I would like to argue about .999

  • Thread starter Curd
  • Start date
  • #1
78
1
In response to the below quote I have written this. The idea that arguing is "spreading misinformation" is ridiculous as argument is the prime mover behind the evolution of the general intelligence of humanity as a whole. If that's the sort of attitude this forum carries with it though, fine, I'll leave. I came here to become educated, not merely indoctrinated. (education being something highly dependent on the dialogue created by argument and indoctrination being something highly dependent on the lack of dialogue created by argument). So, I submit the below in an attempt to meet out my expectations of education. Whether I get such education by means of knowing more about the world or more about the current subject, .999.... will depend on the reaction I get.

So, here goes...


(moderator's note: I've removed someone asserting that 0.999... is not equal to one, along with the followups)


decimals of fractions never accurately equal those fractions. for example .111111111111111...never fully equals 1/9 (which applies directly to a certain proof) but for practical purposes you can assume it does.

.999999.... never equals one. (i still find it hard to believe that saying that warrants an infraction). as the decimal points increase onto infinity the space between 1 and .9999... becomes smaller, but the two never do connect to equal the exact same number. there is an infinite number of spaces to which decimals can extend (as noted by pie), so there is no reason for .999999999.... ever to change to one.

you can argue that .9999999 REPRESENTS one in the language system you happen to be using, but .99999... will never actually equal one in real life.

let's say you want to count all of the entities (particles, doodads, whatever) in the universe in relation to the whole. we will represent the whole with 1. once you have accounted for every item within the universe, lets represent each such item with a 9 after the decimal point, you have counted the whole of it and achieved accounting for 1 universe. since the universe expands on forever, the actual ability to do such a thing does not exist. therefore you can continue on counting 9's in .99999... without ever actually achieving 1.

If you were to say otherwise you would have to also say that infinity is impossible, and ergo pie would be impossible (along with other things). as it is a fact that the 9's after the decimal point can go on into infinity it is also a fact that .999... never equals one except for practical purposes in which such fine details are not important.

and as for that argument by Hurkyl that i'm "spreading misinformation"

and spreading misinformation? so, you're saying that the losing party to any argument is "spreading misinformation"? how ever is one to have conversations on topics if one is to always be correct (i'm not saying i'm wrong by arguing this by the way), when being incorrect is "spreading misinformation"?

who is the owner of this forum? i would very much like to take this up with them.
 

Answers and Replies

  • #2
Char. Limit
Gold Member
1,204
14
Here's a proof that .999...=1.

.999... can be written as the infinite sum as follows:

[tex].9 + .09 + .009 + .0009 + .00009 + ... = .9 \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n[/tex]

Now, evaluating the sum on the right, we use the fact (proven below) that...

[tex]\sum_{n=0}^\infty r^n = \frac{1}{1-r}[/tex]

for all r with a magnitude less than 1. Using this fact, we find that...

[tex].9 \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n = .9 \frac{1}{1-.1} = \frac{.9}{.9}=1[/tex]

Now, to prove that fact that we used, note the proof below:

[tex]S = \sum_{k=0}^{n-1} a r^k = a + a r + a r^2 + a r^3 + a r^4 + ... + a r^{n-1}[/tex]

[tex]rS = a r + a r^2 + a r^3 + a r^4 + a r^5 + ... + a r^n[/tex]

[tex]S - rS = a - a r^n = a (1 - r^n)[/tex]

[tex]S(1-r) = a (1 - r^n)[/tex]

[tex]S = \frac{a (1-r^n)}{1-r}[/tex]

Now let n go to infinity. For r with a magnitude less than 1, r^n tends to 0 as n tends to infinity. Thus...

[tex]\lim_{n \rightarrow \infty} S = \lim_{n \rightarrow \infty} \frac{a (1-r^n)}{1-r} = \frac{a}{1-r}[/tex]

Q E D
 
Last edited:
  • #3
jhae2.718
Gold Member
1,161
20
More correctly, that holds as long as [itex]|r|<1[/itex].

Might also help to add the step [tex]S = \sum_{k = 0}^{n-1} ar^k[/tex] or similar...
 
  • #4
78
1
Here's a proof that .999...=1.

