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.999999... == 1? (Somewhat OT)

by Sherlock (Deacon)
on Jul 25, 2001 at 00:37 UTC ( #99473=perlmeditation: print w/ replies, xml ) Need Help??

Hello all,
This post is somewhat off topic, but I've begun to notice that there are enough math buffs around here that some of you may get some enjoyment out of this (if you haven't seen this already, that is).

A math professor of mine came to me one day with a proof. Although I can't remember the exact format, I can recall the general concept. This proof stated that .9 repeating was equal to the number 1. Following are the arguments to support this statement:

1. By Substitution

3*(1/3) = 1
1/3 = .3 repeating, therefore
3*(.3 repeating) = 1
However, 3 * .3 repeating = .9 repeating
Therefore, .9 repeating must be equal to 1


2. By the Interval Between Numbers

It is give that, for any two distinct numbers, there is an infinite number of values that fall between the two. For example, between 2 and 3, there is an infinite number of values that are greater than 2 and yet less that 3.

If that's the case for all distinct numbers, give me just one number that falls between .9 repeating and 1 (after all, if they're distinct, there should be an infinite number of values between them - I only ask for one, as that would disprove the above statement).

Challenge

For a long time, I found this argument impossible to refute. One day, while in the shower, no less, I think I came upon a reasonable argument against this statement. Of course, I, unlike the professor that proposed this to me, do not have a PhD in mathematics and I'm sure my proof could be somewhat more sound.

Rather than simply giving what I consider a "fairly legitimate" answer, I'd like to see what arguments I can get from the other monks either supporting or refuting the above statement. I'm really looking forward to seeing what types of arguments I see. Assuming that the rest of the monks don't give my answer verbatim, I'll post my answer in a day or so and you can all have fun ripping it apart.

Have a little fun. Good luck!

- Sherlock

Skepticism is the source of knowledge as much as knowledge is the source of skepticism.

Comment on .999999... == 1? (Somewhat OT)
Re (tilly) 1: .999999... == 1? (Somewhat OT)
by tilly (Archbishop) on Jul 25, 2001 at 00:48 UTC
    This is very OT.

    I direct you to the FAQ for sci.math. 0.9999... is 1.

    The long and the short of it is that our decimal system is just a representation of an underlying number system. In the underlying system, if two numbers differ by 0, then they are the same point. It is just an artifact of our representation that we happen to have 2 ways to represent 1 and only one way to represent pi.

Re: .999999... == 1? (Somewhat OT)
by Masem (Monsignor) on Jul 25, 2001 at 00:54 UTC
    0.99999... = 9/10 + 9/100 + 9/1000 + ... = 9*( 1/10 + 1/100 + 1/1000 + ... ) = 9*sum( (1/10)^n, n=1..inf ) = 9*(1/(1 - 1/10)-1) #geometric series sum = 9*(10/9 - 1) = 9*(1/9) = 1 QED

    -----------------------------------------------------
    Dr. Michael K. Neylon - mneylon-pm@masemware.com || "You've left the lens cap of your mind on again, Pinky" - The Brain

Re: .999999... == 1? (Somewhat OT)
by arhuman (Vicar) on Jul 25, 2001 at 00:58 UTC
    I remember I was stunned by my teacher demonstration (I was 14)
    It was SO simple :

    X=.999999999...
    10X=9.9999999999...
    10X-X=9
    9X=9
    X=1

    Needless to say that since this day there were no doubts for me...
    .999999... is 1

    UPDATE : I should have read the tilly's link first.
    My demonstration is presented there, apologizes for the redundant information...

    "Only Bad Coders Code Badly In Perl" (OBC2BIP)
Re: .999999... == 1? (Somewhat OT)
by arashi (Priest) on Jul 25, 2001 at 01:27 UTC
    I'm not a math major, and I haven't really taken much math in college, but here goes anyways:

    The proof of this problem hinges on the assumption that 1/3 = .3 repeating. Is this a correct assumption?

    I can punch 1/3 into my calculator and it says .3 repeating, but is that really equal to 1/3, or is it just infinitely close to 1/3? 1/3 of something means it is broken into three equal parts that add up to 1, but since .3 repeating + .3 repeating + .3 repeating adds up to .9 repeating, wouldn't .9 repeating just be infinitely close to 1? Because of the infinity problem, there isn't a number between .9 repeating and 1, is this because .9 repeating isn't really a number, I mean you can't just write it out, because is keeps going, wouldn't that make it a limit function, more of a number concept than an actual number?

