It went missing attempting to prove 1.0 = 0.9 recurring. Thank you for finding it again. Smart.
Such a small part of a fraction was lost in the math and you found it again. =D>
Maybe it was never really lost in the first place, just forgotten, and you figured out where it got put.
Let’s face it :
$$ 0.\dot{0}1=0 $$
Is only logical. 
Geez, it looks more like you’ve repeated the same mathematical error.
Silly, repeatedly making the same mistake and expecting a different result is not logical.
And this >0.(\dot9)< + >.(\dot0)1< - >1< = >0<
Sorry missed this.
Can you show me the long division you used? I personally find some difficulty performing long division on an infinite string of recurring digits. Are you certain you didn’t take a short cut somewhere?
Do you think 1/3 + 1/3 + 1/3 = 1, and there is not one recurring digit in the equation.
Then you come back with but a third + a third + a third = 0.9 recurring.
Do me a favor and divide 1 into 3 using >long division<?
See ya back here when you’re done, no short cuts now…
You’re doing the long division improperly. That’s the problem.
(\frac{1}{3} \neq 0.\dot3)
Rather, (\frac{1}{3} = 0.\dot3 + \text{remaining value}) where (\text{remaining value} \neq 0).
You can dismiss the remaining value if and only if it’s equal to (0). But in this case, it is not.
Let’s see:
( \frac{1}{3} = \frac{3}{10} + (\frac{1}{3} - \frac{3}{10}) )
( \frac{1}{3} = \frac{3}{10} + (\frac{10}{30} - \frac{9}{30}) )
( \frac{1}{3} = \frac{3}{10} + \frac{1}{30} )
( \frac{1}{3} = \frac{3}{10} + \frac{3}{100} + (\frac{1}{3} - (\frac{3}{10} + \frac{3}{100})) )
( \frac{1}{3} = \frac{3}{10} + \frac{3}{100} + (\frac{1}{3} - \frac{33}{100}) )
( \frac{1}{3} = \frac{3}{10} + \frac{3}{100} + (\frac{100}{300} - \frac{99}{300}) )
( \frac{1}{3} = \frac{3}{10} + \frac{3}{100} + \frac{1}{300} )
( \frac{1}{3} = \frac{3}{10^1} + \frac{3}{10^2} + \frac{3}{10^3} + \cdots + \frac{3}{10^n} + \frac{1}{3 \times 10^n} )
( \frac{1}{3} = \frac{3}{10^1} + \frac{3}{10^2} + \frac{3}{10^3} + \cdots + \frac{3}{10^{\infty}} + \frac{1}{3 \times 10^{\infty}} )
( \frac{1}{3} = 0.333\dotso + \frac{1}{3 \times 10^{\infty}} )
The remaining value is ( \frac{1}{3 \times 10^{\infty}} ), not (0).
You’ll have to demonstrate that (\frac{1}{3 \times 10^{\infty}} = 0) before you can say that (\frac{1}{3} = 0.333\dotso).
They don’t 
Math is built on logic - ever heard of Zermelo-Fraenkel set theory?
It’s so funny to see another mathematician try to speak a word of sense here 
Your input is some nice fresh air, phyllo.
This new (“\frac1{3\times10^\infty}”) is also pretty funny, as it too must have infinitely more 3s “at the end of endless 0s” in its decimal representation just as all (\frac1{3\times10^n}) do.
So either you can always divide a term by 10 one more time, showing (“\frac1{3\times10^\infty}”) to hardly be some “ending remainder” to endlessness, or the higher order of magnitude 0s really do never get to these fictitous 3s that never ever appear as “already there” - as would be consistent with the (\infty) being erroneously used AGAIN as a value rather an operator. When will this finally sink in? It will take an infinite amount of time no doubt.
All there really is here is an objection to decimal representations, and as phyllo correctly says, as I too said earlier in the thread, either all math is consistent or it’s all broken.
(\frac9{3}=3) with absolutely no problem so either (\frac{10}3=3.\dot3) using exactly the same math, or division itself is broken, making math inconsistent and even (\frac9{3}=3) invalid…
There’s just no consideration of the consequences of trying to make out like there’s an end to endlessness for the sake of some illusory remainder…
Absolutely correct. You literally never reach the end of the endless 0s to get to that infinitely elusive “1”.
Any smallest number will always be inconsistent, anything you need it to be, undefined - just the same as (\infty), simply because either you could always divide it smaller - or there’s an ending limit to its smallness.
This logic in that simple sentence is as exhaustive as it is unequivocally definitive, as well as backing up the consistency of math from its grounding in logical roots, all the way up - for us to be able to have this conversation at all in the first place(!)
If something doesn’t work in some situations, does it mean it doesn’t work in all situations? Of course not.
Saying “If long division doesn’t work in a limited number of cases then it follows that it never works (and all of mathematics is therefore broken)” is being overly dramatic.
But it does work. It’s just that you’re dismissing the remaining value.
