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#1
06-01-2003, 07:12 PM
 Lint6 Guest Join Date: Jan 2003 Posts: 2
Why doesn't .9999~ = 1?

Someone asked this question on another message board I go to. I'm asking it here because the other message board is mostly about video games and I figured someone here would have a better idea.

1/3 = .3333~
2/3 = .6666~
3/3 = .9999~

But
3/3 = 1

So why doesn't .9999~ = 1?
#2
06-01-2003, 07:16 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
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#3
06-01-2003, 07:23 PM
 Orbifold Guest Join Date: Oct 2000 Location: Melbourne, VIC, Australia Posts: 2,090

(Curse you and your fast typing skills, Q.E.D.!!!)
#4
06-01-2003, 07:58 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,124
#5
06-01-2003, 08:10 PM
 Azael Guest Join Date: Nov 2002 Location: Houston, Texas Posts: 780

It does.
#6
06-01-2003, 08:15 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,124
I sent the wrong link and I can't find the one I meant to send, but .999... will always be smaller than 1.

And 1/3 does not equal .333...
#7
06-01-2003, 08:16 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Might want to read your cites a bit more thoroughly, x-ray vision:
Quote:
 You can show (using calculus or other methods) that with a large enough number of 9s in the expansion, you can get arbitrarily close to 1, and here's the key: THERE IS NO OTHER NUMBER THAT THE SEQUENCE GETS ARBITRARILY CLOSE TO. Thus, if you are going to assign a value to .9999... (going on forever), the only sensible value is 1.
#8
06-01-2003, 08:17 PM
 Azael Guest Join Date: Nov 2002 Location: Houston, Texas Posts: 780
Quote:
 but .999... will always be smaller than 1.
Really? By how much?
#9
06-01-2003, 08:19 PM
 TBone2 Guest Join Date: Oct 1999 Location: Mansfield, OH Posts: 610
Re: Why doesn't .9999~ = 1?

Quote:
 Originally posted by Lint6 3/3 = .9999~ But 3/3 = 1

The first of your statements that I quoted is false, and can only be discussed using the principles of "The Calculus." Your second quoted statement is true, following the basic rule of math that states that any number divided by itself equals unity, or 1.

While 1/3 can be said to be .333333..... and 2/3 can be said to be .6666666......, it is silly to simply multiply the infinite decimally-expressed value of 1/3 by 3 to arrive at a statement like "3/3 = .9999~."
#10
06-01-2003, 08:20 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Quote:
 Originally posted by Azael Really? By how much?
By 1/infinity.
#11
06-01-2003, 08:23 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Re: Re: Why doesn't .9999~ = 1?

Quote:
 Originally posted by TBone2 While 1/3 can be said to be .333333..... and 2/3 can be said to be .6666666......, it is silly to simply multiply the infinite decimally-expressed value of 1/3 by 3 to arrive at a statement like "3/3 = .9999~."
No it's not. You agree that 1/3 = .33333... and 2/3 = .66666. Ergo:

Since 1/3 + 2/3 = 3/3

and .33333... + .66666.... = .99999

Therefore 3/3 = .99999

Q.E.D.
#12
06-01-2003, 08:25 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Put .... after both instances of .99999 in my above post.
#13
06-01-2003, 08:25 PM
 Achernar Guest Join Date: Aug 1999 Location: 23 male, Boston Posts: 5,791
Anyone who thinks it is untrue that 0.999... = 1 does not understand limits. (Which is not a terrible thing; lots of Calculus students don't understand limits.)

Imagine you had a blackboard with 0.999 on it. And every second you added another 9 to the end. The number you would have at the "end of time" would not be the same as 0.999....

Also, it is quite possible for every entry in a sequence to be less than the limit of that sequence. That doesn't mean that its limit is not its limit.

