b$. More briefly $\frac{1}{\infty_a} = 0$ suggests $\infty_a = \frac{1}{0} = \infty_1$.

I doubt you intend this, so at some stage the distinctions you are trying to create or the manipulations you hope to preserve are lost.

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What you want to look at is called nonstandard analysis. You don't divide by zero but by infintesimals. See http://en.wikipedia.org/wiki/Non-standard_analysis

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could you give an example of how it would work, and perhaps how it would handle Andre's counter example (if that is applicable here). – picakhu Dec 2 '11 at 21:33
Infinitesimals do not define divison by zero. They are an invention to get around the problem of not being about to define division by zero for the use of calculus, so they don't really answer the question in general. Even in nonstandard analysis, you can't divide by zero. – tomcuchta Dec 3 '11 at 0:18
Nonstandandard analysis has little to do with the issues discussed here. Moreover, nonstandard analysis requires much more than infinitesimals in order to yield a sufficiently rich theory. See my answer here for more on this point. – Gone Dec 27 '11 at 18:52

is it defined in some branch of mathematics?

http://en.wikipedia.org/wiki/Surreal_number

http://en.wikipedia.org/wiki/Hyperreal_number

and maybe Non-Standard Analysis.

The point is how to make a system that can have division by 0 defined without encountering any inconsistencies. But another question to ask is are the all the 0's the same? for example is 0 bananas same as 0 apples? what about limit $x$ going to 0 and $x^2$ going to zero ? are both the limits the same thing in every context?

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