.999... can be written as the infinite sum as follows:

[tex].9 + .09 + .009 + .0009 + .00009 + ... = .9 \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n[/tex]

Now, evaluating the sum on the right, we use the fact (proven below) that...

[tex]\sum_{n=0}^\infty r^n = \frac{1}{1-r}[/tex]

for all r with a magnitude less than 1. Using this fact, we find that...

[tex].9 \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n = .9 \frac{1}{1-.1} = \frac{.9}{.9}=1[/tex]

Now, to prove that fact that we used, note the proof below:

[tex]S = \sum{k=0}^{n-1} a r^k = a + a r + a r^2 + a r^3 + a r^4 + ... + a r^{n-1}[/tex]

[tex]rS = a r + a r^2 + a r^3 + a r^4 + a r^5 + ... + a r^n[/tex]

[tex]S - rS = a - a r^n = a (1 - r^n)[/tex]

[tex]S(1-r) = a (1 - r^n)[/tex]

[tex]S = \frac{a (1-r^n)}{1-r}[/tex]

Now let n go to infinity. For r with a magnitude less than 1, r^n tends to 0 as n tends to infinity. Thus...

[tex]\lim_{n \rightarrow \infty} S = \lim_{n \rightarrow \infty} \frac{a (1-r^n)}{1-r} = \frac{a}{1-r}[/tex]

Q E D

could you explain the above in layman's terms (i haven't seen anything like that in several years)?

and does it account for the possibility of an infinity of something in a finite space?
 
  • #5
Char. Limit
Gold Member
1,204
14
More correctly, that holds as long as [itex]|r|<1[/itex].

Might also help to add the step [tex]S = \sum_{k = 0}^{n-1} ar^k[/tex] or similar...
Edited those in. Thanks!
 
  • #6
Char. Limit
Gold Member
1,204
14
could you explain the above in layman's terms (i haven't seen anything like that in several years)?

and does it account for the possibility of an infinity of something in a finite space?
Actually, you often get a sum of infinite elements giving a finite number. Well, I'm assuming that's what you mean. And as for layman's terms, I'll try, but I'm not very good at explaining stuff like this, so bear with me.

Basically, I just converted our number .999... into an infinite sum. Then I proved, using the fact that this sum was a geometric series, that such a sum is equal to 1. Therefore, .999 is equal to 1.
 
  • #7
jhae2.718
Gold Member
1,161
20
Here's an explanation of the proof:
Here's a proof that .999...=1.

.999... can be written as the infinite sum as follows:

[tex].9 + .09 + .009 + .0009 + .00009 + \dots = .9 \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n[/tex]
We can write 0.999... as a geometric series. A geometric series is a series of the form [tex]\sum_{k=0}^N ar^k[/tex], where a is a constant and r is a constant ratio.

Now, evaluating the sum on the right, we use the fact (proven below) that...

[tex]\sum_{n=0}^\infty r^n = \frac{1}{1-r}[/tex]

for all r with a magnitude less than 1. Using this fact, we find that...

[tex].9 \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n = .9 \frac{1}{1-.1} = \frac{.9}{.9}=1[/tex]
The sum of a geometric series is given by [tex]\frac{a}{1-r}[/tex], where [itex]|r|<1[/tex]. So, this series sum to exactly 1. (I.e., the infinitude of partial sums is one.)

Now, to prove that fact that we used, note the proof below:

[tex]S = \sum_{k=0}^{n-1} a r^k = a + a r + a r^2 + a r^3 + a r^4 + ... + a r^{n-1}[/tex]
We begin by examining a more general case of a geometric series which takes the sum of [itex]n-1[/tex] terms.

[tex]rS = a r + a r^2 + a r^3 + a r^4 + a r^5 + ... + a r^n[/tex]
We can multiply by the constant ratio r to get an expression that will allow us to get rid of most of the terms, shown below:
[tex]S - rS = a - a r^n = a (1 - r^n)[/tex]

[tex]S(1-r) = a (1 - r^n)[/tex]

[tex]S = \frac{a (1-r^n)}{1-r}[/tex]
Factoring and solving for S, we get the expression for the sum of an arbitrary geometric series.
Now let n go to infinity. For r with a magnitude less than 1, r^n tends to 0 as n tends to infinity. Thus...
For our case, we want the special case where [itex]n \to \infty[/itex]. To ensure this limit exists, we need to make the [itex]r^n[/itex] term go to zero. We do this by saying that [itex]|r|<1[/itex].

