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The Indeterminate Nature of 0/0

Date: 12/21/2000 at 16:02:44
From: Rob
Subject: Ludicrous new 0/0 theory?

I have been fiddling around with dividing 0 by 0, and have come up 
with an interesting theory. I have not yet seen any proof that this 
theory cannot work, so as of now I think it is valid. It goes as 

My theory is that:

     0/0 = any number

To start, use the definition of division equation where:

     a/b = c   because   c*b = a

In the situation of 0/0 = c, then c*0=0 for any real value of c (I am 
not familiar with how to work with infinity or non-real numbers so I 
have left them out.)

Some have argued that this would simply be an indeterminate form, but 
I counter with this; the equation of the line y = 1 in standard form 

     y = 0x+1

If you were to solve for x, then

     x = (y-1)/0

By my theory only 0/0 can satisfy every value of x, thus y must always 
be equal to 1 (otherwise it is undefined). In this situation, where y 
always equals one, x can be any number of values. If you say that 0/0 
is indeterminate, then the equation y = 1 would also be indeterminate 
because there are an infinite number of x values for that single y 

That is, unless 0/0 can be equal to any one of the infinite numbers at 
any given time. For example at point x = 5 in the x = (y-1)/0 
equation, the value 0/0 = 5. This theory would radically change 
certain views of math.

Consider the equation y = 0/[(x-2)(x+3)] using my theory. The points x 
= 2 and x = -3 in this equation would stretch forever straight up and 
down, since y would equal 0/0. Under traditional math laws, these 
points would be asymptotes.

Also consider the equation y = (0/0)x. Using traditional math this 
could simplify to 0y = 0x, making the graph cover every point. This 
corresponds exactly to my original theory.

I hope my writing has made sense. If you can think of a way to tear 
down this theory, I am anxious to hear it. Please keep in mind that I 
am not arguing 1/0, because that is still undefined, I am only arguing 
for 0/0.

Thank you,

Date: 12/21/2000 at 17:35:39
From: Doctor Ian
Subject: Re: Ludicrous new 0/0 theory?

Hi Rob,

This is an interesting theory, but there is a fly in the ointment. 
Let's say for a moment that we accept that 0/0 = any number. Then we 
can do this:

   0 = 0     Trivially true

 0/0 = 0/0   Dividing both sides by the same thing preserves equality

   1 = 2     Substitute for 0/0

Oops! It appears that your definition leads to a contradiction. This 
is a problem, because once you can 'prove' a contradiction, you can 
use that contradiction to prove anything, which means that the concept 
of proof becomes meaningless.

Perhaps the most common type of proof used in mathematics is "proof by
contradiction." To prove something by contradiction, you assume that 
the thing is false; then you show that the assumption leads to a 
contradiction, which means that the thing can't be false; which means 
that it must be true.

The classic example is the following proof that the square root of 2 
is irrational:

  1. Assume that it is rational, of the form a/b, where a/b is reduced 
     to lowest terms.

  2. Then (a/b)*(a/b) = 2, or a^2 = 2b^2, so a^2 must be even.

  3. Since a^2 is even, a must be even, say a = 2c.

  4. Then (2c)^2 = 2b^2, or 4c^2 = 2b^2 or 2c^2 = b^2, so b must also 
     be even.

  5. So in a/b, both a and b are even, but we assumed we'd reduced the 
     fraction to lowest terms. We have a contradiction, so sqrt(2) 
     must be irrational.

With the addition of your new definition, proof by contradiction 
becomes trivial, regardless of what we want to prove. For example, 
let's prove that the square root of 2 is rational.

  1. Assume that it is irrational.

  2. Let 0/0 = 0/0, where 0/0 is any number.

  3. Then 1 = 2, which is a contradiction. Therefore, sqrt(2) must be 

Do you see the problem? If 0/0 = any number, then the square root of 2 
is both rational and irrational; pi is the ratio of circumference to 
diameter, and it's also not that ratio; there is no largest integer, 
and there is a largest integer; every number is prime, and no number 
is prime...

