# 1=0.999... infinities and box of chocolates..Phliosophy of Math...

Discussion in 'General Philosophy' started by Quantum Quack, Nov 2, 2013.

1. ### hansdaValued Senior Member

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2,424
Good post. Good observation.

Also observe that (from sl no 4 to 40) as the string of digit 9's is increasing in the Total, the string of 0's is also increasing in the last number to be added with this infinite geometric series.

3. ### BaldeeeValued Senior Member

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2,226
Do you accept that 0.999... x 10 = 9.999... ?
Or are you going to insist on there being a zero at the end?

5. ### someguy1Registered Senior Member

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727
Not in the least. What makes you think you can multiply a convergent infinite series term-by-term by a constant, such that the new series converges to the constant times the original series?

If you check out the axioms for real numbers, you'll see that you have only the 2-element distributive law a(b+c) = ab + ac. From that you can use induction to generalize the distributive law to any finite number of terms inside the parens. But there is absolutely no fundamental rule that allows you to put infinitely many terms inside the parens. You have to prove you can do it by carefully defining the real numbers, the limit of a sequence, and the limit of a series. Then you have to prove the theorem on term-by-term multiplication.

Without doing all this, you can't use this fact. And the proof of term-by-term multiplication is already more mathematically sophisticated than the mere fact that .999... = 1. So this is a bogus proof. It's really just a heuristic argument for high school students, not an actual proof.

It's one thing to be a crank. But it's another thing to not be a crank yet not realize you are posting a completely unjustified "proof." I don't care about the cranks; but people who should know better need to understand that term-by-term multiplication of a convergent infinite series by a constant is a theorem that must be proven.

Same remarks for the equally bogus 3 x .333... = .999... proof. Anyone who can write down a formal proof to legitimize that fact would have no doubt that .999... = 1. So again, this is just a cheap story for beginners, not a legitimate mathematical proof.

7. ### rpennerFully WiredValued Senior Member

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4,833
Because the series is absolutely convergent.

If $a_n = \frac{9}{10^{k+1}}$ and $b_n = 10 \delta_{0n}$ then the Cauchy product of the series is
$\left( \sum_{k\geq 0} a_k \right) \times \left( \sum_{k\geq 0} b_k \right) = \sum_{k\geq 0} c_k = \sum_{k\geq 0} \left( \sum_{\ell=0}^{k} a_{\ell} b_{k-\ell} \right) = \sum_{k\geq 0} \left( \sum_{\ell=0}^{k} \frac{9}{10^{\ell+1}} \times 10 \delta_{k \ell} \right) = \sum_{k\geq 0} \frac{9}{10^{k}} = 9 + \sum_{k\geq 1} \frac{9}{10^{k}} = 9 + \sum_{k\geq 0} a_n$ ​
which obvious has the same convergent properties as the original.

As for
$\left( \sum_{k\geq 0} b_k \right) \times \left( \sum_{k\geq 0} a_k \right) = \sum_{k\geq 0} \left( \sum_{\ell=0}^{k} b_{\ell} a_{k-\ell} \right) = \sum_{k\geq 0} \left( \sum_{\ell=0}^{k} 10 \delta_{0 \ell} \frac{9}{10^{k-\ell+1}} \right) = \sum_{k\geq 0} \left( 10 \frac{9}{10^{k+1}} \right) = \sum_{k\geq 0} \frac{9}{10^{k}} = 9 + \sum_{k\geq 1} \frac{9}{10^{k}} = 9 + \sum_{k\geq 0} a_n$ ​
which is the same result.

Identical results happen if $b_n = \left{ \begin{array}{lcl} 1 & \quad \quad \quad & \textrm{if} \; n \lt 10 \\ 0 & & \textrm{otherwise} \end{array} \right.$

Finally,
$\left( \sum_{k\geq 0} a_k \right) \times \left( \sum_{k\geq 0} a_k \right) = \sum_{k\geq 0} \left( \sum_{\ell=0}^{k} a_{\ell} a_{k-\ell} \right) = \sum_{k\geq 0} \left( \sum_{\ell=0}^{k} \frac{9}{10^{\ell+1}} \times \frac{9}{10^{k-\ell+1}} \right) = \sum_{k\geq 0} \frac{81 (k+1)}{10^{2+k}} = \lim_{n\to\infty} \frac{10^{2+n} - 9n - 19}{10^{2+n}} = 1$ ​

It's not rocket science -- it's analysis.

Last edited: Feb 25, 2014
8. ### Motor DaddyValued Senior Member

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5,425
Do you accept that 10=9.999.....?

It's not rocket science. See the 1 in the tens position? See how the 9.999...doesn't have a 1 in the tens position? See how it has a 9 in the ones position?

Clueless, and then defend it steadfast. Unbelievable!