Consider a fraction [tex]\frac{a^n}{b^n}[/tex]. If [tex]\frac{a^n}{b^n} < 1 \Rightarrow a^n < b^n[/tex].
If we take the limit as [itex]n \to \infty[/itex] of this, bn grows faster than an, which implies the limit goes to zero.

[tex]\lim_{n \rightarrow \infty} S = \lim_{n \rightarrow \infty} \frac{a (1-r^n)}{1-r} = \frac{a}{1-r}[/tex]

Q E D
We then take the limit and end up with the sum for an infinite geometric series, which we used to find [tex].9 \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n[/tex]

As for the concept of a sum of an infinite amount of terms converging to a finite number, a lot of people have difficulty accepting this.

Let's consider the geometric series [tex]\sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^n[/tex]. If we use the formula to find the sum, we get:
[tex]\sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^n = \frac{1}{1-\frac{1}{2}}=\frac{1}{\frac{1}{2}} = 2[/tex]

To visualize this, consider an empty rectangle. We add a square that is half of the rectangle. We then add another rectangle that is half of the remaining empty rectangle, and so on...
350px-Geometric_progression_convergence_diagram.svg.png

Image from Wikipedia
As you can see, although there are an infinite number of terms, they never get outside the bounds of the first rectangle.
 
Last edited:
  • #8
78
1
Actually, you often get a sum of infinite elements giving a finite number. Well, I'm assuming that's what you mean. And as for layman's terms, I'll try, but I'm not very good at explaining stuff like this, so bear with me.

Basically, I just converted our number .999... into an infinite sum. Then I proved, using the fact that this sum was a geometric series, that such a sum is equal to 1. Therefore, .999 is equal to 1.
layman as in a man who hasn't done calculus in over 5 years, took college algebra in a 20 day course 7 years ago (and therefore never had a solid foundation in it), and is now about a third of the way through going through his college algebra book again.
 
  • #9
34,521
6,211
Actually, you often get a sum of infinite elements giving a finite number. Well, I'm assuming that's what you mean. And as for layman's terms, I'll try, but I'm not very good at explaining stuff like this, so bear with me.

Basically, I just converted our number .999... into an infinite sum. Then I proved, using the fact that this sum was a geometric series, that such a sum is equal to 1. Therefore, .999 is equal to 1.
Correction: What Char. Limit proved was that .999... is equal to 1. He inadvertently omitted the ellipsis (...).

layman as in a man who hasn't done calculus in over 5 years, took college algebra in a 20 day course 7 years ago (and therefore never had a solid foundation in it), and is now about a third of the way through going through his college algebra book again.
 
  • #10
jhae2.718
Gold Member
1,161
20
layman as in a man who hasn't done calculus in over 5 years, took college algebra in a 20 day course 7 years ago (and therefore never had a solid foundation in it), and is now about a third of the way through going through his college algebra book again.
Are you familiar with sequences and series?
 
  • #11
Char. Limit
Gold Member
1,204
14
Are you familiar with sequences and series?
And limits. Don't forget limits.
 
  • #12
78
1
And limits. Don't forget limits.
not any longer. the only one that i remember vaguely is limits.
 
  • #13
statdad
Homework Helper
1,495
35
".999999.... never equals one. (i still find it hard to believe that saying that warrants an infraction). as the decimal points increase onto infinity the space between 1 and .9999... becomes smaller"

Yes: each time you stop at a finite number of decimals you define another number, and the differences between those numbers and 1 gets smaller.

, but the two never do connect to equal the exact same number. there is an infinite number of spaces to which decimals can extend (as noted by pie), so there is no reason for .999999999.... ever to change to one.

"you can argue that .9999999 REPRESENTS one"

I'm reasonably sure you made a typing error here and really mean ".999999..." instead of "0.999999" BUT, in case you didn't, nobody is saying 0.9999999 represents one.


"in the language system you happen to be using, but .99999... will never actually equal one in real life."

You are stuck in the idea (whether you realize it or not) of thinking in finite blocks of 9s. The notation [itex] 0.999 \cdots [/itex] is a single object: the [itex] \cdots [/itex] informs us that the 9s march on forever.

