A Conversation for Logical Completeness

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Post 1

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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Post 2

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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Post 3

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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Post 4

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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Post 5

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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Post 6

DMarsh3000

x = 0.99999...
10x = 9.99999...

Subtracting equations, 9x = 9

So x = 1, that is, 0.9999... = 1.

DM


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Post 7

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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Post 8

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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