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Beating an old horse .(9) question

  1. Mar 11, 2010 #1
    I know, and conform to the ideal that .9 infinitely repeating does in fact equal 1 but I have a thought that keeps bothering me.
    How can.9999999999999999999999999999999999999999999999999999999(repeating forever) multiplied by 2, equal 2?
     
  2. jcsd
  3. Mar 11, 2010 #2
    What do you think it should be?
     
  4. Mar 11, 2010 #3
    Is that the right question? Because, if .9999... equals 1, then .9999... multiplied by 2 equals 1 multiplied by 2.

    Maybe you're trying to ask how is multiplication of periodic decimals defined; or if it is defined at all, other than by the procedure in the first paragraph. (Not that I know the answer, but I'm interested in the question.)

    Edit: sorry, I was answering the original post, not the second; the second poster was (again) faster than me.
     
  5. Mar 11, 2010 #4
    Well I guess look at it this way. .(9)^2 =1? while 9^2 =81... That just confuses me.
     
  6. Mar 11, 2010 #5
    It's an optical illusion. :) You may want to try a few more nines, like 99^2, 999^2 or 99999999^2.
     
  7. Mar 11, 2010 #6
    Yea, they always execute with the length of the original number but with 8 at the end then the length of the number, with a 1 at the end terminating it.
    ie. 9*9=81
    99*99=9801
    999*999=998001
    9999*9999=99980001
    99999*99999=9999800001
     
  8. Mar 11, 2010 #7
    And this decoration looks like an artifact caused by the number of nines being finite.
     
  9. Mar 12, 2010 #8

    Hurkyl

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    0.99 * 0.99 is less than 0.99(9) * 0.99(9), right?

    But 0.99(9) * 0.99(9) cannot be greater than 0.99(9), right?

    So, can you tell me what the first digit after the decimal point of 0.99(9) * 0.99(9) is?



    0.999 * 0.999 is less than 0.999(9) * 0.999(9), right?

    But 0.999(9) * 0.999(9) cannot be greater than 0.999(9), right?

    So, can you tell me what the second digit after the decimal point of of 0.999(9) * 0.999(9) is?
     
  10. Mar 12, 2010 #9
    This is a (very) small ambiguity in positional systems, but first, to answer your question:

    [tex]0.9\left(9\right) = \sum^{+\infty}_{n=1}\frac{9}{10^n}=9\frac{0.1}{1-0.1}=1[/tex]

    Therefore:

    [tex]2\times0.9\left(9\right) = 2\sum^{+\infty}_{n=1}\frac{9}{10^n}=18\frac{0.1}{1-0.1}=2[/tex]

    But this is a general feature of any positional numerical system: for example, [itex]0.1\left(1\right) = 1[/itex] in base 2. In general, given a base b > 1, its highest digit is d = b - 1 and you have:

    [tex]0.d\left(d\right) = 1[/tex]

    Note that 1 has the same representation relative to every basis > 1.
     
  11. Mar 13, 2010 #10
    Allow me to demonstrate on how the multiplication can be done.

    Now, traditionally when multiplying the following (the *s are used for space-filling, please adjust fonts or copy-paste into a simple text editor to line up characters) (I'm a newbie and can't figure out the LaTeX codes that would make this look even semi-coherent):

    *0.999
    *x 2
    *------

    we were taught to start from the right and work our way left:

    *0.999
    *x 2
    *------
    *0.018
    *0.18
    +1.8
    --------
    *1.998

    However, since "2 x 0.999" is really just "2 x (0.009 + 0.09 + 0.9)", we can use the commutative property of addition/multiplication to change the order to "2 x (0.9 + 0.09 + 0.009)" and still get the same answer.

    Therefore we can calculate from left to right, instead:

    *0.999
    *x 2
    *------
    *1.8
    *0.18
    +0.018
    --------
    *1.998

    Using this method, we can multiply 0.999... by 2 and get the following:

    *0.999999999999...
    *x 2
    *-----------------
    *1.8
    *0.18
    *0.018
    *0.0018
    *0.00018
    *0.000018
    *0.0000018
    *0.00000018
    ..
    ...
    .....
    +
    ------------------
    *1.999999999999999....

    2 x 0.999... = 1.999... = 1 + 0.999... = 1 + 1 = 2


    Now, onto 0.999... x 0.999... (in my next comment)!
     
