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

1. 0.5 = 1 / 2 Total = 0.5
2. 0.25 = 1 / 4 Total = 0.75
3. 0.125 = 1 / 8 Total = 0.875
4. 0.0625 = 1 / 16 Total = 0.9375
5. 0.03125 = 1 / 32 Total = 0.96875
6. 0.015625 = 1 / 64 Total = 0.984375
7. 0.0078125 = 1 / 128 Total = 0.9921875
8. 0.00390625 = 1 / 256 Total = 0.99609375
9. 0.001953125 = 1 / 512 Total = 0.998046875
10. 0.0009765625 = 1 / 1024 Total = 0.9990234375
11. 0.00048828125 = 1 / 2048 Total = 0.99951171875
12. 0.000244140625 = 1 / 4096 Total = 0.999755859375
13. 0.0001220703125 = 1 / 8192 Total = 0.9998779296875
14. 0.00006103515625 = 1 / 16384 Total = 0.99993896484375
15. 0.000030517578125 = 1 / 32768 Total = 0.999969482421875
16. 0.0000152587890625 = 1 / 65536 Total = 0.9999847412109375
17. 0.00000762939453125 = 1 / 131072 Total = 0.99999237060546875
18. 0.000003814697265625 = 1 / 262144 Total = 0.999996185302734375
19. 0.0000019073486328125 = 1 / 524288 Total = 0.9999980926513671875
20. 0.00000095367431640625 = 1 / 1048576 Total = 0.99999904632568359375
21. 0.000000476837158203125 = 1 / 2097152 Total = 0.999999523162841796875
22. 0.0000002384185791015625 = 1 / 4194304 Total = 0.9999997615814208984375
23. 0.00000011920928955078125 = 1 / 8388608 Total = 0.99999988079071044921875
24. 0.000000059604644775390625 = 1 / 16777216 Total = 0.999999940395355224609375
25. 0.0000000298023223876953125 = 1 / 33554432 Total = 0.9999999701976776123046875
26. 0.00000001490116119384765625 = 1 / 67108864 Total = 0.99999998509883880615234375
27. 0.000000007450580596923828125 = 1 / 134217728 Total = 0.999999992549419403076171875
28. 0.0000000037252902984619140625 = 1 / 268435456 Total = 0.9999999962747097015380859375
29. 0.0000000018626451492309570312 = 1 / 536870912 Total = 0.9999999981373548507690429687
30. 0.0000000009313225746154785156 = 1 / 1073741824 Total = 0.9999999990686774253845214843
31. 0.0000000004656612873077392578 = 1 / 2147483648 Total = 0.9999999995343387126922607421
32. 0.0000000002328306436538696289 = 1 / 4294967296 Total = 0.9999999997671693563461303710
33. 0.0000000001164153218269348145 = 1 / 8589934592 Total = 0.9999999998835846781730651855
34. 0.0000000000582076609134674072 = 1 / 17179869184 Total = 0.9999999999417923390865325927
35. 0.0000000000291038304567337036 = 1 / 34359738368 Total = 0.9999999999708961695432662963
36. 0.0000000000145519152283668518 = 1 / 68719476736 Total = 0.9999999999854480847716331481
37. 0.0000000000072759576141834259 = 1 / 137438953472 Total = 0.9999999999927240423858165740
38. 0.000000000003637978807091713 = 1 / 274877906944 Total = 0.9999999999963620211929082870
39. 0.0000000000018189894035458565 = 1 / 549755813888 Total = 0.9999999999981810105964541435
40. 0.0000000000009094947017729282 = 1 / 1099511627776 Total = 0.9999999999990905052982270717

Good post. Good observation.

looks like 1/2 + 1/4 + 1/8 + ... is going to equal 0.999... ?

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.
 
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  • Do you accept that 0.999... x 10 = 9.999... ?

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

    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!
     