Q.E.D., please be a little more careful with your ellipses. They're more important when discussing this problem than any other time, in my estimation.
#14
06-01-2003, 08:27 PM
 Jabba Guest Join Date: Mar 2002 Posts: 886
TBone2 asserted:
Quote:
 The first of your statements that I quoted is false, and can only be discussed using the principles of "The Calculus." Your second quoted statement is true, following the basic rule of math that states that any number divided by itself equals unity, or 1. While 1/3 can be said to be .333333..... and 2/3 can be said to be .6666666......, it is silly to simply multiply the infinite decimally-expressed value of 1/3 by 3 to arrive at a statement like "3/3 = .9999~."
Why is it silly?
Using "The Calculus" ( or analysis, if you prefer;specifically, properties about infinite series):
0.333... = S3/10i = 1/3 ( sum of an infinite G.P.)
and so
1 = 3/3 = 3*1/3 = 3S3/10i = S9/10i = 0.999...
#15
06-01-2003, 08:27 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Quote:
 Originally posted by Achernar Q.E.D., please be a little more careful with your ellipses. They're more important when discussing this problem than any other time, in my estimation.
#16
06-01-2003, 08:35 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,124
here is another web-page purporting that .999... doesn't = 1, but check out the last two paragraphs (bolding mine).

Quote:
 The numbers in the "Answer" column do indeed get closer and closer to some number, namely 1. The fact that they never get there is irrelevant. 1 is by definition the exact answer to the infinite addition problem. So in conclusion, Point-Nine-Repeating is indeed equal to one. In order to see it, we needed to know how to add up infinitely many small numbers. By the way, this knowledge itself is probably more important that the little fact that .999...=1.
How the hell is the fact that they never get their irrelevant? The fact that they never get there is exactly why it doesn't equal 1!
#17
06-01-2003, 08:38 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
It's an infinite series. When they talk about it never reaching one, they are referring to an increasing, but still finite, number of 9s being tacked onto the end.
#18
06-01-2003, 08:39 PM
 Achernar Guest Join Date: Aug 1999 Location: 23 male, Boston Posts: 5,791
Quote:
 Originally posted by Q.E.D. I've addressed that, schmuck!
You'll notice that the two posts in question are less than a minute apart. So nyah. And anyway, you missed one. The first 0.6666....
#19
06-01-2003, 08:40 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Those are just upside-down 9s.
#20
06-01-2003, 08:43 PM
 Jabba Guest Join Date: Mar 2002 Posts: 886
No, the paragraohs you quote are quite correct. It is an important feature of limits that they do noy have to reach their limiting value after any finite number of steps.

For a ( usually) less controversial example, consider the sequence 1/n. Its terms are
1, 1/2, 1/3, 1/4, 1/5, ...
Most people would agree that this sequence tends to 0 as n tends to infinity, and they would be correct. The fact that no term of the sequence is actually 0 is wholly irrelevant.

Similarly, when we say that
0.999... = 1 we mean that 1 is the ( unique) limit of the sequence
0.9, 0.99, 0.999, 0.9999, ...
#21
06-01-2003, 08:45 PM
 Orbifold Guest Join Date: Oct 2000 Location: Melbourne, VIC, Australia Posts: 2,090
x-ray vision...in what way does that web page purport that .999... doesn't equal one? The entire page is devoted to showing that it does equal one! Just like your first cite.

Anyway, your more recent cite contains my favourite argument as to why 0.999...=1, namely:

Quote:
 If it's not 1, there must be a number between it and 1. What number would that be?
I have yet to see anyone identify a number between 0.9999... and 1, nor do I expect anyone to do so. Not that this convinces anyone, of course.
#22
06-01-2003, 08:48 PM
 RealityChuck Charter Member Join Date: Apr 1999 Location: Schenectady, NY, USA Posts: 40,687
We can always bring out the formula for those not bothering with the other threads. We use the basic method for turning a repeating decimal into a fraction.

let x = 0.9999...
then 10x = 9.99999...

Now we subtract
10x - x = 9.99999... - 0.99999...

