# The Infinity Theory

1. May 6, 2005

### lawtonfogle

First, i would have put this in the theory development, but i can't. And since this is pure mathmatical, but i do not know what type, I have put it here.

Second, you need to keep an open mind and do not call me crazy.

Third, the person who came up with imaginary numbers was probally called crazy at first, but now he isn't.

Fourth, i will use 'in.' to represent infinity.

Fifth, the math rule of law 'Any number times zero equals zero' must be changed to 'Any real, imaginary, or complex number times zero equals zero', which is (as of right now) just as true as the first rule.

OK, now down to business.

Does not (X/1)*(a/X)=A/1

this should work for anynumber, true

What about X=0 then.

It will if three new laws exist.
(note, the 'in.' stands for the infinity sign, and that the powers apply to the infinity sign, not a.)

1) (a'in.'^x)/0 = a'in.'^(x+1)
2) (a'in.'^x)*0 = a'in.'^(x-1)
3) 'in.'^0 = 1

KEEP AN OPEN MIND

so

(x/1)*(a/x) if x=o
then
(0/1)*(a/0)
which equals
0*(a/0)
which equals due to rule number 1)
0*(a'in.'^1)
which equals due to rule number 2)
a'in'^0
which equals due to rule number 3)
a*1
which equals
a
which equals
a/1

so

(0/1)*(a/0)=(a/1)

so

(x/1)*(a/x)=(a/1) even if x=0

This is a theory because a theory is a set of rules which are always proven true. And the rules 1),2),3) are proven true in all equations i put them in.

So this must be a true theory, not an idea

please respond if you find this false so i can try to modify these rules.

2. May 6, 2005

### matt grime

That isn't an issue, though the fact that you think *number* is well defined without qualification is. 0*x=0 in any Ring.

again, that is meaningless without context. In a ring, if X is invertible, then it is true, but so what?

why? what is a *number* in the sense you're using it? We'd take it to mean usually in the Ring of reals, or complexes and so on.

can you please rewrite that in mathematical notation so I can decipher it? what are the ' marks for? what is a, ie in what set is a?

what follows is also hard to read.

look, it's very simple, learn some ring theory, define the things yo'ure talking about properly, and state all the laws of the operations, such as commutativity, or associativity, prove they're consistent, and then look up the extended real and complex systems to see why, if we declare 1/0 to be infinity, the result isn't a well defined ring.

I suspect you don't even know what a ring is, or what associativity is, nor have you adequately, for my money, defined infinity, whatever that may be.

3. May 6, 2005

### Zurtex

First, lets looks at your fifth rule:

'Any number times zero equals zero' must be changed to 'Any real, imaginary, or complex number times zero equals zero',

I think you will first find that there is no rule which says "Any number multiplied by 0 is 0". But yes it is true that all complex numbers multiplied by 0 does equal 0.

Next:

"(X/1)*(a/X)=A/1"

It is true that:

$$\frac{x}{1} \frac{a}{x} = \frac{a}{1} \quad \forall \, x \, \in \mathbb{C} \backslash \{ 0 \}$$

This means it is true for all complex numbers excluding 0 as a/0 is not defined under complex numbers.

Next you seem to have defined 3 axioms to govern your system of numbers:

$$\begin{gather} \frac{a \infty^x}{0} = a \infty^{x+1} \\ a \infty^x \cdot 0 = \infty^{x-1} \\ \infty^0 = 1 \end{gather}$$

I think it would of been nicer if you'd made it a separate axiom that:

$$a \infty = \infty \quad \forall a \neq 0$$

Here is the basic problem with your system:

$$\frac{1}{0} \cdot 0 = 1$$

$$\frac{1}{0} \left( 2 \cdot 0 \right) = 1$$

$$\frac{1}{0} \cdot 2 \cdot 0 = 1$$

$$\frac{1}{0} \cdot 0 = \frac{1}{2} = 1$$

Contradiction! Hence there must have been a false assumption somewhere.

Last edited: May 6, 2005
4. May 6, 2005

### lawtonfogle

You are right, im only in high school, but i'll go study it.

5. May 6, 2005

### lawtonfogle

Zurtex, thanks for putting it in symbols.

I'll try to work on your problems.

6. May 6, 2005

### lawtonfogle

Im trying to do the special symbols. So this might take a while

$$\frac{1}{0} \cdot 2 \cdot 0 \ = 1$$

false
it equals two

If
$$\frac{1}{0} \cdot 2 \cdot 0$$
then
$$\ 1 \infty^1 \cdot 2 \cdot 0$$
then
$$\ 2 \infty^1 \cdot 0$$
then
$$\ 2 \infty^0$$
then
$$\ 2 \cdot 1$$
then
$$\ 2$$

Last edited: May 6, 2005
7. May 6, 2005

### Zurtex

Don't put spaces after the \ and there is no need for a \ before a number, I think you were going for:

You are in fact right, it does equal 2 under your number system, but another way of looking at is that:

$$\frac{1}{0} \cdot 2 \cdot 0 = 2 \frac{1}{0} \cdot 0$$

Now, 2 is just some number, lets just call it a for the moment:

$$a \frac{1}{0} \cdot 0 = a \infty^1 \cdot 0$$

Which by your 2nd axiom is the same as:

$$a \infty^1 \cdot 0 = 1$$

So basically we have that 1 = 2, hence there must be some logical error and unfortunately the logical error is in your mathematical system. Don't worry, lots of people make this mistake, as you learn more and more about mathematics you'll begin to appreciate the nightmare this sort of system is and why it doesn't work on so many levels.

Another problem with your system:

$$0 = 0 \cdot 0$$

Therefore it stands that:

$$\frac{1}{0} = \frac{1}{0 \cdot 0} = \frac{1}{0} \frac{1}{0}$$

And hence:

$$\infty^1 = \infty^2$$

Multiplying both sides by 0 we get:

$$1 = \infty^1$$

Last edited: May 6, 2005
8. May 6, 2005

### lawtonfogle

Yay i can latex.

9. May 6, 2005

### lawtonfogle

$$a \infty^1 \cdot 0 = 1$$
no sorry

$$a \infty^1 \cdot 0 = a$$

10. May 6, 2005

### lawtonfogle

im trying to solve this problem.

11. May 6, 2005

### Hurkyl

Staff Emeritus
How about, instead of trying to talk about division by zero, you introduce numbers that are infinitely close to zero, which you can invert to produce an infinitely large number. In this way, you can retain all of the nice properties of arithmetic.

12. May 7, 2005

### Zurtex

Again the problem remains:

$$a \infty^1 \cdot 0 = \infty^1 (a \cdot 0) = \infty^1 \cdot 0 = 1 = a$$

13. May 10, 2005

### lawtonfogle

naaa
that would be too simple :rofl:

really. I want zero to work for X

14. May 10, 2005

### lawtonfogle

now i see what you are saying.

I have worked on it and have thought that a new order of operations is needed.

some base lines are like this
1) $$a \infty^x \cdot 0$$
does not equal
$$\infty^x (a \cdot 0)$$
which is kinda like 1-2-3 does not equal 1-(-1)

maybe it is the comunitive property.

or maybe it is that $$1 \cdot 0 = 1 \infty^0$$
like all nonimaginary numbers are really imaginary numbers that are $$i^4$$

and all known numbers are in the $$\infty^1$$

side note: they need a thinking icon

15. May 10, 2005

### matt grime

You are indeed now requiring your "extended" system to fail to be commutative and associative wrt multiplication. Have you looked up any of the well defined extended systems there are?