I think your problem is based in your (admitted) lack of familiarity with these topics as much as it is skepticism of the usefulness and applicability of infinite decimal expansions.
 
  • #14
34,521
6,211
In response to the below quote I have written this. The idea that arguing is "spreading misinformation" is ridiculous as argument is the prime mover behind the evolution of the general intelligence of humanity as a whole. If that's the sort of attitude this forum carries with it though, fine, I'll leave. I came here to become educated, not merely indoctrinated. (education being something highly dependent on the dialogue created by argument and indoctrination being something highly dependent on the lack of dialogue created by argument). So, I submit the below in an attempt to meet out my expectations of education. Whether I get such education by means of knowing more about the world or more about the current subject, .999.... will depend on the reaction I get.

So, here goes...


decimals of fractions never accurately equal those fractions. for example .111111111111111...never fully equals 1/9 (which applies directly to a certain proof) but for practical purposes you can assume it does.
Since you are asserting that .11111... is not the same as 1/9, please tell me how far apart they are.

Your statement "decimals of fractions never accurately equal those fractions" doesn't match your example. The decimal fractions .1, .11, .111 are successively closer to 1/9, but as you point out, none of them is equal to 1/9. The notation .111..., means an infinitely repeating pattern of 1's, and this shorthand notation does equal 1/9.
.999999.... never equals one. (i still find it hard to believe that saying that warrants an infraction). as the decimal points increase onto infinity the space between 1 and .9999... becomes smaller, but the two never do connect to equal the exact same number.
They aren't equal at any finite decimal place, true, but we're talking about 9's extending infinitely far to the right of the decimal point.
If they aren't equal, their difference must be nonzero. Care to do the subtraction for me?
there is an infinite number of spaces to which decimals can extend (as noted by pie), so there is no reason for .999999999.... ever to change to one.

you can argue that .9999999 REPRESENTS one in the language system you happen to be using, but .99999... will never actually equal one in real life.
You're being sloppy here. No one is arguing that .9999999 is equal to 1. We are saying however, that .9999999.. is exactly equal to 1. Note the ellipsis.
let's say you want to count all of the entities (particles, doodads, whatever) in the universe in relation to the whole. we will represent the whole with 1. once you have accounted for every item within the universe, lets represent each such item with a 9 after the decimal point, you have counted the whole of it and achieved accounting for 1 universe. since the universe expands on forever, the actual ability to do such a thing does not exist. therefore you can continue on counting 9's in .99999... without ever actually achieving 1.

If you were to say otherwise you would have to also say that infinity is impossible, and ergo pie would be impossible (along with other things). as it is a fact that the 9's after the decimal point can go on into infinity it is also a fact that .999... never equals one except for practical purposes in which such fine details are not important.

and as for that argument by Hurkyl that i'm "spreading misinformation"

and spreading misinformation? so, you're saying that the losing party to any argument is "spreading misinformation"?
It is a fairly common occurrence around here that someone asks whether .999... is different from 1. It is easy to prove that the two are equal.
how ever is one to have conversations on topics if one is to always be correct (i'm not saying i'm wrong by arguing this by the way), when being incorrect is "spreading misinformation"?

who is the owner of this forum? i would very much like to take this up with them.
 
  • #15
jhae2.718
Gold Member
1,161
20
not any longer. the only one that i remember vaguely is limits.
Ok, I'm going to start with sequences.

A sequence is basically a list of numbers. We usually write sequences in the form [tex]\left\{a_n\right\}_{n=b}^{c}[/tex].

What this means is that for some term [itex]a_n[/itex], we plug in some number b, add it to our list, then plug in some number b+1, add it to our list, and do that until we reach c.

So, for example:
[tex]\left\{n\right\}_{n=1}^5 = \{1,2,3,4,5\}[/tex]
A more complicated example:
[tex]\text{Find }\left\{n^2-n\right\}_{n=2}^6[/tex]
The first term, [itex]a_2[/tex], is:
[tex]a_2 = 2^2-2 = 2[/tex]
We do this for 2-6, to yield:
[tex]\left\{n^2-n\right\}_{n=2}^6 = \{2,6,12,20,30\}[/tex]

Then we have infinite sequence; a general infinite sequence is given by:
[tex]\left\{a_n\right\}_{n=0}^\infty = \{a_0, a_1, a_2, \dots\}[/tex]

This goes on forever. Now that we have the concept of an infinite sequence, we need to introduce convergence. When we say something converges we mean than, as [itex]n \to \infty[/itex], each sequence term gets closer to a finite number. If each term gets bigger or terms don't approach the same number, we say the sequence diverges.

