# Defining division by zero?

HallsofIvy
Homework Helper
Since I've started learning division, it is learnt that division by zero is undifined. But in the book "Introduction to Analytic Number Theory" by Tom M. Apostol, I've found that "Zero divides only zero" [Theorem 1.1 (h)]. What does this statement mean?
I haven't read that book (or any book on "Analytic Number Theory") but this might be relevant.

$$\frac{a}{0}$$, for a non-zero is not defined, in the real number system, because if we set a/0= b we get a= b(0)= 0 which is not true for any b. But $$\frac{0}{0}$$ which is also not defined is a separate situation- if we set 0/0= b we get 0= b(0)= 0 which is true for all b. I rather suspect that this theorem is connected to that distinction.

Since I've started learning division, it is learnt that division by zero is undifined. But in the book "Introduction to Analytic Number Theory" by Tom M. Apostol, I've found that "Zero divides only zero" [Theorem 1.1 (h)]. What does this statement mean?
There is a small imprecision here: Apostol is not talking about division; he is talking about divisibility, which is a diffrent concept in number theory.

Divisibility is a binary relation, defined on pairs of integers (or elements of a more general ring), defined by:

a|b iff there is an integer k, such that b = ka

With this definition, it's easy to see that 0|0 and, if x|0, then x = 0. In particular, there is no mention of invertible elements.

Division, on the other hand, is (usually) a function defined on a ring by:

$$a/b \equiv a \times b^{-1}$$

Here, b must be invertible, otherwise, as was pointed out by Hurkyl, if I'm not mistaken, if we allow 1/0 to be an element of the ring and keep the other operations unchanged, we end up with 1 = 0, and the entire eing collapses to {0}.

CRGreathouse
Homework Helper
In theoretical computer science there is a generally accepted convention
that uses the symbol $$\bot$$ to denote undefined.
Be cautious, though; that's also used in logic to denote falsity.

i think 0-1=x.
x is every point on the complex plane, including the infinite and finite in magnitude.
and
0/0=0*any number on the complex plane
0/0=0*infinite=finite
0/0=0*infinetsimal=0
0/0=0*finite=0

so 0/0 is every point on the complex plane with a noninfinite magnitude, including zero.

Last edited:
Mentallic
Homework Helper
i think 0-1=x.
x is every point on the complex plane, including the infinite and finite in magnitude.
and
0/0=0*any number on the complex plane
0/0=0*infinite=finite
0/0=0*infinetsimal=0
0/0=0*finite=0

so 0/0 is every point on the complex plane with a noninfinite magnitude, including zero.
and what about $$lim_{x -> \infty} \frac{x^2}{x}=\frac{0}{0}=lim_{x -> \infty}x=\infty$$

and how about the function $$f(s)=0^{-s}$$ many textbook take f(s) to be alwyays 0 by analytic regularization

I do wish there was an accepted symbol for undefined though.
There is, depending on context of course. In recursive function theory, undefined is denoted by ↑ , to indicate that the function diverges (meaning its program/algorithm runs in an infinite loop and never stops) on a particular value, and thus is undefined on that input. I don't see why a similar notation couldn't be adopted for general math.

As to the original question, as those before me have stated, if you work with the more common systems of numbers, in order to define division by zero and maintain the 'nice' algebraic properties of these monoids/rings/fields, you wind up with the trivial ring, of which there is only one element, namely 0 (which is fairly boring).

Along with your question, you cited the fact that we give a definition to √-1, which seems like a rather arbitrary, silly thing to define. I'd just like to point out that this definition is actually very natural and makes a lot of sense within the context of other areas of math and science. As far as I'm concerned, i is no more 'imaginary' than √2 is. :)

Gib Z
Homework Helper
Along with your question, you cited the fact that we give a definition to √-1, which seems like a rather arbitrary, silly thing to define. I'd just like to point out that this definition is actually very natural and makes a lot of sense within the context of other areas of math and science. As far as I'm concerned, i is no more 'imaginary' than √2 is. :)
No really so much a definition as it is assigning it a symbol.