Solving both sides
9x = 9.00000

Reducing
x = 9/9 = 1

Therefore 0.9999... = 1

No calculus involved.
__________________
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#23
06-01-2003, 08:50 PM
 Jabba Guest Join Date: Mar 2002 Posts: 886
RC: If you take calculus as the study of limits, then strictly you are using calculus as soon as you write
0.999...
and also when you multiply it by 10 to obtain
9.999...
#24
06-01-2003, 08:51 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,124
Quote:
 originally posted by [b]Jabba[/b For a ( usually) less controversial example, consider the sequence 1/n. Its terms are 1, 1/2, 1/3, 1/4, 1/5, ... Most people would agree that this sequence tends to 0 as n tends to infinity, and they would be correct. The fact that no term of the sequence is actually 0 is wholly irrelevant.
Again, how is this irrelevant? If a number isn't actually zero (and never will be), then it isn't zero.
#25
06-01-2003, 08:53 PM
 wolf_meister Guest Join Date: May 2003 Location: Where the owls say "Whom" Posts: 5,484
Response to Tbone2

You stated that ...
"following the basic rule of math that states that any number divided by itself equals unity, or 1."
Zero divided by zero is not 1 nor is it zero for that matter. I believe mathematicians call the answer either "indeterminate" or "undefined".
And Achernar, the proper phrase for "..." is ellipsis and not ellipses.
#26
06-01-2003, 08:53 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,124
Orbifold, I meant to write does, not doesn't. I need some sleep!
#27
06-01-2003, 08:56 PM
 Achernar Guest Join Date: Aug 1999 Location: 23 male, Boston Posts: 5,791
The limit of the sequence:

{ 0.9, 0.99, 0.999, 0.9999, ... }

is 1. This is really the nomen ludi.

You realize that "limit" isn't just a term mathematicians throw around, right? It has a strictly-defined meaning, and it's not ambiguous. It may not be the most intuitive meaning to you, which could be causing the confusion. For instance, it is not correct to think of the limit as the last term in an infinite sequence. The limit does not have to be found in the sequence at all. Anywhere.
#28
06-01-2003, 08:57 PM
 TBone2 Guest Join Date: Oct 1999 Location: Mansfield, OH Posts: 610
This entire discussion hangs on a pretty thin thread -- the limitations of the decimal number system.

Q.E.D. and Jabba, adding the decimal values for 1/3 and 2/3 makes no more sense than multiplying the decimal value of 1/3 by 3. Why strain yourself? 1/3 * 3/1 = 3/3 = 1. (Duh.) 1/3 + 2/3 = 3/3 = 1. (Duh again.)
#29
06-01-2003, 09:00 PM
 TBone2 Guest Join Date: Oct 1999 Location: Mansfield, OH Posts: 610
Nicely done, RealityChuck!!!
#30
06-01-2003, 09:05 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Quote:
 Originally posted by TBone2 ...adding the decimal values for 1/3 and 2/3 makes no more sense than multiplying the decimal value of 1/3 by 3.
Why not? It's perfectly valid. Since .9999... - 1 saying .33333... x 3 = .99999... = 1 or .33333... + .66666... = .99999... = 1 are both entirely correct.
#31
06-01-2003, 09:05 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,124
Quote:
 If it's not 1, there must be a number between it and 1. What number would that be?
That's logical?
#32
06-01-2003, 09:07 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Quote:
 Originally posted by x-ray vision That's logical?
Perfectly. mathematically, it is asking you to solve for 1 - .9999... = x. What would you have the value of x be?
#33
06-01-2003, 09:08 PM
 Jabba Guest Join Date: Mar 2002 Posts: 886
None of the terms of the sequence is 0, but the limit of the sequence is. Sometimes the limit of a sequence is one of the terms. For example,
1, 1, 1, ...
certainly tends to 1 and so does
1, 1/2, 1, 2/3, 1, 3/4, 1, ...
But it is by no means necessary that this be so.

Perhaps a look at the formal definition of a limit will help. For a sequence of real numbers, we say that an tends to the limit l if:
Given any real e>0, we can choose a natural number N such that |an - l| < e whenever n>N.
In other words, if you give me any positive margin of error, I guarantee that all the terms of the sequence from some point on will approximate the limit number l to within that margin of error.

