1. Originally Posted by Chokfull
Which rule does it not obey? I looked them over again...

And look at my previous post about undefined... it is about the same with indefinite
How about 6a for one, but why not check them yourself.

CB

2. Originally Posted by HallsofIvy
What rules of math are you referring to? If you mean things like "you can't take the square root of a negative number" or "you can't divide by 0", there are no "rules of math" that say you can.

The set of real numbers forms a "complete ordered field". The "rules of math" for the real numbers are precisely those of a complete, ordered, field:
1) The real numbers form a commutative group under addition.
_____a) (x+ y)+ z= x+ (y+ z)
_____b) x+ y= y+ x
_____c) there exist a number, 0, such that x+ 0= x for all x
_____d) for every x there exist a unique y such that x+ y= 0

2) xy= yx
3) there exist a number, 1, such that 1x= x for all x
4) For every x except 0, there exist y such that xy= 1
5) x(y+ z)= xy+ xz

6) There exist an order x< y such that
____a) if x< y then x+ z< y+ z
____b) if x< y and 0< z then xz< yz
____c) for any two number x and y, one and only one of
_______i) y= x
______ii) x< y
_____iii) y< x
is true.

7. All Cauchy sequences converge.

Which of those do negative numbers or 0 "defy"?
you missed this:

(xy)z=x(yz)

and(maybe)
number 0 and number 1 are not same(additive and multiplicative identities are different).

3. $3<\infty$
$3+2<\infty+2$
this seems to fit the rule...

4. Originally Posted by Chokfull
Which rule does it not obey? I looked them over again...

And look at my previous post about undefined... it is about the same with indefinite
Given any real number, x, there exist a real number, called "-x" such that x+ (-x)= 0. (If x= 0, then -x= 0 also. For any other x, -x is not equal to x).

There is no number that, added to infinity gives 0. No, $\infty+ (-\infty)$ is NOT equal to 0!

Another is "if x is a real number, then x+ 1> x. That is not true for x= infinity.

5. Originally Posted by HallsofIvy
Given any real number, x, there exist a real number, called "-x" such that x+ (-x)= 0. (If x= 0, then -x= 0 also. For any other x, -x is not equal to x).

There is no number that, added to infinity gives 0. No, $\infty+ (-\infty)$ is NOT equal to 0!

Another is "if x is a real number, then x+ 1> x. That is not true for x= infinity.
uh... yes $\infty + (-\infty)$ is 0....
But assuming that $\infty$ is a real number, that does not mean it is the last one... look up an interesting puzzle about a hotel with infinite rooms.
I seriously thought someone would bring up #4, but you were too smart 4 me... $1/\infty$

6. Originally Posted by Chokfull
uh... yes $\infty + (-\infty)$ is 0....
No. That is wrong.

Suppose for the sake of contradiction that,

$\infty + (-\infty) = 0$

$1 + (\infty + (-\infty)) = 0 + 1$

Then Rules 1a and 1c:

$(1 + \infty) - \infty = 1$

Then using a rule you yourself agreed with, $1+\infty = \infty$, leaves us with:

$\infty - \infty = 1$

Giving $0 = 1$ $\Rightarrow \Leftarrow$

This is similar to what opalg said, which was pretty much the only piece of evidence needed to refute what you are saying.

It's all very well asking questions and pushing ideas, that's what this site is for and that's why I love it, but when people who really do know what they are talking about are throwing counterproof after counterproof at you, I think it's time to step back and assess the situation.

Read around the subject, read what has been suggested untill you fully understand it. If there is then something you struggle to understand, ask for help. When help is given to you on good authority, do not dismiss it with shoddy maths, try instead to use it to better your understanding.

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