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Posted Jan 23, 2001 by Wayfarer -MadForumArtist, Keeper of bad puns, Greeblet with Goo beret, Tangential One
there is no fraction for .9 repeating (that's .9999999999999999999999999999999999999999....... and so on forever)
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Posted Jan 24, 2001 by Decaf Silicon
Our math class keeps that as a running joke, as we've been told that .9 repeating = 1, through some formula, since it's 9/9. (5/9=.55555..., etc)
In a way, it does equal 1, as it's infinitely near 1.
-- Dmitri
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Posted Jan 24, 2001 by HenryS
I *does* equal one.
To prove it properly, you need to write 0.999999... as an infinite sum of the form:
Sum(n=1 to oo) { 9*(10)^(-n) }
then show that this sum converges to 1. But you need first year undergraduate mathematics to do that.
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Posted Jan 24, 2001 by Wayfarer -MadForumArtist, Keeper of bad puns, Greeblet with Goo beret, Tangential One
really? you do? i have used that formula in class before, but i wasn't a graduate anything then. it is near enough to one as makes no difference, but it isn't, quite. i think its 1-(.000......1(zero repeating with a one after all the zeros which is as close as you can get to zero and still not be))
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Posted Jan 25, 2001 by HenryS
Well, you need the university maths to define precisely what 'converges' means. Intuitively yes, it keeps getting nearer to 1, but thats not enough to be sure.
"1-(.000......1(zero repeating with a one after all the zeros which is as close as you can get to zero and still not be))"
This isnt a number. There is no place to put that one after all the zeros, because the zeros never end.
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Posted May 31, 2001 by DMarsh3000
x = 0.99999...
10x = 9.99999...
Subtracting equations, 9x = 9
So x = 1, that is, 0.9999... = 1.
DM
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Posted Jun 4, 2001 by Researcher 178849
This discussion lacks a definition of what is meant by writing 0.999... One cannot just use symbols arbitarily.
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Posted Jun 10, 2001 by HenryS
I'm assuming 0.999,,,, means:
sum from i = 1 to infinity of 9/(10^i)
With that definition its possible to show this sum converges to 1.
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