9. ### BaldeeeValued Senior Member

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2,226
So you don't think 0.999... x 10 = 9.999... ?
"Proof"?
There was no "proof" posed in the question at all.
Just a question to see what BdS thought of an infinite series multiplied by 10, and whether they would blindly stick a 0 on the end (not that there would be an end).
Anything else is unfortunately you reading far too much into the issue.

The case has already been proven (rpenner does such a good job at those) that it's no longer a matter of needing to prove anything, but in seeing why people still don't accept it.
And for that one needs to understand what their views are of other aspects.

10. ### Motor DaddyValued Senior Member

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5,425
So because stupid people count sheep as 1,2,3... and the really smart math people count sheep .999... ,1.999..., 2.999..., because that's how it's done? You don't even believe your own BS!

11. ### BaldeeeValued Senior Member

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2,226
That wasn't the question.
The question was merely what one considered 10 x 0.999... to be.
Whether they thought it was 9.999... or whether they would think it would be 9.999...0

It's not a difficult question, as it didn't even ask for the factual answer, just whether they accepted that 0.999... x 10 = 9.999... or not.

Do you think it does, or not?

12. ### Motor DaddyValued Senior Member

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5,425
That was my question to you. Seriously, do you really believe the BS that you're spreading?

13. ### someguy1Registered Senior Member

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727
You're more clueless than all the cranks here. Re-read what I wrote. I said:

Anyone who can write down a formal proof to legitimize that fact would have no doubt that .999... = 1.

You totally misconstrued my point.

But what's really funny is that you got the math wrong!!!! Yes for all the sexy LaTeX you made a crucial mistake in your first line. Term-by-term multiplication by a constant is valid for any convergent series, not just an absolutely convergent one. And the proof is far far simpler than what you made it out to be. To be fair you proved the more general fact that you can multiply two convergent series term-by-term. That's nice, but it's far more than what's needed to show that 10 x .999... = 9.999... Your response is more designed more to impress than to enlighten. I'm not impressed.

But you can LaTeX with the best of 'em, I'll grant you that. Very inspiring.

14. ### BaldeeeValued Senior Member

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2,226
It's not a matter of belief.
10 really does = 9.999...
The maths shows it to be.

15. ### Motor DaddyValued Senior Member

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5,425
No, your BS math shows it to be BS, and BS=BS, so your claim is good, BS does in fact equal BS!

1=100%
.999... is not even complete. The only thing that it says is that, "I am trying to build a 100% piece by cutting in half a 100% piece, continuously, and adding those halves together, forever." Seriously, that's what it says. And you are claiming that if I build a brick wall and complete it, that your incomplete (never to be finished) wall is the same as my complete wall.

That's how FOS you are.

16. ### James RJust this guy, you know?Staff Member

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39,421
Hmmm...

hansda:

You can't find the infinity-th term, because there isn't one. An infinite number of terms means the terms never end. There's no last term.

Your posts are starting to look an awful lot like trolling.

Have you found a flaw of some kind in post #915, above?

Do you, for that matter, understand post #915? Or is this too complicated for you?

Yes. Of course it is.

9.999... = 9 + 0.999... = 9 + 1 = 10.

That's just a difference in notation. The same number is represented two different ways.

Did you see my post above on numbers in bases other than 10? Did you understand that one? Any comments?

Sorry, you'll have to do better than that.

Please point out the flaw in post #915.

If you have nothing to offer other than empty claims that something is BS, then you'd better stop commenting now.

Yes it is.

someguy1:

That's a pretty big call you're making.

Have you found a flaw in post #915?

Ok. Please write one down for the benefit of Motor Daddy, hansda and the others. I'd like to see your simple proof, too.

That's not a crucial mistake. An absolutely convergent series is convergent. It's only the converse that is not necessarily true.

So, how about you show us your easy way to show that 10 x 0.999... = 9.999... ?

Seeing as you know the easy method, why not show us?

17. ### James RJust this guy, you know?Staff Member

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39,421
Let's consider a different example for a minute.

What is 1/7?

Plug it into your calculator and you'll get an answer like 0.1428571429.

That's rounded, though. The actual answer is:

0.142857 142857 142857 ...

where the pattern repeats to infinity.

I am wondering whether Motor Daddy, hansda or others think that what I have said here is actually wrong. That is, do you think that pattern stops at some point? If so, what is the last digit in the decimal expansion of 1/7?

Once you've answered that, I want you to multiply the decimal number by 7 and tell me what you get as a decimal.

In particular, what is the last digit in the result of 7 x 0.142857142857...etc.?

See, because the problem I have is that if I say the last digit is, say, 2, then when I multiply the whole thing by 7 then the last digit should be a 4. And yet, the answer should be 7.000000....

So, Motor Daddy, hansda and others, what's going wrong in this case?

18. ### James RJust this guy, you know?Staff Member

Messages:
39,421
And if that one's too hard, consider this:

What is 1/9, expressed as a decimal?

I say it's 0.1111...