Let us apply this to the two sequences we have been discussing. I have claimed that
1, 1/2, 1/3, 1/4, ... converges to 0
and that
0.9, 0.99, 0.999, 0.9999, ... converges to 1.
As an example take e, the magin of error, to be 0.001
For the first sequence, any term after the 1000th is less than 0.001 and so is within 0.001 of the proposed limit.
For the second, every term after the 3rd is within 0.001 of 1.

If instead we take e = 0.000001, then we just have to go further along the two sequences: the 1,000,000th term will do for the first sequence; the 6th term for the second.

In general, whatever positive e you choose, I can guarantee that all terms from some point on will be within e of my claimed limit. That is what we mean by convergence, and in particular by the value of an infinite sum.
#34
06-01-2003, 09:08 PM
 JRootabega Guest Join Date: Sep 2001 Posts: 527
.000~1

#35
06-01-2003, 09:11 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Quote:
 Originally posted by JRootabega .000~1
I'll accept that. And since .0000...1 = 1/infinity = 0, you're correct.
#36
06-01-2003, 09:12 PM
 JRootabega Guest Join Date: Sep 2001 Posts: 527
Now now, I'm with you Q.E.D., but performing arithmetic on infinity isn't going to strengthen our cause.
#37
06-01-2003, 09:17 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,124
Q.E.D., what does 2 - .9999... = x. What would you have x be?
#38
06-01-2003, 09:18 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
1
#39
06-01-2003, 09:19 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,124
Scratch the "what does".
#40
06-01-2003, 09:31 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,124
1 - .9 = 0.1

1- .99 = 0.01

1 - .999 = 0.001

I can go on forever and never get 0, however, 1 -1 =0.

My conclusion is 1 cannot be the same as .999...
#41
06-01-2003, 09:56 PM
 Orbifold Guest Join Date: Oct 2000 Location: Melbourne, VIC, Australia Posts: 2,090
Quote:
 Originally posted by x-ray vision That's logical?
Q.E.D. has already expounded on this a bit, but let me elaborate as well.

Given any two distinct real numbers, there is always a third real number lying strictly between the first two. Between 0 and 1, there is 1/2. Between 1 and 1/2, there is 3/4. Between x and y, there is (x+y)/2.

By the same token, if there is no such third number then the first two numbers can't be distinct. There's no number x such that 1 < x < 1, which is why 1 and 1 are the same number.

Now, what about 0.999... and 1? I claim there is no number between them. Indeed, suppose 0.999... < x < 1; I claim this leads to a contradiction.

For starters, it must be true that 0 < x-0.999... < 1-0.999... . But 1-0.999... is clearly less than 1-0.9=0.1. It's also less than 1-0.99=0.01, 1-0.999=0.001, 1-0.9999=0.0001, and so on. In fact, given any positive number y I can show that 1-0.999... is less than y by writing out enough 9's. But somehow 1-0.999... isn't less than x-0.999..., even though x-0.999... is a positive number. This is the contradiction.

So there does not exist any such number x.

The same thing happens with the sequence 0.1, 0.01, 0.001, ... . No term in the sequence is 0. But the limit must be zero. After all, the limit certainly isn't negative. And the limit can't be postive either, for if it were then it would be smaller than 0.000......00001 for some number of zeroes, and how can the limit of a decreasing sequence be smaller than any of the terms in the sequence? Zero is the only possibility remaining.
#42
06-01-2003, 09:57 PM
 Exapno Mapcase Charter Member Join Date: Mar 2002 Location: NY but not NYC Posts: 29,348
X-Ray Vision, the whole point behind infinite arguments is that is you go on forever you do get 0.

These threads always come down to this one basic factor. Either you get the idea of infinity and understand this equality or you don't. What's ironic is that your argument - that we keep getting smaller and smaller but never reach zero - is arguing for an infinitesimal. But infinitesimals and limits are incompatible ideas in our arithmetic.