The long division never gives us an infinite sequence of zeroes when dividing (1) by (3). Therefore, you can’t act as if it does.
Nah, infinity is not an operator.
There’s no doubt a huge amount of background that non-mathematicians are lacking that could make them think that only choice parts of math is broken…
For example, how division is an extension of subtraction, which is an extension of addition, which is an extension of logical operators - it’s all different forms of the same thing, each relying on the consistency of the previous step, and each being only as consistent as as the next step - and the most basic step is logic. So to try to cherry pick only the parts you like is literally illogical.
And by the way, if anyone is going to say that (\frac{1}{3 \times 10^{\infty}}) is meaningless, note that that’s not a proof that it is (0). You have to prove it’s zero – and not merely by approximation. Not sure how you can do that if you think that infinity is undefined and at the same time an operator (:
It’s not cherry-picking.
By the way, your posts would have more value if you tried to counter existing arguments or provide your own instead of constantly complaining about non-mathematicians.
Here’s another way to put it:
The long division ends when we reach the point at which every subsequent digit is equal to (0). That is when we can discard the remaining value, since that’s the point at which it attains (0). But with the long division of (1 \div 3) this never happens. There is no point at which every subsequent digit is equal to (0). Hence, there is no point at which we can discard the remaining value. We can’t even do it at the point of infinity – the digits that follow the infinity-th digit are not (0)s, they are (3)s.
And yet the arguments have remained unaddressed. Go figure.
What fraction of decimal notation is being ignored?
Where do you find a statement of truth within a infinite string of false results.
When has 1/3 been resolved to a definitive result?
Okay, long division of 1/3.
The pattern can easily be seen to repeat : [attachment=0]Division1over3.JPG[/attachment]
What’s been unaddressed?
No fraction of decimal notation is being ignored.
In every discipline, if you can’t “get there to see it with your own eyes”, logical extrapolation with zero margin for error is the best you can do and in many ways better than seeing it with your own eyes because it shows any other answer to be necessarily impossible no matter how far you looked with your own eyes. How things seem to look at first with your own eyes is the whole problem here, and that’s what maths is for - a surgical tool to unequivocally surmise beyond the superficial. This is exactly what philosophical argument does too - it’s basic Epistemology.
Nothing is considered to be mathematically proven until the logic is laid out exhaustively and completely to cover every possible permutation that could ever exist. Just as calculus proves gradients and areas in this way, its use of limits extrapolate to conclusions with zero margin for error. It’s no coincidence that all attempts here to show that some remainder “looks like it ought to be there” are falling foul to simple logical mistakes/contradictions.
Calculus isn’t “merely an approximation” that “might be a little bit bigger or smaller we dunno”. It’s as precise as anything can get and certainly more precise than how things might appear to a non-mathematician.
Just like with this topic it uses infinity as an operator to arrive at precise results. As I covered a seemingly infinite amount of times already, the undefined is being used exactly for its property of not being defined, without defining it, and never as a value except to communicate that any resulting value would be undefined. For infinity, the best you can do is define the way it is constructed without actually completing the construction - as in the ZFC axiom of infinity at the foundation of mathematics. Pointing out the precise beginning of a definite path isn’t the same as completely defining all of the path.
And yes it is cherry-picking to try to use and talk about math to prove non-mathematician intuitions about how things seem to appear, only to conclude that math must be wrong in some other way than how you’re trying to use it - just to fit your suspicions - when all math is inextricable interlinked. Criticising the parts you don’t like while using the parts you do is practically the definition of cherry-picking (:
And now I will cut to the chase:
1 = 0.9 + .01 = 0.3 - .01 This looks true.
1 = 0.99 + .001 = 0.33 - .001 This too
1 = 0.999 + .0001 = 0.333 - .0001 and this too.
You can get “here from there” by being a very meticulous accountant.
This >1 = (0.\dot9)< is false.
and this > (.\dot0)1 = 0< is also false.
And that is how math is performed with recurring digits.
ILP’s very own mathematical proof that 1 is not equal (0.\dot9)
At any point in the decimal notation you realize that 1/3 will always leave a remainder that is part of “1 whole” too much which is exactly the same part of the “1 whole” the fraction (0.\dot9) is missing which is accounting for all the infinite parts of 1 whole, such that the sum of all “of the parts” is always equal to the whole.
OK Phyllo, I’m done editing (I think) at least for now.
Who the heck knows what you’re trying to say?
Each section evaluates to a different value : 1, 0.91, 0.29
Phyllo, give me a moment to account for what has/is taken/taking place.
I’m “checking” the math …
Check back to that post in a few minutes, I paused before I was finished to make some breakfast. I can’t do it all at once. Well the math takes place all at once, but the checking in combination with preparing some breakfast takes some time.
As an editor, I’m also checking how well the words chosen match up with the idea I’m trying to convey, and attempting to insure I haven’t included any typos.
As an additional thought that took place somewhere between a bite of toast, a bite of egg and a sip of coffee; I got someone else to do the math and added the logic that was missing.
Thanks for all your help.