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Simple maths but i don't understand why (about recurring decimals or repeating)

  1. Aug 17, 2006 #1
    Simple maths but i don't understand why (about recurring/repeating decimals)

    When calculating recurring decimals, we let X to be that number to calculate it for example:

    0.4* = x
    4.4* = 10x

    10x - x = 4.4* - 4*
    9x = 4
    x = 4/9

    Therefore 0.4* = 4/9

    But I when I calculate 0.9* this, i get
    0.9* = X
    9.9* = 10X

    10X - X = 9.9* - 0.9*
    9X = 9
    X = 1

    Therefore 0.9* = 1

    but we know that 0.9* = 0.999999999............................. is close to 1 and not equal to 1
    but how can we prove 0.9* is not equal to 1?
    Last edited: Aug 17, 2006
  2. jcsd
  3. Aug 17, 2006 #2
    We can't because it IS equal to 1. Search for other threads like this on this forum, there have been a ton of them.
  4. Aug 17, 2006 #3
    The value 0.9* does not have any real meaning. Infinite numbers are irrelevant to mathematics as we can only deal with rational numbers (that are certain time approximation of irrational numbers). Letting this aside, 0.9* can be interpreted as the limit of a the geometrical sum:

    9*(1/10) + 9*(1/10)^2 ...9*(1/10)^n

    As n increases beyond all bounds (thus creating an "infinite" number). The limit is in fact 1.
  5. Aug 17, 2006 #4
    What? Infinite numbers seem, to me at least, to definitely be relavent to mathematics, and are you seriously saying that we only deal with rational numbers? Are pi and e rational?
  6. Aug 17, 2006 #5
    PI and e aren't rational, but we can only work with rational approximations. But the concept of "infinity" has no real meaning in mathematics. 0.9* dosen't mean anything if you don't take it as the previously mentionned limit.
  7. Aug 17, 2006 #6

    Yeah, but how often when working a calculus problem let's say do you really work with 3.14, or 2.718 instead of just calling it pi or e and having an exact answer?
  8. Aug 17, 2006 #7
    Would you care to explain this a little more in depth?
  9. Aug 17, 2006 #8
    There is no single awnser. For example, the function 1/x has irrational values of rational x, but its integral from 1 to x will give an exact awnser for x = e^a.
  10. Aug 17, 2006 #9
    Umm.. Ok. That didn't really answer the question I asked though.
  11. Aug 17, 2006 #10
    How to explain? Mathematics has roots in the study of nature. From that point of view, we can only deal with rational approximations. 0.9* is simply another notation for:

    C = 9(1/10) + 9(1/10)^2... 9(1/10)^n

    Where C is always closer to reality as n grows. Thus the notation 0.9* for a value that C is never equal to, but always closer to it. It can also be written;

    C = B + a

    Where B = 1, and a can be as close to 0 as we wish, as long as n is big enough. Finally, the value 0.9* is thus equal to 1, since 1 is the value that "C is never equal to, but always closer to it".
  12. Aug 17, 2006 #11
    How often cannot be awnsered in mathematics :tongue:

    An exact awnser will only occur if two inverse operations, involving e or pi, are made. For logs, trig and roots operations such as addition and multiplication can give a rational number (example: 3^1/2 * 3^1/2).
  13. Aug 17, 2006 #12
    Ok. But that's a very experimentalist view in my opinion, and I will say again that we often deal with pi and e in problems and proofs as just that and take those to represent the irrational number.

    I understand exactly why .999... is equal to 1 so you don't need to explain this to me or convince me of this fact, but the work you have here makes little sense and is certainly not rigorous. And the way you set this up makes it seem that you should have

    C + a = B where B is 1 and a is as close to zero as you would like depending on the value of n using the above.
  14. Aug 17, 2006 #13
    Give me more credit than that:rolleyes:

    I know how we can obtain exact answers using operations similar to those above, my question was asking you to justify your claim that infinite numbers are irrelevant to mathematics.
  15. Aug 17, 2006 #14
    For your comment on how e an pi are used in proofs and problems, I would never go against that. But my point is e an pi are merely notations, when dealing with real data we can only use rational numbers. For the

    C = 1 + a

    What I meant to say is that, if a = a(x)

    0<a(n+1) < a(n)

    Also for how infinite numbers are irrelevant: they're not irrelevant as if they don't exist, they're irrelevant as in they are only notations of a limit. The dealing of infinity in mathematics only has a meaning when we look at the limit of the expressions.
    Last edited: Aug 17, 2006
  16. Aug 17, 2006 #15
    Again I agree with this, but I would think that in most cases when working specifically in pure mathematics the data is unlikely to be so experimental in nature that it is approximated by rational numbers.