The limit as .999999... goes to infinity is 1.
#43
06-01-2003, 10:03 PM
 3_14159265358979323846 Guest Join Date: Aug 2002 Posts: 35
Quote:
 Originally posted by x-ray vision 1 - .9 = 0.1 1- .99 = 0.01 1 - .999 = 0.001 I can go on forever and never get 0, however, 1 -1 =0. My conclusion is 1 cannot be the same as .999...
No, you can't. That's just it. You'll die, eventually.
0.999... has an infinite number of 9s.
Using .9, or .99, or .999 is an approximation, which is why you don't get exactly zero. But the approximation gets closer by a factor of 10 for each digit you add. So, what happens when you use an infinite number of digits?
#44
06-01-2003, 10:05 PM
 Donut Guest Join Date: Aug 2001 Location: Slightly left of centre Posts: 96
Quote:
 My conclusion is 1 cannot be the same as .999...
Your conclusion is based on an intuitive understanding of what 0.999... means, an intuitive understanding which is simply wrong. 0.999... is generally understood to mean the limit of the sequence 0.9, 0.99, 0.999, 0.9999, ... and so on. That limit is 1, as inarguably as any mathematical fact, no matter how strange it seems to you.
#45
06-01-2003, 10:31 PM
 3_14159265358979323846 Guest Join Date: Aug 2002 Posts: 35
Let me expand on my previous response.

.9 differs from 1 by (1-.9)/1, which is .1 or 1/10.
.99 differs from 1 by (1-.99)/1, which is .01 or 1/100.
.999 differs from 1 by (1-.999)/1, which is .001 or 1/1000.

Let n be the number of 9s you are using in the approximation. Thus, the difference between 1 and .9999.... is 1/(10n).

For n=infinity, the difference is 1/(10infinity) or 1/infinity.

Since 1/infinity equals zero*, the difference between .999... and 1 is zero. Saying that the difference between two numbers is zero is equivalent to saying that they are the same number. QED.

*Yes, I know that it isn't technically correct to say that 1/inf equals 0, but rather limx-->inf 1/x = 0. Small technicality.
#46
06-01-2003, 10:50 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,124
Quote:
 Originally posted by Orbifold In fact, given any positive number y I can show that 1-0.999... is less than y by writing out enough 9's. But somehow 1-0.999... isn't less than x-0.999..., even though x-0.999... is a positive number. This is the contradiction.
That's a contradiction I just can't grasp.

Quote:
 Originally posted by 3_14159265358979323846 No, you can't. That's just it. You'll die, eventually.
C'mon now.
#47
06-01-2003, 10:55 PM
 ultrafilter Guest Join Date: May 2001 Location: In another castle Posts: 18,988
Let's leave infinite arithmetic out of this, and not cloud the issue with technicalities.

Others have already said most of what I would, and I've been there and done that on this topic, so let me just add that the fact that .9... = 1 is a direct consequence of the most important property of the real numbers, and if they're not equal, everything falls apart.
#48
06-01-2003, 11:07 PM
 InLikeFlynn BANNED Join Date: Jan 2003 Location: I'm a NYC Boyie!! Posts: 194
If you went .999 of the distance to 1.0, then guess what?
You never got there!!
Terribly obviously wrong again!!
If you went .99999999999~~ way around the world you never reached your starting point so .99999~ will never equal 1!!
I don't care who tells you it does!!
WRONG!!

#49
06-01-2003, 11:18 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Quote:
 Originally posted by InLikeFlynn Q.E.D. your wrong again!! If you went .999 of the distance to 1.0, then guess what? You never got there!! Terribly obviously wrong again!! If you went .99999999999~~ way around the world you never reached your starting point so .99999~ will never equal 1!! I don't care who tells you it does!! WRONG!! got your back again x-ray!!
Well, I tell you what. If you can find ONE source with a correct proof that .9999... != 1, i'll eat my computer.
#50
06-01-2003, 11:28 PM
 dejahma Guest Join Date: Apr 2002 Posts: 81
To all the mathematicians out there give it up; those who don't understand limits and calculus and choose not to study will never understand. To all of the math challenged out there let's discuss something a bit more on your level: How many angels can dance on the head of a pin?

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