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.99999 repeating = 1

Name: Anonymous 2007-02-18 15:23

x= .999 repeating
10x = 9.999 repeating
10x-x = 9.999 repeating - .999 repeating
9x= 9
x=1

Name: Anonymous 2007-02-22 1:22 ID:kjwB4oM4

>>40
0*10^n will always be zero, so you can add as many zeroes as you want to a number without changing its value (and therefore without changing it).

As for 0.000...1, that is a malformed number and has no meaning. Consider .123...4, where the ellipsis still represents an infinite number of digits. Each digit is multiplied by 10 to some n proportional to its distance and direction from the decimal point, so .123... can be represented as (1*10^-1)+(2*10^-2)+(3*10^-3)+.... What is n for the 1 in .000...1? The correct application of that hotel thing from earlier tells us that it can't be infinity.

Alternatively, I challenge you to write this number - write an infinitely long string of zeroes, and then put a 1 at the end of it. If you manage to do so, well, that wasn't a very infinite string, was it?

>>43
Troll. That crap where the teacher supposedly tried to create a number for the reciprocal of zero does nothing to resolve the conflict, since 0*(1/0) would equal 0*(nullity), which would still be required to equal both 0 and 1 by 0x=0 and n/n=1.

>>45
Where did you come up with that 1, boy?

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