<p>Legend, you're right that it would make more sense to question 1/3 = 0.3333...</p>
<p>What we really are talking about when we use a decimal expansion is .a1a2... = [sum] ai 10-i, where an infinite sum is defined as the limit of partial sums, etc. (definition.) Then, showing that the things we do with infinite decimals are justified requires a bit of work. Since showing 1/3 = [sum] 3 * 10-i isn't any easier than showing 1 = [sum] 9 * 10-i, most proofs given don't really shorten anything. Similarly for the 10x - x = 9 approach - you're subtracting two infinite sums, and you need to know that's justified when the two sums are absolutely convergent. So the easiest rigorous proof is probably just straghtforwardly showing the limit is 1 (or using any number of other calculus related proofs for people who can't fathom the validity of the limit because they are using their own idea of an infinite process rather than Cauchy's strict definition).</p>
<p>On the other hand, it's easier to give the naive proofs to doubters, if it will shut them up! Odds are they will tune out if you try to explain limits to them.</p>
<p>(paraphrased from a post on the thread I linked)</p>
<p>Basically, if you believe that 1/3 = 0.333.. you should believe that 1 = 0.999...
It is straightforward to prove (and obviously impossible to demonstrate to be false, since it is in fact true).</p>
<p>I know, I read it and went..."oh crap, of course!"</p>
<p>Psh, intuition is often not the answer it seems. I had forgotten the definition of a repeating decimal and contorted it in my head for some odd reason.</p>
<p>Anywho, there you are. I believe that is the definitive answer. Again, correct me if I'm wrong.</p>
<p>"Basically, if you believe that 1/3 = 0.333.. you should believe that 1 = 0.999...
It is straightforward to prove (and obviously impossible to demonstrate to be false, since it is in fact true)."</p>
<p>This is what I mean. People don't question that .3 repeating is 1/3 but they question that .9 repeating is 1? Doesn't make sense to me.</p>
<p>Aiie. The point of the series was that it tends to both 1 and .9999999.... The series itself does not equal to .999999... It tends to it by definition. But as we take the limit it tends to one. And the limit is unique.</p>
<p>You have to be kidding me ".0000....1" If there is an infinte amount of zeros before the 1, where does the 1 come from. By adding a 1 to the end, there can't be an infinite number of zeros before it. .99999 is equal to 1 because all math scholars say that it is. period</p>
<p>I agree. Thinking about adding a one to the end of all of those zeroes is not a good idea. Rather, the limit that defines such a number would be the Sum of 10^-n starting with n=1. Clearly this series of partial sums is .1 .01 .001 .0001 and tends to .00000...0001 whatever that means. But it also tends to 0. Thus what we think of as .0000..00001 is actually zero.</p>
<p>
[quote]
If you wish to talk about 0.0000....1, first rigorously define the number 0.0000....1, and then see what you get. ie, It's perfectly reasonable to define 0.0000....1 as the number which is greater than 0 but less than any positive real number; but it won't help you prove that 0.999... is not equal to 1.
[/quote]
i was never trying to prove that .999... was not equal to 1.
I was just saying that the two questions were equivalent to eachother - that is, the answers to both rely on the same logic. it was part of a response to another poster...</p>
<p>"You have to be kidding me ".0000....1" If there is an infinte amount of zeros before the 1, where does the 1 come from. By adding a 1 to the end, there can't be an infinite number of zeros before it. .99999 is equal to 1 because all math scholars say that it is. period"</p>
<p>
[quote]
*0.999...999
is a little below 1, but 0.999999... doesn't fall short of 1 <em>until</em>
you stop expanding it. But you never stop expanding it, so it never
falls short of 1. *
[/quote]
</p>
<p>That is the definitive answer from a previous post. Now please stop arguing people!</p>
<p>atomic fusion, are you insane? of course they cancel out and my arithmetic is correct. i don't know though, maybe you know more than my calculus (PhD in mathematics) teacher or me. Both of which dominate you in math. or maybe you know more than my 7th grade algebra teacher (masters degree). or maybe you know more than the entire mathematics community! or, more likely, you're wrong.</p>
<p>i'll break it down for you:
1+3=4
2+8=10
(1+3)-(2+8)=(4)-(10)
i.e.
A+B=C
D+E=F
A+B-D-E=C-F
i.e.
x=0.99999999999999999...
10x=9.999999999999...
subtract the two:
9x=9
x=1
qed</p>
<p>oh and here. since we know that 0.9... is a number, can you please give me its rational equivalent? unless, of course, you are going to prove that 0.9... is irrational (and then the burden of proof lies with you). so you've got some (futile) work to do! let me know when you get back.</p>
<p>oh god y'all done sprung out the mathematicans....RUN FOR YOUR LIFE!</p>
<p>
If there's one thing I've learned from my very gracious teachers: NEVER, EEEEEVER, ASSUME SOMETHING IS TRUE SIMPLY BECAUSE SOMEONE SAYS IT IS. Question it, question everything...that is the true essence of knowledge and wisdom.</p>
<p>8,000,000,000 peoplewould agree that I am mathematically retarded</p>