# .999... = 1?

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• Apr 4th 2005, 03:34 PM
Yankee77
.999... = 1?
I have heard that .999 with a repeating 9 equals exactly 1.

Can someone please discuss this here?
Why is it true?
• Apr 4th 2005, 04:09 PM
Math Helper
Try this
Here is a cool way to come to terms with it though it is not a proof (I'll let the experts handle that).

1/9 = .111111111111...
2/9 = .222222222222...
.
.
.
8/9 = .888888888888...

so this would imply that .9999999999... = 9/9 = 1

Math Helper
• Apr 4th 2005, 04:33 PM
Math Wiz
This will help
Here is another way to look at it:

Let n = 0.9999...
10n = 9.9999...
10n - n = 9.9999... - 0.9999...
9n = 9
n = 1 = 0.9999...

I don't know whether this constitutes a proof but it convinces me.

Math Wiz
• Apr 5th 2005, 10:27 AM
MathMan
wow that is pretty cool, I have never seen that before.
• Apr 8th 2005, 05:00 AM
theprof
Quote:

Originally Posted by Yankee77
I have heard that .999 with a repeating 9 equals exactly 1.

Can someone please discuss this here?
Why is it true?

1.
Performing a devision you'll never get a decimal periodic number with an indefinite sequence of 9 or 0 digit. So the problem is only formal.

2.
Remember how decimal periodic number are transformed in fractions?
For instance 2.(3) - where () means indefinite sequence of 3 as decimal digit

2.3 = (23-2)/9=21/9=7/3

This applies to all cases

0.(9) = 9/9 = 1

3.
1/3 is 0.333333333333...........
3 times 1/3 is 1
but 3 times 0.333333333333........... = 0.99999999999...........

4.

0.(9)=0.99999.....
develops as the infinite sum

9/10+9/100+9/1000

that is 9(1/10+1/10^2+1/10^3+...)
the infinite sum in the parenthesis is the geometric serie with ratio 1/10 and gives
9[1/(1-1/10) -1] = 9[1/(9/10)-1] = 9[10/9-1] = 9(1/9)=1

bye
• Apr 14th 2005, 08:02 PM
beepnoodle
Quote:

Originally Posted by Yankee77
I have heard that .999 with a repeating 9 equals exactly 1.

Can someone please discuss this here?
Why is it true?

If you look at the sequence, .9,.99,.999, ... and so on, you see that each term in the sequence is a closer approximation to the number 1. Obviously, the elements of the sequence will never reach 1, but instead get closer and closer to 1. This said, the sequence converges to 1, but none of the elements are ever exactly equal to 1. But I like the idea of looking at it as an infinite geometric series better. Very clever.
• Apr 15th 2005, 07:44 AM
Shmuel
Quote:

Originally Posted by beepnoodle
If you look at the sequence, .9,.99,.999, ... and so on, you see that each term in the sequence is a closer approximation to the number 1. Obviously, the elements of the sequence will never reach 1, but instead get closer and closer to 1. This said, the sequence converges to 1, but none of the elements are ever exactly equal to 1. But I like the idea of looking at it as an infinite geometric series better. Very clever.

I know you are refering above to a sequence that converges at one. However it is important to note that .999' is exactly equal to 1.
• Apr 16th 2005, 09:14 PM
beepnoodle
Of course, I didn't realize whether we were talking about an infinite number of 9's or a finite number. If the number of 9's is infinite then yes it is equal to 1.
• Jul 24th 2005, 01:32 PM
Rebesques
?
Here 's a little proof, that doesn't move far into sequence territory:

Consider the difference a=|0.(9)-1|. (the parentheses mean period).

Let this number be positive. Observe that a is non-negative, and less than any number of the form 0.0(etc)01, with a random number of zeros. This contradicts a>0 (*), and so a=0.

(*) 'Cause, if a>0, there ought to be a number of the form 0.0...01 < a. :eek:
• Aug 5th 2005, 01:39 PM
rgep
Perhaps a helpful way to think about this is to separate the concepts of 'real number' and 'decimal expansion'. We can set up the system of real numbers in various ways, some of the more high-brow being Dedekind cuts in the real numbers or the completion of the rationals via Cauchy sequences modulo null sequences. Once you have a notion of real number, you can then describe a real number by a decimal expansion. It turns out that decimal expansions have this awkward property that sometimes two different decimals express the same real number: and this happens when one of them ends in 9 recurring. Of course one way of setting up the real number system is to start with decimals expansions.
• Aug 6th 2005, 01:46 PM
rgep
For a further discussion of this point, see the page by Tim Gowers (author of the excellent "Mathematics: a Very Short Introduction") on What is so wrong with thinking of real numbers as infinite decimals? on his website http://www.dpmms.cam.ac.uk/~wtg10/decimals.html
• Aug 28th 2005, 04:29 AM
hoeltgman
The most simple way in convincing someone that 0.99999.... = 1 is to calculate the difference=

1- 0.999999.... = 0

If the number of 9 is infinite, you never get the chance to write something else than 0.
• Aug 28th 2005, 05:50 PM
Dr. de Seis
Sum 0.9 / ( 10^i ) from i = 0 to infinity. The sum is geometric, so the total sum equals the first term in the sequence over 1 - r where r = 1 / 10.

0.9 / ( 1 - 0.1 ) = 1
• Nov 14th 2005, 07:25 PM
ThePerfectHacker
The answer is basic, DEFINITON: THE REPETION OF A DECIMAL EQUALS THE VALUE OF ITS CONVERGENT. Thus, what most people do not relized is this is not paradoxical (nothing in math is paradoxical) but rather it is a DEFINITION and not a theorem. It is proven by looking at it geometric sum.
• Nov 16th 2005, 09:51 AM
CaptainBlack
Quote:

Originally Posted by Rebesques

(*) 'Cause, if a>0, there ought to be a number of the form 0.0...01 < a. :eek:

This is an assertion that there are no infinitesimals. But there
are models of the reals which do have infinitesimals, but 0.999..
is still equal to 1 in these models (or is it?).

RonL
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