  12. Mar 13, 2010 #11
    Here is a (long and convoluted) way to multiply 0.999... by 0.999...:

    **0.999...
    x 0.999...
    -------------

    I will break this into separate components that will all be added back together at the end. Since 0.999... can be expressed as the geometric series "0.9 + 0.09 + 0.009 + 0.0009 +...", let's multiply each of those components by 0.999...:

    **0.999...
    x 0.900...
    -------------
    *0.81
    *0.081
    *0.0081
    *0.00081
    *0.000081
    .
    ..
    ...
    +
    ------------------
    *0.8999999999....

    and

    **0.9999...
    x 0.0900...
    -------------
    *0.081
    *0.0081
    *0.00081
    *0.000081
    *0.0000081
    .
    ..
    ...
    +
    ------------------
    *0.08999999999....

    and

    **0.99999...
    x 0.00900...
    -------------
    *0.0081
    *0.00081
    *0.000081
    *0.0000081
    *0.00000081
    .
    ..
    ...
    +
    ------------------
    *0.008999999999....

    and so on and so forth...


    Yeah, I hope you recognize how the pattern goes. Anyway, let's try adding the multiplied components up:

    *0.89999999...
    *0.08999999...
    *0.00899999...
    *0.00089999...
    *0.00008999...
    .
    ..
    ...
    +
    --------------------

    Eeek! Guess I'll have to break even this one down a bit. Remember, the above is the result of "0.999... x (0.9 + 0.09 + 0.009 +...)".

    Let's add one row at a time, shall we? Here are the first two rows added together:

    *0.89999999...
    +0.08999999...
    ------------------

    Let's see...since we know that there are only 9s from the 1000th place onward to the right for both rows, we can reaarange this from "(0.89 + 0.00999...) + (0.08 + 0.00999...)" to "(0.89 + 0.08) + (0.00999... + 0.00999...)".

    So, broken down:

    *0.89
    +0.08
    --------
    *0.97

    plus

    *0.00999...
    +0.00999...
    ------------
    *0.01999... = 0.00999... x 2

    (Remember the "0.999... x 2" thing? Same stuff.)

    And then we add the two results together:

    *0.97000000...
    +0.01999999...
    -------------
    *0.98999999...

    Phew! First two rows done. Now to add the third row (look back if you've lost track of what we were originally doing):

    *0.98999999...
    +0.00899999...
    ---------------

    Okay, repeat of what we did in the previous steps, except this time it's from the 10000th place:

    *0.989
    +0.008
    ---------
    *0.997

    and

    *0.000999...
    +0.000999...
    ------------
    *0.001999... = 0.000999... x 2

    Which added together would be:

    *0.99700000...
    +0.00199999...
    -------------
    *0.99899999...


    Adding fourth row:

    *0.99899999...
    +0.00089999...
    ---------------

    You should know the procedure by now:

    *0.9989
    +0.0008
    ---------
    *0.9997

    and

    *0.00009999...
    +0.00009999...
    ------------
    *0.00019999... = 0.0000999... x 2

    Which added together would be:

    *0.99970000...
    +0.00919999...
    -------------
    *0.99989999...


    Fifth row:

    *0.99989999...
    +0.00008999...
    ---------------
    *0.99998999...

    Okay, so I skipped typing out the work here, but this comment is getting way too long. Just to refresh your (and my) memory, this is what the 0.999... x 0.999... equation looked like when converted it into an addition equation:

    *0.89999999...
    *0.08999999...
    *0.00899999...
    *0.00089999...
    *0.00008999...
    .
    ..
    ...
    +0.00000000...
    --------------------

    I then started adding up the rows one by one (albeit I started by adding the first two rows). The results of each addition are as follows:

    1st row: 0.89999999...
    2nd row added: 0.98999999...
    3rd row added: 0.99899999...
    4th row added: 0.99989999...
    5th row added: 0.99998999...
    6th row added: 0.99999899...
    etc.

    Well, the pattern should be obvious here, with that "8" going further to the right with the addition of each row. And since this pattern will go on without end, the result would be:

    *0.89999999...
    *0.08999999...
    *0.00899999...
    *0.00089999...
    *0.00008999...
    .
    ..
    ...
    +
    --------------------
    *0.99999999...

    And thus I show that 0.999... x 0.999... = 0.999...! Go me!
     
  13. Mar 13, 2010 #12

    Borek

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    The more I look, the more nines I see. I wonder if that means you are all my creation.
     
  14. Mar 13, 2010 #13
    Dodo's right: if you believe that .(9)=1, then 2*.(9)=2*1=2, end of story. That's the only answer you need...anything else just boils down to more arguments that .(9)=1, which you claim to believe already.
     
