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### The Indeterminate Nature of 0/0

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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
follows:

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
is:

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
value.

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,
Rob
```

```
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

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
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.

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
rational.

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...

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.

more, or if you have any other questions.

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```

```
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

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:

http://mathforum.org/dr.math/problems/sheryl3.4.98.html

"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
that

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

- 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
possible?

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.

Thanks,
Rob
```

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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)]

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
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
solution

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
asymptotes.

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
http://mathforum.org/dr.math/
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Associated Topics:
High School Logic
High School Number Theory

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