The revised infinity theory.

[lawtonfogle]

For those who were on the last thread concerning this, I have started a new one over since the last one is dead. I have thought on the rules for a while, and now have made a few new rules and changed the otherones.

For those who have not read the last thread, what I am trying to do is create a system so that \frac {a} {x} \cdot \frac {x} {1} = \frac {a} {1} even if x = 0

It is finished.

Rules
First the order of operations must be revised.
The order will go
1)All workable exponents
2)All workable parentesis
3)Multiply and Divide with zeros
4)Solve for any \infty ^ 1
5)Normal multiplication and division
6)Normal addition and subtraction
7)Solve for any \infty ^ 0

Second the comunive property cannot be used until step 4 is applied.

Third a \infty ^ 1 = a and a \infty ^ 0 = 0 and to avoid complication 0 \neq 0 \infty ^ 0 . This does allow a \cdot 0 = 0

Fourth comes the basic rules I started with \frac {a \infty ^ {x}} {0} = a \infty ^ {x + 1}
and a \infty ^ x \cdot 0 = a \infty ^ {x-1}

These are the basic rules.
Now for the other rules.

First, how to squareroot a \infty number.
\sqrt {4} = 2 need someone to show me how to do +/- sign.
so \sqrt {4 \infty ^ 1} = 2 \infty ^ 1
which is done by saying that \sqrt {a \infty ^ x} = \sqrt {a} \infty ^ {\sqrt{x}}

Second of infinity numbers with different infinity powers. The rule goes a \infty ^ x \cdot b \infty ^y = a b \infty ^ {xy} . This does work if x = y

Third is addition and subtraction of numbers with different powers. The rule goes a \infty ^ x + b \infty ^ y = a \infty ^ x + b \infty ^ y .
Now if x = y then a \infty ^ x + b \infty ^y = \left( a + b \right) \infty ^ x

[lawtonfogle]

Ok i think the rules are finished. If you want to know how i came up with any rule, please ask.

[Hurkyl]

First off, "order of operations" is a notational convenience... NOT part of an algebra.

Secondly, parentheses are not operations -- they're a necessary part of notation to indicate the operands of applying a function.


Have you played much with your algebraic operations? You also seem to be missing 0/0.

But! We can figure it out: for any a, we have:

0/0 = (a ∞0) / 0 = a ∞1

So, we've run into a problem: a ∞1 = b ∞1 for all a and b.

[lawtonfogle]

so you go from \frac {0} {0} to \frac {a \cdot 0} {0}

but you only multiplied one side by a, it should be \frac {a \cdot 0} {0} = a
which would work out to
\frac {a \infty ^ 0} {0} = a
a \infty ^ 1 = a
a = a

I have seen many people times one side by a but not the other, why.

[quasar987]

\pm (click on the symbol to get the code)

I see you found about \neq already. Just download the two pdf files chroot's first post tells you to download in the Latex Tysetting post. It's all in there.

[Zurtex]

But the thing is if you multiply any finite number by 0 it is equal to 0. Or more simply 0 + 0 = 0, hence:

1 = \infty^1 \cdot 0 = \infty^1 \cdot (0 + 0) = \infty^1 \cdot 0 + \infty^1 \cdot 0 = 1 + 1 = 2

[Hurkyl]

but you only multiplied one side by a, it should be

I didn't do any multiplying -- I only applied your identities.

0/0 = (a ∞0) / 0 is true because of the identitiy a ∞0 = 0.

(a ∞0) / 0 = a ∞1 is true because of the identity (a ∞x) / 0 = a ∞x+1.

Therefore, 0/0 = a ∞1 is true.


Also, when one writes 0/0 = (a*0)/0, they are not multiplying by a -- they're applying the identity 0 = a*0 to replace one of the 0's with a*0.

[lawtonfogle]

I didn't do any multiplying -- I only applied your identities.

0/0 = (a ∞0) / 0 is true because of the identitiy a ∞0 = 0.

(a ∞0) / 0 = a ∞1 is true because of the identity (a ∞x) / 0 = a ∞x+1.

Therefore, 0/0 = a ∞1 is true.