    Arashi

    I'm sure Edison turned himself a lot of colors before he invented the lightbulb. - H.S.
Re: .999999... == 1? (Somewhat OT)
by fenonn (Chaplain) on Jul 25, 2001 at 01:29 UTC

    1/3 = .3 repeating. 2/3 = .6 repeating. Since .3 repeating and .6 repeating = .9 repeating, and 1/3 and 2/3 = 1, it stands to reason that .9 repeating = 1 also. The latter is true, since if A = C, and B = D, then if A + B = E, and C + D = F, then E = F.

    Since 1 minus .9 repeating = .0 repeating, and .0 repeating = 0, then 1 = .9 repeating since any A - B that = 0, necessitates that A = B.

    Fenonn.

      Actually, the statement that .3 repeating == 1/3 is not *entirely* correct. The more you repeat the 3, the closer it gets to actually being 1/3, but it is NEVER EXACTLY 1/3. Try this: divide 9 by 3. The answer is 3. Now multiply 9 by .3. You get 2.7. Multiply by .33 and you get 2.97. Keep adding 3s and you keep adding 9s in the product. What you never get, however, is 3. It may get so close to 3 as to make a distinction meaningless (depending on what you're doing), but 2.99...7 != 3. For the same reason, .33... != 1/3. Apply that to .99... and you get very close to 1, but never quite there. So I chalk the whole thing up to a floating point rounding error. =-)

      ~Cybercosis

      nemo accipere quod non merere

        This is really what I was hoping someone would come up with. I expected the barrage of mathematical proofs that prove, without a doubt the .9 repeating is indeed equal to 1. However, in a computer, you can't repeat forever. This concept, although true in the word of theoretical mathematics, does not hold true in the world of computers. Attempt this simple script:

        #!/usr/bin/perl -w use strict; my $num1 = .9999999999999999; my $num2 = 1; print "num1 = $num1\n"; print "num2 = $num2\n"; print "The numbers are equal" if $num2 == $num1; print "The numbers are not equal" if $num2 != $num1;
        You'll notice that these two numbers are indeed unequal. You're probably thinking big deal, obviously, $num1 doesn't repeat forever so they shouldn't be equal. Try the script again, but add a single 9 to the end of $num1. You should find the results more interesting. (I'm by no means a Perl guru and am not sure if this will work the same from one Perl installation to another. I am using Perl 5.6.1 on a Win32 platform.) In that case, the two values, even though $num1 still does not repeat forever, are considered equal.

        What I had come up with, as a refutation to the argument, was that whether .9 repeating is equal to 1 is dependent upon your context. In a mathematical world, it is obviously true. Within a computer, however, it is an entirely different case.

        - Sherlock

        Update: Please note, I am not attempting to say that .9 repeating is not equal to 1 simply because a computer can't represent .9 repeating. I was simply trying to point out that there is some level of discrepency between theoretical mathematics and the mathematics done in a computer/calculator. I was really just hoping this post would cause a few of you (and myself) to think about how you use floating point numbers and make you more aware of the rounding errors that can take place.

        Skepticism is the source of knowledge as much as knowledge is the source of skepticism.
(MeowChow) Re: .999999... == 1? (Somewhat OT)
by MeowChow (Vicar) on Jul 25, 2001 at 01:32 UTC
    Pshaw! And I suppose next you'll tell me that 0.0000... repeating is equal to 0 ;-)
       MeowChow                                   
                   s aamecha.s a..a\u$&owag.print
      Proving that 0.00000 ... (infinite zeros) ... 00001 = 0 is more of a challenge.

      ____________________
      Jeremy
      I didn't believe in evil until I dated it.

        Yes, especially since it's not. :)

        Consider:

        0.00000 ... 00001 > 0.00000 ... 000009

        Now if A > B and A = 0 than 0 > B so

        0 > 0.00000 ... 000009

        and we've found a positive number which is strictly less than zero.<blink><blink><twitch>

        The problem comes from specifying a last digit since one can always specify one more after that which isn't zero and get a 'really' different number.

Re: .999999... == 1? (Somewhat OT)
by Superlman (Pilgrim) on Jul 25, 2001 at 10:51 UTC
    One simple proof which I'm surprised hasn't been brought up is to consider the difference between the left hand and the right hand sides of the equation.

    If you have one decimal, (.9 vs 1) then the difference is 0.1
    With two, it is 0.01
    With n, it is 10^(-n)

    The question is: How many decimals do we have? Well, since it is repeating FOREVER, we take the limit as n approaches infinity. As n approachs infinity, 10^(-n), and hence the difference, approaches zero.

    If the difference between the numbers is zero, they are the same.

    **POOF**

    For my next trick, I need a volunteer from the audience...

Re: .999999... == 1? (Somewhat OT)
by scott (Chaplain) on Dec 01, 2001 at 05:36 UTC

    I just found this.

    UPDATE: Fixed the URL. Thx blakem and tilly.

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