MadSci Network: Other |
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....
Try the links in the MadSci Library for more information on Other.