  15. Mar 14, 2010 #14
    n*2 -n = n
    .9(9) * 2 -.9(9) = 1.9(9) -.9(9) = 1 = .9(9)
     
  16. Mar 14, 2010 #15

    HallsofIvy

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    By the definition of our decimal number system, .99999... means the limit of .9, .99, .999, .9999, ....

    Multiply each of those by 2:
    1.8, 1.98, 1.998, 1.9998, ...

    The limit of that is 2.
     
  17. Jun 19, 2010 #16
    CjStaal wrote "I know, and conform to the ideal that .9 infinitely repeating does in fact equal 1 but I have a thought that keeps bothering me. How can.999...(repeating forever) multiplied by 2, equal 2?"

    It can't, and doesn't. Simple. Why do you conform to the ideal, since it is clearly false? You've noticed the obvious. Those values aren't equal. It is illogical to pretend they are equivalent. Calculator programmers know the answer is wrong even though it's close to the right one, and they therefore program calculators to return a 1, which is the right answer, instead of the .999... that is actually calculated.

    Decimal fractions are good for giving us fast close estimates in most cases, but this system is incapable of accurately representing most fractional values. Specifically, any values that are not evenly divisible by ten or its factors of two and five, are to a greater or lesser extent inaccurately represented in decimal fraction format. Some inaccurately represented values terminate, but are in the wrong "place value" positions; the others are the endless repeating or non-repeating representations for values. These "infinite" representations of values are clear indications of a falsely represented value. The value is exact and finite. The attempted representation for it is not, and can never be accurately arrived at through this calculation technique.

    The more use of decimal fractions, the greater the likelihood of increased error and deviation in representing any values which include fractions. Those who use them a lot, like engineers, and some scientists, etc. know this, and adjust their decimal fraction calculation to be closer to accurate. There are established techniques for doing this. This still doesn't necessarily fully correct the value representation, rendering it accurate, however. The calculation to correct the error inherent in the decimal fraction calculation system would be more complex, and more work than just doing the calculations accurately in the first place, which can be done using "vulgar" old fashioned fractions.

    The decimal fraction system is a buggy computing system, but useful enough for quickly estimating values that are pretty close to the value sought, if that's all that's required. If accuracy is important, old fashioned "vulgar" fractions will do the job with perfect accuracy, but are far more difficult and time-consuming to use. Also, the accurate statement of some important values remains unknown, like pi, for instance. We only have the inaccurate estimate that includes a decimal fraction. This presents difficulties and an impediment, but hopefully some diligent mathematicians will apply themselves to the problem.

    Sorry if I'm tardy for the party, but hope this helps.
     
  18. Jun 19, 2010 #17

    Borek

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    Yes it does. 0.(9) = 1, It can be proven in zillion ways. Search the forums for proofs as they were posted on many occasions. The key is in word "forever".
     
  19. Jun 19, 2010 #18

    HallsofIvy

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    No, it doesn't help at all. Pretty much everything you have said is wrong. "Diligent mathematicians" have applied themselves to the problem (hundreds of year ago) and, in particular, defined exactly what is meant by an "infinite series" and how to calculate them.

    In particular, "0.999999.."= [itex].9+ .9(.1)+ .9(.01)+...= .9(1+ .1+ .1^2+ .1^3+ ...)[/itex] is a "geometric series" which has sum
    [tex]\frac{.9}{1- .1}= \frac{.9}{.9}= 1[/tex]
    exactly.

    I suggest you learn some mathematics before you make statements like "Also, the accurate statement of some important values remains unknown, like pi, for instance." We know exactly what pi is equal to, thank you. The fact that it cannot be written in decimal form with a finite number of decimal places is irrelevant. The decimal numeration system is only one way of writing numbers- and not a particulary important one.
     
  20. Jun 19, 2010 #19
    Multiplication by two is a continuous mapping. To get someting less trivial (or ambiguous), you need to consider a mapping that is not continuous. E.g., define the function

    f(x) = x for |x|<1,

    f(1) = 0

    Then what is f(0.999999999999999999999999999....) ?


    This of course depends on whether you should take the limit first and then evaluate the function or the other way around. My opinion is that it is not natural to take the limit inside the argument of the function first.
     
