### Subject: Higher than infinity? Problem where infinity is too small....

Date: **Tue Dec 28 17:55:56 1999**

Posted by **Somebody**

Grade level: **10-12**
School: **No school entered.**

City: **No city entered.** State/Province: **No state entered.**
Country: **No country entered.**

Area of science: **Other**

ID: **946425356.Ot**

**Message:**

I know I sound 3th gradish by asking is there a number higher than
everything, but there is a problem which I can't solve.
Mathematically, infinity can be defined as 1/0, right? Anyway, when you
multiply 0 * 1/0, you get 0/0, which is any real number.
If infinity is everything, then shouldnt it be able to break free of
normal bounds put on numbers? For example, 1^x is always 1.However, if we
insert 1/0 as x, we get 1^1/0, which is also the 0 root of 1. (I am going
to use @ to mean radical sign...)
0@1 = x
1 = x^0
This means x can be any number because any number ^ 0 is 1. However, when
I did 0^1/0, here is what I got: (@ means radical sign:)
0^1/0 = x
0@0 = x
0 = x^0
0 = (x^1)/(x^1)
0 = x?
I am saying here that x^0 = (x^1)/(x^1), which is true, right? And since
0/0 = 0, because 0 = 0*0, doesnt this mean that 0^(1/0) equals 0? What
then, can break the "bounds" on 0 and put 0 to the power of a number is 1,
or 2, or whatever... Infinity, I conclude, does not work in this case...
Would you need an number higher than infinity, in this case? I know it
sounds meaningless, but would you...? Or is my basis that 0/0 wrong and
0^0 is 0/0 wrong? Im not sure which....

Re: Higher than infinity? Problem where infinity is too small....

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