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0.9999 = 1

  1. May 6, 2010 #1
    0.9999.... = 1

    i want to show 0.9999....=1

    by showing 0.9999... is a supremum of numbers smaller than 1,

    i would be able to prove it.

    how can i show 0.999.... is a supremum of numbers smaller than 1?
     
  2. jcsd
  3. May 6, 2010 #2
    Re: 0.9999....=1?

    This is a common question on the forums. Do a search.

    If you want to create a hard proof of the statement: 0.999.... = 1, you first need to give a rigorous definition for a repeating decimal. The usual definition is it's the LIMIT of the sequence created by appending digits one at a time:

    So 0.9999..... is the limit of the sequence 0, 0.9, 0.99, 0.999, 0.9999, ...

    Using this definition, the proof is simply to show that that sequence converges to 1.
     
  4. May 6, 2010 #3
    Re: 0.9999....=1?

    i want to show 0.9999... is strictly same as 1,

    not sequence 0.9, 0.99, 0.999, ... converges to 1.
     
  5. May 6, 2010 #4
    Re: 0.9999....=1?

    1.00..01 > 0.999.... > 0.99...9 for all finite decimals. Both the limits tend to 1. By the Dedekind's cut principle, 0.999... = 1. Q.E.D.
     
  6. May 6, 2010 #5
    Re: 0.9999....=1?

    Yes, but what is 0.999....?

    It's the limit of the sequence 0.9, 0.99, 0.999, .... .

    Therefore, showing that sequence converges to 1 is the proof.
     
  7. May 6, 2010 #6
    Re: 0.9999....=1?

    can i show 0.9999....=sup{x|real number less than 1} ?
     
  8. May 6, 2010 #7
    Re: 0.9999....=1?

    See my above post.
     
  9. May 7, 2010 #8

    Landau

    User Avatar
    Science Advisor

    Re: 0.9999....=1?

    What could you possibly mean by this?
     
  10. May 7, 2010 #9
    Re: 0.9999....=1?

    [tex]
    1 + \frac{1}{10^{n}}
    [/tex]
     
  11. May 7, 2010 #10
    Re: 0.9999....=1?

    1/3 = .3333333...
    Multiply both sides by 3 and see what you get.

    By the way, this is assuming that you have established that 1/3 = .33333... in the first place.
     
  12. May 7, 2010 #11
    Re: 0.9999....=1?

    And also, [itex]3 \times 0.333... = 0.999...[/itex]
     
  13. May 7, 2010 #12
    Re: 0.9999....=1?

    ^I don't get the point of your post, but okay.
     
  14. May 7, 2010 #13
    Re: 0.9999....=1?

    We start multiplying from the rightmost digit. What is the rightmost digit of 0.333...?
     
  15. May 7, 2010 #14
    Re: 0.9999....=1?

    You only start from the rightmost digit because it's convenient not to have the carries messing up what you've already written down. I quite often multiply from left to right.

    But really its just the result that if [itex]\Sigma a_n[/itex] converges, [itex]\Sigma ka_n[/itex] converges to [itex]k\Sigma a_n[/itex].
     
  16. May 7, 2010 #15
    Re: 0.9999....=1?

    lol.
     
  17. May 7, 2010 #16
    Re: 0.9999....=1?

    It saves working out how much space to leave before you start writing it down. Mind you I do quite often get it wrong, but then that applies if I go the other way as well.
     
  18. May 7, 2010 #17
    Re: 0.9999....=1?

    Please refrain from off-topic posting. The OP wanted a rigorous proof and you are talking about your arithmetical adventures.
     
  19. May 7, 2010 #18
    Re: 0.9999....=1?

    Sorry.

    The comment about convergence is a proof. That is [itex]0.\={3}[/itex] is defined as [itex]\sum_{0}^{\infty}3\times 10^{-r}[/itex], so if you've proved this converges to [itex]\frac{1}{3}[/itex] you can say [itex]\sum_{0}^{\infty}3\times 3\times 10^{-r}=\sum_{0}^{\infty}9\times 10^{-r}=0.\={9}[/itex] (by definition) converges to [itex]3\times \frac{1}{3}=1[/itex].

    This doesn't mean that showing [itex]0.\={3}=\frac{1}{3}[/itex] is any easier than showing [itex]0.\={9}=1[/itex] in the first place, so I wouldn't personally recommend this route.
     
    Last edited: May 8, 2010
  20. May 8, 2010 #19
    Re: 0.9999....=1?

    Is this proof acceptable?
    [tex] 0.\={9} = 0.999... = 0.9 + 0.09 + 0.009 + ... = \sum_{k=1}^{\infty} \frac{9}{10^k} = \frac{9/10}{1-1/10} = 1[/tex]
    Using the infinite geometric series formula, of course.
     
  21. May 8, 2010 #20
    Re: 0.9999....=1?

    That would obviously depend on losin.

    The formula for the sum of an infinite gp depends on some basic results about limits, e.g. [itex]|x|<1[/itex] implies [itex]\lim_{n\rightarrow \infty}x^n=0[/itex].

    It sounds to me as if losin may be working from axioms for the reals at a stage before much in that way has been proved, in which case he may prefer a direct proof from the axioms using the fact that the [itex]n^{th}[/itex] partial sum for [itex]0.\={9}[/itex] is [itex]1-\frac{1}{10^n}[/itex]. This seems to fit in more with his posted ideas.

    When I said that showing [itex]0.\={3}=\frac{1}{3}[/itex] is no easier than showing [itex]0.\={9}=1[/itex], I didn't in any way mean to imply either was particularly hard, so I'm sure losin will soon settle on a proof he finds satisfactory even if it doesn't necessarily correspond with any of the suggestions on the forum.
     
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