  21. Jun 19, 2010 #20

    Hurkyl

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    Why not? That's what you wrote. If you meant something else, you should have written something else! :wink:

    I've seen this notation:
    f(1-)​
    when someone wants the value of f(x) as x approaches 1 from the left, rather than the value of f at 1. Such ideas can even be built into a number system (although it's somewhat subtle and I don't remember how exactly it works). Of course, you could replace use hyperreals or operator algebras to get the effect too.
     
  22. Jun 21, 2010 #21
    Borek wrote: "Yes it does. 0.(9) = 1, It can be proven in zillion ways. Search the forums for proofs as they were posted on many occasions. The key is in word 'forever.'"

    Thanks for your response. Perhaps you could recommend how to go about the search you suggest. There are many thousands of posts of varying content and titles.

    Concerning the key "forever," in the current case it would seem to be equivalent to "never," wouldn't it? That being the case I don't see how any proofs could be logically acceptable, actually. They would only be acceptable if we pretend that something that never happens and can't ever and doesn't ever exist, is equivalent to something that either occurs, or has the potential to occur and be actual.

    There's "forever" a little bit missing from every iteration of the calculated value, and never a point when the value actually recoups the missing bit, rendering the value of one. An infinity of infinities can't correct that. There's never a time or an occurrence of the value, in all infinity, when it possesses that little missing bit. Do fairies or angels add that little missing bit somewhere in eternity, since infinity forever remains lacking it? Where in any of the proofs is that missing bit added, exactly, rendering the value valid as an exact equivalency for one?

    It seems to me that the point of numbers is they quantify. That's why the number system starts with an absolute quantum, that being represented by the number one, and the value of each and every other number is logically and consistently derived from, and therefore divisible by that value.

    If quantity has no consistent meaning, and could be some particular thing, except when we feel it might be fun for it to be something else, then what's the point of quantification through use of numbers? If 0.(9) can equal 1, that system has no valid quantum.

    Having no valid and absolute quantum for a constructive basis invalidates the entire number system, as a system of quantification, which is its purpose. None of the values in such a system can be considered valid. Or do you opine otherwise? Is violation of the quantum, and a variable quantum (and therefore random and variable numeration) perfectly okay, for a number system of consistently quantized values?
     
  23. Jun 21, 2010 #22

    Borek

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    Built in search is not as effective as Google, but this search yields:

    https://www.physicsforums.com/showthread.php?t=66671

    https://www.physicsforums.com/showthread.php?t=124382

    https://www.physicsforums.com/showthread.php?t=47827

    https://www.physicsforums.com/showthread.php?t=2849

    https://www.physicsforums.com/showthread.php?t=215017

    "Forever" works, stating that it doesn't you are repeating ancient Greek arguments (Zeno paradox), long shown to be incorrect. And I have a feeling you are mistaking number representation and its value - these are not the same. Pi is precisely defined, even if we can't write its decimal representation. Same about 1/3. 3*1/3 = 1, even if the decimal representation (nothing else but 0.(9)) seems to say different.
     
  24. Sep 14, 2010 #23
    I think the easiest way to prove that 0.99999.... = 1 is:

    Let X = 0.999999....
    Then 10X = 9.99999..... (moving the decimal point one place to the right)
    So 10X - X = 9.999999.... - X (subtracting the same number from both sides of the equation)
    We then get 9X = 9.99999.... - 0.99999....
    but 9.99999.... = 9 + 0.99999....
    so 9X = 9
    and X = 1

    Also, you can show that
    1/9 = 0.11111....
    2/9 = 0.22222....
    3/9 = 0.33333....
    4/9 = 0.44444....
    5/9 = 0.55555....
    6/9 = 0.66666....
    7/9 = 0.77777....
    8/9 = 0.88888....
    9/9 = 0.99999....

    Further,
    1/3 * 3 = 1
    but 1/3 = 0.33333....
    so 0.33333.... * 3 = 0.99999.... = 1
     
  25. Sep 24, 2010 #24
    Dear zgozvrm:

    Thanks for your response. My post was removed, so I stopped visiting the site since my honest logical discussion of this fairly simple subject matter was censored, and I am surprised to receive your response. It's a sad day when dogma is enforced as a substitute for logic and discussion of it in fields of math or science, especially at a venue supposedly devoted to teaching and learning in such fields. A good teacher should consider, learn, and be responsive to students and their communications, not merely require recitations of what they insist others thoughtlessly recite. Otherwise, how are teachers and experts any better than mere access to a textbook? No wonder the United States is falling below nearly every other developed country in these fields.

    I appreciate your response but find your proof lacking. It's no proof because one of the values is never actually stated. I'll demonstrate.