It's great that you're thinking about this stuff! What you've actually 
stumbled onto is one of the deepest issues in mathematics, which is 
the irresolvable tension between completeness and consistency, a 
subject explored by the mathematician Kurt Godel, who showed that you 
have to choose one or the other, because you can't have both in the 
same system. 

With that in mind, I strongly recommend that you take a look at a book 
called _Godel, Escher, Bach_, by Doug Hofstadter. You should be able 
to find it at any bookstore or library. Ignore the chapters the first 
time through, and just read the dialogs in between the chapters. Then 
go back and read the chapters. It's a fabulous book, which will give 
you a lot of new material to think about.

I hope this helps. Let me know if you'd like to talk about this some 
more, or if you have any other questions.

- Doctor Ian, The Math Forum

Date: 12/28/2000 at 01:27:03
From: Rob
Subject: Re: Re: Ludicrous new 0/0 theory?

Thanks for writing back, Doctor Ian.

I hope I don't sound (too) stubborn, but I have a couple of follow-up 
questions about your answer.

First a summary of my points:

   ->  0/0 = any number

   ->  a/b = c because c*b = a, likewise 0/0 = any number because 
       (any number)*0 = 0

   -> for the equation y = 1:

      table of values for x     table of values for y
        x   y                   (only y = 1 will work)
       -2   1                        y   x
       -1   1                        1   ?
        0   1 
        1   1 
        2   1

There is no single x value for this point it has many values. If 
having many values makes 0/0 indeterminate, then it should also make y 
= 1 indeterminate, which it isn't.

If y = 1 is a legitimate graph, then 0/0 should be good also. 
Conversely, if 0/0 is not legitimate, then neither should is the 
equation y = 1. You might say that my conclusion is not based on my 
facts, but examine this:

     y = 1 --> y = 0x+1 --> y-1 = 0x --> x = (y-1)/0

According to my theory, 0/0 = any number, the only way to satisfy the 
equation for any value of x is for the other side of the equation to 
be 0/0, which can only happen if y = 1. This seems to be a logical 
extension of my theory and the manipulation of equations.

Consider also this, from "Zero Laws and L'Hopital's Rule" in your 
archives at:


  "Say we have two functions, f(x) and g(x). If you take their 
  quotient, f(x)/g(x), you get another function. Let's call this h(x). 

  "You can also take limits of functions. For example, the limit of a 
  function f(x) as x goes to 2 is the value that the function gets 
  closest to as it takes on x's that approach 2. We would write this 
  limit as:

        lim{x->2} f(x) 

  "This is a pretty intuitive notion. Just draw a graph of your 
  function, and move your pencil along it. As the x values approach 2, 
  see where the function goes.

  "Now for our example. Let:

        f(x) = 2x - 2
        g(x) = x - 1

  and consider their quotient:
        h(x) = f(x)/g(x)

  "What if we want the lim{x->1} h(x)? There are theorems that say 

        lim{x->1} h(x) = 
        lim{x->1} f(x)/g(x) =
        (lim{x->1} f(x))/(lim{x->1} g(x)) = 
        0 / 0 !!!

   "So we now have:
        lim{x->1} h(x) = 0/0

   "But I can employ another theorem, called l'Hopital's rule, that 
   tells me that this limit is also equal to 2. So in this case, 
   0/0 = 2

   -Doctors Sonya and Celko,  The Math Forum"

You wrote:

    0 = 0     Trivially true

  0/0 = 0/0   Dividing both sides by the same thing preserves equality

    1 = 2      Substitute for 0/0

   "Oops! It appears that your definition leads to a contradiction."

   - Doctor Ian, The Math Forum

To rebut your contradiction theory, I would say that it is not the 
0/0 = any number that is wrong, but the substitution. If you may 
substitute any value in for 0/0 arbitrarily, then why wouldn't this be 

       2x = 3y
     2(5) = 3(6)
       10 = 18

or even

b       3x-1 = 2x+4
     3(-7)-1 = 2(-7)+4
         -22 = -10

Can you arbitrarily substitute in numbers that make no sense in the 
equation? It was also my theory that the value of 0/0 is dependent on 
the situation. 0/0 is equal to 1 and 2, but not in an equation at the 
same time necessarily. In the equation y = 1, can you not have the 
point P(4,1) and the point P(-3,1)? You just can't get them by 
substituting a single x coordinate, that is all.

