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.

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

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...

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;
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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                                   
               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.

I just found this.

UPDATE: Fixed the URL. Thx blakem and tilly.