Also, when one writes 0/0 = (a*0)/0, they are not multiplying by a -- they're applying the identity 0 = a*0 to replace one of the 0's with a*0.

ill take care of this one first. I have the rule 0 \neq \infty ^0

also, changing a a \infty ^ 0 = 0 is done last. it is the same with
2 \cdot 2 + 1
they can be done without order of ops.
4 + 1 = 5 or 2 \cdot 3 = 6

must go

[matt grime]

ill take care of this one first. I have the rule 0 \neq \infty ^0

also, changing a a \infty ^ 0 = 0 is done last. it is the same with
2 \cdot 2 + 1
they can be done without order of ops.
4 + 1 = 5 or 2 \cdot 3 = 6

must go

That makes no sense.

If a \infty ^ 0 = 0 , then dividing by a and usign the rules of arithmetic that surely we want to keep if must follow that \infty^0=0, which you claim is false.

[Zurtex]

In arithmetic there is no rules that things must be carried out in a certain order, that would make no sense, I think you are confusing up the generalization of binary operations on real numbers.

With a binary operations like * (multiply) and + (add) you can only perform these operations with 2 numbers, like so: 2 + 3 = 5 and 2*3=6. However, because we often need to use many of these operations at once we generalize and say things like 2+3*4=14, what we actually mean is: 2 + (3*4) = 14, it could just as easily be standard notation that it 2+3*4=20 as in: (2+3)*4=20. But that's all it is, standard notation, we need to choose one so we can use less brackets, it's not actually some law of numbers.

[lawtonfogle]

so 14 = 2 + 3 \cdot 4 = 20
put 2 + 3 \cdot 4 = x
so that 14 = x = 20
flip the postions of x and 20 14 = 20 = x
so we have 14 = 20
which does not work.

[lawtonfogle]

That makes no sense.

If a \infty ^ 0 = 0 , then dividing by a and usign the rules of arithmetic that surely we want to keep if must follow that \infty^0=0, which you claim is false.

maybe i should clear up. I was saying that 0 \neq 0 \infty ^ 0
because that would equal \infty ^ {-1} and that can be applied. I will cange it. Any ways, you should only go form \infty ^ 0 to zero, when there is nothing else to be done.

[matt grime]

So, that's another rule, actually several, then that makes the symbol \infty different from all real numbers then. How on earth can that be said to be a good thing, given your premise.

[lawtonfogle]

That makes no sense.

If a \infty ^ 0 = 0 , then dividing by a and usign the rules of arithmetic that surely we want to keep if must follow that \infty^0=0, which you claim is false.

i meant that 0 \neq 0 \infty ^ 0

[lawtonfogle]

also, i was basing the off of an idea that 0 should never be turned into a \infty ^0

[Hurkyl]

I can safely say, then, that what you're describing is certainly not mathematics. The law of substitution is one of the most important things, and it's what makes the language of mathematics useful. If two things are equal, then substituting one for the other does not change the meaning of a statement.

[Zurtex]

so 14 = 2 + 3 \cdot 4 = 20
put 2 + 3 \cdot 4 = x
so that 14 = x = 20
flip the postions of x and 20 14 = 20 = x
so we have 14 = 20
which does not work.
NO!

My point was there is no order of doing things as all addition and multiplication are binary operations and all you can do is stick 2 numbers in and get a single number out.

2+3*4 does not make sense at all! Not unless you define some standard notation for it to mean something. Please try and read what people put more carefully.

[lawtonfogle]

now i think i see what you mean by a binary operation.

Does that means that 2+3*4 is a trinary operation

i was taught that order of operations must always be followed so the same answer can be optained.

[lawtonfogle]

im going ahead and printing out my fist post. thanks for helping me develop the rules better, and i can now latex type yay. im not going to speak on this more unless for some reason i get a big break thruogh, if that is possible. I'll post what my teacher says about this during end of august.

Thanks again.

[Zurtex]

im going ahead and printing out my fist post. thanks for helping me develop the rules better, and i can now latex type yay. im not going to speak on this more unless for some reason i get a big break thruogh, if that is possible. I'll post what my teacher says about this during end of august.

Thanks again.
To be honest this area of mathematics has been studied on a great level of detail by some of the greatest mathematicians and they have found ways to generalise it and look at it all at once and prove what is and isn't possible with this sort of thing. But you need to learn a great deal of maths first.

[lawtonfogle]

all i know is algerbra two, learning pre-cal next year