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  • 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?
    So you don't think 0.999... x 10 = 9.999... ?
    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.
    "Proof"?
    I merely asked a question.
    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.

    But thanks for your contribution.
     
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  • 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.

    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!
     
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  • Do you accept that 10=9.999.....?
    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?
     
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  • Because the series is absolutely convergent.

    (silliness omitted)

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

    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.
     
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  • Hmmm...

    hansda:

    Let us stop at k = infinity and try to find the infinity-th term of this infinite geometric series: 0.9 + 0.09 + 0.009 + 0.0009 + 0.00009 + 0.000009 + ... upto infinite terms.

    Can you explain about the infinity-th term of this infinite series? What will be the decimal format of this infinite-th term? or do you think that the infinity-th term of this infinite series does not exist?

    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.


    Motor Daddy:

    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?

    Do you accept that 10=9.999.....?

    Yes. Of course it is.

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

    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?

    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?

    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!

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

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

    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.

    1=100%
    .999... is not even complete.

    Yes it is.


    someguy1:

    You're more clueless than all the cranks here.

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

    Have you found a flaw in post #915?

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

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

    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.

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

    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.

    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?
     
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  • 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?
     
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  • 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.
     
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  • 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
     
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  • 0.999 + 0.999 = 1.998
    $$ \begin{eqnarray} \left( \sum_{k=1}^n \frac{9}{10^k} \right) + \left( \sum_{k=1}^n \frac{9}{10^k} \right) & = & \sum_{k=1}^n \left( \frac{9}{10^k} + \frac{9}{10^k} \right) \\ & = & \sum_{k=1}^n \frac{18}{10^k} \\ & = & \sum_{k=1}^n \left( \frac{10}{10^k} + \frac{8}{10^k} \right) \\ & = & \left( \sum_{k=1}^n \frac{10}{10^k} \right) + \left( \sum_{k=1}^n \frac{8}{10^k} \right) \\ & = & \left( \sum_{k=0}^{n-1} \frac{1}{10^k} \right) + \left( \sum_{k=1}^{n-1} \frac{8}{10^k} \right) + \frac{8}{10^n} \\ & = & \frac{1}{10^0} + \left( \sum_{k=1}^{n-1} \frac{1}{10^k} \right) + \left( \sum_{k=1}^{n-1} \frac{8}{10^k} \right) + \frac{8}{10^n} \\ & = & 1 + \left( \sum_{k=1}^{n-1} \frac{9}{10^k} \right) + \frac{8}{10^n} \end{eqnarray}$$
    So for n=3
    0.999 + 0.999 = 1.998
    0.999... + 0.999... = 1.999...8
    $$ \begin{eqnarray} \left( \sum_{k\geq 1} \frac{9}{10^k} \right) + \left( \sum_{k\geq 1} \frac{9}{10^k} \right) & = & \sum_{k\geq 1} \left( \frac{9}{10^k} + \frac{9}{10^k} \right) \\ & = & \sum_{k\geq 1} \frac{18}{10^k} \\ & = & \sum_{k\geq 1} \left( \frac{10}{10^k} + \frac{8}{10^k} \right) \\ & = & \left( \sum_{k\geq 1} \frac{10}{10^k} \right) + \left( \sum_{k\geq 1} \frac{8}{10^k} \right) \\ & = & \left( \sum_{k\geq 0} \frac{1}{10^k} \right) + \left( \sum_{k\geq 1} \frac{8}{10^k} \right) \\ & = & \frac{1}{10^0} + \left( \sum_{k\geq 1} \frac{1}{10^k} \right) + \left( \sum_{k\geq 1} \frac{8}{10^k} \right) \\ & = & 1 + \left( \sum_{k\geq 1} \frac{9}{10^k} \right) \end{eqnarray} $$

    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.
     
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  • 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 ...
     
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  • Motor Daddy:

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

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


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

    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.
     
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  • 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. Rememebr 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 ...
    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.
     
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