Anyway could we make a 'o ring' i mean the ring with elements

$$a+ b.(1/0)$$ with a and b considered real numbers , so this set have RING properties

Hurkyl
Staff Emeritus
Gold Member
What is the sum and product you propose?

the sum would be the usual

$$a+b.(1/0)+c+d.(1/0)= (a+b)+(c+d).(1/0)$$

the problem with the product is that you make it worse , in the product a term $$(1/0).(1/0)$$ would appear

Hurkyl
Staff Emeritus
Gold Member
Well, then, with those operations, the set of {a + b (1/0)} is not a ring.

Of course, the set of all polynomials in one variable is a ring, and we can name that variable 1/0 if we choose. But does that have anything to do with division by zero?

Now, since complex numbers require the special definition of √(-1), could you just define division by zero (arbitrarily, say, as "1/0 = m") and make an even more general number group?

Consider defining a number m and also redefining zero with it as a way of extending the number system and thus defining division by zero. By defining both zero and "m" through division instead of subtraction, they can be multiplicative inverses. How might this be carried out?

Let's follow your example of the special definition of the √(-1) where i indicates an imaginary number when written together with a Real number, and choose a symbol that will indicate nothingness when written with a Real number. In other words, instead of the single number 0, there will be a set of absent numbers represented by Real numbers together with an absence symbol.

So let m be n$$\infty$$ where n is a Real number, and let the absence symbol be $$\overline{\infty}$$. These numbers of nothing would then be n$$\overline{\infty}$$. In general, division by zero numbers would be $$\frac{x}{n\overline{\infty}} = (x/n) \infty$$. As with a reciprocal under a fraction bar, the inverse inverts. A zero number divided by another zero number would equal the quotient of their Real parts.

Note that the example of the imaginary numbers can only be followed so far. Normally at least, the additive identity property would preclude forming numbers made up of Real numbers plus absent numbers in a way analogous to the complex numbers.

The bar in the absence symbol is an inverse bar. It indicates absence and functions somewhat like a fraction bar except there is never a numerator. (This prohibition entails an exception for the division rule of equality although one much less onerous than the lack of definition of division by 0.) The infinity symbol here represents an array of the Real numbers. (Other definitions are possible.) To see some of the advantages of this notation, look to pp. 379-382, 580-2, 897-907, 916-25 in the book The Road to Reality by the physicist and mathematician Roger Penrose. He uses it for n-real dimensional space, without, however, a multiplicative inverse. Also note the compatibility of the new numbers with the complex extended plane or Riemann sphere referred to on this thread by g_edgar on Jul30-09. Indeed they may be seen as a generalization of it.

A number of issues not addressed here such as implications for physics, defining the absence symbol more fully, defining subtraction, an unsigned zero, arithmetic with the inverses of zero numbers, multiplication of zero numbers, a geometric interpretation, the empty sets unsuitability as a rigorous basis for the absent numbers, as well as other issues, are dealt with in the attached paper. Also not addressed here is the question "Does this number system form a field or not?"

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We don't imagine -- we construct a new number system (e.g. the projective numbers) in which 1/0 is defined, and use that one instead of the real numbers.

By the way, both of these limits are of the form 0/0:
$$\lim_{x \rightarrow +\infty} \frac{x}{x^2}$$

$$\lim_{x \rightarrow +\infty} \frac{x^2}{x}$$​
Assuming you meant in the limit as x goes to zero (as Mentallic mentioned) then can 0/0 is undefined even in the projective system, so can we not apply L'Hopital's rule and get:

$$\lim_{x \rightarrow 0} \frac{x}{x^2} = \lim_{x \rightarrow 0} \frac{1}{2x}=\infty \qquad ?$$

For the case that the limit of x goes to infinity (in the projective system):

$$\lim_{x \rightarrow \infty} \frac{x}{x^2}=\lim_{x \rightarrow \infty} \frac{1}{2x}=0$$

In fact it seems we can note that x/x2 is equivalent to 1/x and obtain the same limits as above even without applying L'Hopital's rule in the projective system.