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0.999999... = 1?

Name: Anonymous 2006-05-25 9:53

What the fuck. Why is that true. They got different numbers in them.

Name: Anonymous 2006-07-18 21:37

The best I can say is that lim(n -> +infinity- [infinity greater than zero, approached from the lesser side]) { sigma(x=1; x <= n) {9/(10**n)}  }== 1.

That is to say, as you increase the number of 9 that follow 0. in the expression 0.999999, the value of that expression approaches 1. 0.999... == 1 only in the case that you have iterated an infinite number of 9 after the decimal.

uhhhh someone else finish what i'm saying if i'm going in the right direction. or am i just blowing air out of my ass

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