I hope I have raised some good points, I'll be grateful for whatever 
you can do for me. 


Date: 12/31/2000 at 18:18:16
From: Doctor Rick
Subject: Re: Re: Ludicrous new 0/0 theory?

Hi, Rob.

I don't know exactly what your theory means, and neither did Dr. Ian. 
You say, "0/0 = any number." The natural way to interpret this phrase 
is, "You can choose any number x, arbitrarily, and the equation 
0/0 = x is true." Dr. Ian read it this way, I believe, and from this 
he derived that 1 = 2. This causes big problems for math, as he said.

Your response is that this isn't what you meant. You say that "the 
value of 0/0 is dependent on the situation." The word I would use for 
this is "indeterminate."

Here is what we mean by "indeterminate." The value of 1/0 is called 
"undefined" because there is NO number x that satisfies the equation 
1/0 = x, or equivalently, 0*x = 1. In contrast, EVERY number x 
satisfies the equation 0/0 = x, or equivalently, 0*x = 0. But this 
does not mean that you could substitute any number arbitrarily for 
0/0. It depends on the context - on the details of the problem in 
which you encountered 0/0.

When you run into a problem whose solution appears to be 0/0, there 
may be a solution, but you'll have to back up and try to find it by 
another approach. When you find it, you could say that 0/0 "equals" 
that value, but only for that particular problem - just as you say.

The example you quote from our website, concerning L'Hopital's Rule, 
is a good example of how a problem in which 0/0 crops up can have a 
valid solution. L'Hopital's Rule arises in calculus, and if you 
haven't seen it yet, I couldn't explain it quickly, but the basic idea 
is that, if you encounter 0/0 in evaluating the limit of the quotient 
of two functions, there is another method by which you can find the 
value of the limit. You can't use the 0/0 to find the solution, but 
you can back up and take another route to find the solution.

Thus, a good way to think of the "indeterminate" 0/0 is as a "detour 
sign". It says in effect, "The road ahead is washed out. The town 
beyond it is still there, but you'll have to find another way to reach 
it." To bring this around to the word that you don't like, it says, 
"The solution can NOT be DETERMINED by this method. There may be a 
solution, but you must determine it some other way."

In your original submission, you cite the equation

     y = 0/[(x-2)(x+3)]

as one in which your math gives a different answer than traditional 
math. You say that traditional math identifies x = 2 and x = -3 as 
asymptotes, whereas your math leads to a graph with vertical lines at 
x = 2 and x = -3. The fact is, your answer is the traditional answer 
to the equivalent equation

     y(x-2)(x+3) = 0

In order for the product to equal zero, at least one of the three 
expressions y, x-2, and x+3 must equal zero. This leads to the 

     y = 0   or   x = 2   or   x = -3

which is graphed as three lines: a horizontal line at y = 0, and the 
two that you call "asymptotes." They are part of the solution set, NOT 

Finally, let's consider the equation y = 1, as you do. You say that, 
"if y = 1 is a legitimate graph, then 0/0 should be good also." I 
think you mean that, since the graph of y = 1 is a line consisting of 
all points with y = 1 and x = any number, then by analogy, the 
solution of the single-variable equation 0/0 = x should be the entire 
number line, x = any number.

I'd say, yes, the solution of the equivalent equation 0x = 0 is the 
entire number line. In this particular problem, it's quite true that 
the solution is x = any number. But this does not mean that 0/0 = any 
number, in itself. The example about L'Hopital's Rule is a case in 
which the solution is NOT any number; it is the particular number 2, 
and no other. As I have said, you must consider each particular 
problem on its own. The fact that 0/0 = any number in one problem 
cannot be used to solve another problem.

Have I made sense yet? I think the key concept you have been missing 
is this problem-specific nature of 0/0. The difference between "any 
number" and "indeterminate" is that "indeterminate" means "it MAY be 
any number, but you can't tell just by looking at 0/0 itself."

- Doctor Rick, The Math Forum
Associated Topics:
High School Logic
High School Number Theory

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