    I understand what you mean, but the way you defined C makes it trivially less than 1 for all n and so I still think it should be

    C + a = 1.

    Ok that makes sense and I will agree with you on that for the most part.
  17. Aug 17, 2006 #16
    Pardon my mistake on a, it should be

    0 < |a(n+1)| < |a(n)|


    Just consider my a to be negative

    Sorry for the confusion.
  18. Aug 17, 2006 #17
    Ok now that makes a lot more sense, sorry about that, and thanks for clearing that up.
  19. Aug 20, 2006 #18


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    OK Werg22, lets assume that irrational numbers dont exist and then look at the problem of finding the zeros of the equation f(x) = x^2 + x - 1.

    Assume that x can be written as p/q, where p and q have no common factors. This gives,

    p^2/q^2 + p/q = 1

    Multiply throughout by q and rearrange to get,

    p^2/q = q - p

    Notice that the RHS is an integer. This implies that q must be +/- 1, other wise LHS cannot be an integer.

    Now take the original equation and multiply throughout by q^2/p. This gives,

    q^2/p = p + q

    Again the RHS is clearly an integer and so p must be equal to +/- 1, otherwist the LHS cannot be integer.

    So what is our conclusion, if the original equation has any solution then it can only be +/-1. Sadly however you can easily verify that neither +1 or -1 is a solution to f(x)=0, so we'd have to conclude that the equation has no solutions! But f(0)=-1 and f(1)=+1, hence the function has to pass though zero somewhere in between. So we have a clear contradiction! You cant just ignore irrational numbers or you'll end up with all sorts of contradictions like this.
  20. Aug 20, 2006 #19


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    Sigh, I miss PF for a couple days, and my punishment is to have to clean up one of these messes. :frown:

    What you "know" is wrong. The decimal number system is defined so that 0.9* = 1, and calculations like the one you performed are the reasons such a definition was chosen.

    Unless your arguing that 1 does not have any real meaning either, this contradiction sums up the primary objection I have to your posts.

    0.9* is the notation for a decimal number. That decimal number has no more and no less meaning than any other decimal number. 1 just happens to be another notation for that same decimal number.

    The secondary objection is that mathematics is a much, much broader subject than the study of the arithmetic of terminating decimal expansions. There is absolutely no trouble with working with numbers like pi or e in their exact form, even to a staunch constructivist.

    A tertiary objection is that nature is full of irrational numbers. So, they better have some meaning, even if we can only approximate them. (As an aside, note that most rational numbers don't even have terminating decimal expansions)

    Furthermore, approximating things with terminating decimals is a convenient course of action for scientific pursuits... not a necessary one. There is absolutely no reason at all why we couldn't approximate measured quantities with irrational values.

    And as a quaternary objection (I love that word), this is IMHO one of the most common fallacies that prevent a person from understanding decimal notation. :grumpy: 0.9* is a notation for one particular number... it is not some ambiguous number with an unspecified, but finite number of 9's. "0.9...9 (n 9's)" is a correct alternative notation for your C. 0.9* is not.

    And, to be precise, you should write something like C(n), to express the fact that what you've written is a function of n.
    Last edited: Aug 20, 2006
  21. Aug 23, 2006 #20


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    No, that's not at all true. A person who has actually studied some "higher" mathematics ("Calculus and Beyond") can work very easily with [itex]\pi[/itex] and e- as easily as with [itex]\sqrt{2}[/itex], notapproximations. "0.9* dosen't mean anything if you don't take it as the previously mentionned limit" is very misleading since 0.999... is the "previously mentioned limit" by definition of the decimal numeral system. Would you say "2 doesn't mean anything unless you take it to be 1+ 1"?
    Last edited: Aug 24, 2006
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