    Let's restate and more clearly identify the value to scrutinize the proof more closely. According to your proof I could use any decimal fraction value really, that is less than one, to represent the value one, and it should be justified since we merely never address the value of the missing portion of the decimal fraction that remains unstated. Let's review. I'll restate your proof differently for comparison:

    First, whether we write 0.99999... or 0.9... it can be considered to represent the same assumed untruncated digitally extending value supposedly. To simplify let's use the latter. Now, let's restate your proof, which I find is not a proof but remains an unsolved problem. We still need to solve for the value Y below, which you have neither stated nor solved for.

    Let X = 0.9 + ( an as-yet unidentified value "..." that we'll call Y)
    Then 10X = 9 + 10Y
    So 10X - X = ( 9 - 0.9) + (10Y - Y)
    so 9X = 8.1 + 9Y
    and (again) X = .9 + Y
    Therefore, Y has to equal 1/10 exactly, otherwise stated as 0.1 for X to be equal to 1.

    The value of 1/10 stated decimally truncates, is closed-ended and exact. But it isn't exact and doesn't truncate, in your proof. Instead, your supposed proof fails to state the value at all. It merely states "..." instead of an honest value "Y" which needs to be solved for. In fact, at whatever point we truncate the decimal fraction to solve for Y, we will have a differing value for Y, since the trailing digital infinity is an artifact of an imperfect calculating system, and has nothing to do with the values being calculated themselves. We still don't know the value of Y, because it still has not been stated, and no operation we have performed serves to do anything to affix and reveal it as any actual and distinct value. We could use the value .1 + Y or 0.111...= X and still know as little about the value we're assuming for X as when we use 0.9.... The proof fails. It is a sophistry, since we must pretend a value has been stated and solved for, when it has not.

    Also, 9/9 is not equal to 0.999.... It is only approximately equal to it, just as 99 is only approximately equal to 100. It's inexact. The other decimal values you listed are similarly approximations, not exact values. 9/9 is exactly equal to one, and only approximately represented by 0.999.... I could say 0.1... = 1 under your proof pretext and it should be just as valid as your statement that 0.9... = 1 since the value of ... has not and never will be actually stated or revealed as any particular value by anyone, ever. A minority of fractional values are accurately and exactly represented by decimal fractions. Most are, instead, approximate representations for the value. This is useful for some purposes, but inexact and sometimes worse than no answer at all when science or math require exact values and measures.

    Now, if you believe that 0.11111111.... or 0...1... or 0.0999... are equal to .1, or that 1/9 = 1/10, or that 99=100 there are those who might like to show you a bridge for sale. ;-) Thanks, though, for considering the subject matter and offering a response.

    -------

    For reference, you wrote:

    I think the easiest way to prove that 0.99999.... = 1 is:

    Let X = 0.999999....
    Then 10X = 9.99999..... (moving the decimal point one place to the right)
    So 10X - X = 9.999999.... - X (subtracting the same number from both sides of the equation)
    We then get 9X = 9.99999.... - 0.99999....=
    but 9.99999.... = 9 + 0.99999....
    so 9X = 9
    and X = 1

    Also, you can show that
    1/9 = 0.11111....
    2/9 = 0.22222....
    3/9 = 0.33333....
    4/9 = 0.44444....
    5/9 = 0.55555....
    6/9 = 0.66666....
    7/9 = 0.77777....
    8/9 = 0.88888....
    9/9 = 0.99999....

    Further,
    1/3 * 3 = 1
    but 1/3 = 0.33333....
    so 0.33333.... * 3 = 0.99999.... = 1
    ***************
     
  26. Sep 25, 2010 #25

    Hurkyl

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    Even the greatest teacher cannot teach someone who is unwilling to learn. :tongue:

    You have one chance to convince me that you truly and genuinely do not understand that the notations 0.(9) and 0.999... are being used to notate a decimal with a '9' in every decimal position to the right of the decimal point, and were instead misinterpreting it as being an unspecified, but finite number of '9's.

    Otherwise, I'm going to give you a misinformation infraction for preaching dogma in a subject you do not really appear to know much about.



    Incidentally, non-standard analysis is no different than standard analysis -- in both of them, 0.999... = 1.

    The difference is that, from the viewpoint of standard analysis, non-standard decimals have more positions -- so there exist non-standard decimals that have finitely many 9's by the non-standard measure, but have infinitely many nines by the standard measure, and so are infinitessimally different from 1, but unequal to 1.
     
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