If you disagree, then what do you say it is?

And what happens when I do this:

0.111... x 9 = ?

It seems we can agree that 0.111... x 9 = 0.999..., but if 0.999... doesn't equal 1, then something's gone horribly wrong somewhere.

So, explain it to me.

19. ### BdSRegistered Senior Member

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512
0.999 + 0.999 = 1.998
1.998 + 0.999 = 2.997
2.997 + 0.999 = 3.996

0.999... + 0.999... = 1.999...8
1.999...8 + 0.999 = 2.999...7
2.999...7 + 0.999 = 3.999...6

It's not right to put the number after the infinite dots, but its owed to the sum.

-----------------------------------------

100 = 100%
99.999... = 99.999...%

-----------------------------------------

If 0.999... = 1 then
0.888... = ?
0.777... = ?
0.666....= ?
0.555... = ?
etc???

fill in the ?'s

20. ### rpennerFully WiredValued Senior Member

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4,833
$\begin{eqnarray} \left( \sum_{k=1}^n \frac{9}{10^k} \right) + \left( \sum_{k=1}^n \frac{9}{10^k} \right) & = & \sum_{k=1}^n \left( \frac{9}{10^k} + \frac{9}{10^k} \right) \\ & = & \sum_{k=1}^n \frac{18}{10^k} \\ & = & \sum_{k=1}^n \left( \frac{10}{10^k} + \frac{8}{10^k} \right) \\ & = & \left( \sum_{k=1}^n \frac{10}{10^k} \right) + \left( \sum_{k=1}^n \frac{8}{10^k} \right) \\ & = & \left( \sum_{k=0}^{n-1} \frac{1}{10^k} \right) + \left( \sum_{k=1}^{n-1} \frac{8}{10^k} \right) + \frac{8}{10^n} \\ & = & \frac{1}{10^0} + \left( \sum_{k=1}^{n-1} \frac{1}{10^k} \right) + \left( \sum_{k=1}^{n-1} \frac{8}{10^k} \right) + \frac{8}{10^n} \\ & = & 1 + \left( \sum_{k=1}^{n-1} \frac{9}{10^k} \right) + \frac{8}{10^n} \end{eqnarray}$
So for n=3
0.999 + 0.999 = 1.998
$\begin{eqnarray} \left( \sum_{k\geq 1} \frac{9}{10^k} \right) + \left( \sum_{k\geq 1} \frac{9}{10^k} \right) & = & \sum_{k\geq 1} \left( \frac{9}{10^k} + \frac{9}{10^k} \right) \\ & = & \sum_{k\geq 1} \frac{18}{10^k} \\ & = & \sum_{k\geq 1} \left( \frac{10}{10^k} + \frac{8}{10^k} \right) \\ & = & \left( \sum_{k\geq 1} \frac{10}{10^k} \right) + \left( \sum_{k\geq 1} \frac{8}{10^k} \right) \\ & = & \left( \sum_{k\geq 0} \frac{1}{10^k} \right) + \left( \sum_{k\geq 1} \frac{8}{10^k} \right) \\ & = & \frac{1}{10^0} + \left( \sum_{k\geq 1} \frac{1}{10^k} \right) + \left( \sum_{k\geq 1} \frac{8}{10^k} \right) \\ & = & 1 + \left( \sum_{k\geq 1} \frac{9}{10^k} \right) \end{eqnarray}$

So for never-ending repeating decimals, 0.999... + 0.999... = 1.999...

Never-ending sums have qualitative differences from sums with just a finite number of terms.

21. ### Motor DaddyValued Senior Member

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5,425
James, Repeating the same mistake in different examples doesn't help your case. Anytime you see ... after a number you know the math broke down and it doesn't divide evenly. Remember when I talked about how it ALWAYS needs to total 100%? If the total is not 100% then the problem brings itself to life as ...

22. ### Motor DaddyValued Senior Member

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5,425
I don't troll, James. You know that. (rolls eyes)

I can't answer that question, as I can't make heads or tails of the syntax. Understanding to me doesn't mean that I can read it, it means that I have a full understanding of the concept involved. I don't place much importance on the syntax. It appears to me that some people care more about the syntax than the actual message.

23. ### BaldeeeValued Senior Member

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2,226
Right, so divide a pizza into 3 equal pieces and then join it up again... That doesn't equal a whole pizza?
You think that the act of dividing somehow removes an infinitesimal from the whole?
If so then why does this only work for certain divisions, then?

You'd surely accept that if you divide a pizza in half and rejoin it you'd get back to the whole, right?
1/2 + 1/2 = 1 and all that.

But because you can write it 0.5 + 0.5 = 1 it somehow doesn't lose the infinitesimal that you think 1/3 + 1/3 + 1/3 does?
0.333... + 0.333... + 0.333... = 0.999... = 1 etc.

Where do you think the infinitesimal goes?

I find it staggering that people can disagree with this notion.