h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

there is no fraction for .9 repeating (that's .9999999999999999999999999999999999999999....... and so on forever)

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

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.

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))

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.

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

Subtracting equations, 9x = 9

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

DM

This discussion lacks a definition of what is meant by writing 0.999... One cannot just use symbols arbitarily.

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.