Since you said you vaguely remember limits, we can define a convergent sequence as any sequence for which [tex]\lim_{n \to \infty} a_n = L[/tex] holds, where L is a finite number. Otherwise, a series is divergent.

With me so far?
 
Last edited:
  • #16
22,089
3,293
OK, the 0.999... question... again

Question for the moderators: isn't it a good idea to put an FAQ in the math forums where such things are explained? So that people who want to post on the issue, have at least heard what we think of it? If you want, I'm willing to write such an FAQ, containing basic questions like 0.999... and division by zero.


you can argue that .9999999 REPRESENTS one in the language system you happen to be using, but .99999... will never actually equal one in real life.
Then the question obviosuly becomes: what is 0.9999... in real life? Can you give me an example what it is?

Here's an easy proof that 1=0.999...
Let x=0.999...
Then 10x=9.999...
Then 10x-x=9.999... - 0.999...=9
Then 9x=9
Then x=1

Of course, this isn't really a proof, it's merely an indication why this should be true. The real proof that 1=0.999... can only be given with the explicit construction of the real numbers, i.e. when working with Dedekind sets or Cauchy fundamental sequences.

The truth is actually that we've CHOSEN 1 to be equal to 0.999... If you want, you can choose it another way, but then there's a lot of arithmetic that won't hold. So in order to keep the nice laws of arithmetic, we have to define 1=0.999... You may not like it, but it's much more beautiful this way.
This paragraph is actually personal opinion, you may find many mathematicians who disagree. But you won't find any mathematician who says that 1=0.999... isn't true.
 
  • #17
34,521
6,211
A couple of fine points...
Ok, I'm going to start with sequences.

A sequence is basically a list of numbers. We usually write sequences in the form [tex]\left\{a_n\right\}_{n=b}^{c}[/tex].

What this means is that for some term [itex]a_n[/itex], we plug in some number b, add it to our list, then plug in some number b+1, add it to our list, and do that until we reach c.

So, for example:
[tex]\left\{n\right\}_{n=1}^5 = \{1,2,3,4,5\}[/tex]
A more complicated example:
[tex]\text{Find }\left\{n^2-n\right\}_{n=2}^6[/tex]
The first term, [itex]a_2[/tex], is:
[tex]a_2 = 2^2-2 = 2[/tex]
We do this for 2-6, to yield:
[tex]\left\{n^2-n\right\}_{n=2}^6 = \{2,6,12,20,30\}[/tex]

Then we have infinite sequence; a general infinite sequence is given by:
[tex]\left\{a_n\right\}_{n=0}^\infty = \{a_0, a_1, a_2, \dots\}[/tex]

This goes on forever. Now that we have the concept of an infinite series,
infinite sequence
we need to introduce convergence. When we say something converges we mean than, as [itex]n \to \infty[/itex], each sequence term gets closer to a finite number. If each term gets bigger, we say the sequence diverges.

Since you said you vaguely remember limits, we can define a convergent sequence as any sequence for which [tex]\lim_{n \to \infty} a_n = L[/tex] holds, where L is a finite number.

Then, a sequence is divergent if: [tex]\lim_{n \to \infty} a_n = \infty[/tex]
True, but a sequence can be divergent without being unbounded; e.g., {(-1)n}.
With me so far?
 
  • #18
jhae2.718
Gold Member
1,161
20
A couple of fine points...infinite sequence
Oops...typo'd. Thanks for catching that.

True, but a sequence can be divergent without being unbounded; e.g., {(-1)n}.
Very true. I forgot to include those cases; I think I've worded it better now.
 
Last edited:
  • #19
841
0
So, I submit the below in an attempt to meet out my expectations of education. Whether I get such education by means of knowing more about the world or more about the current subject, .999.... will depend on the reaction I get.

So, here goes...






decimals of fractions never accurately equal those fractions. for example .111111111111111...never fully equals 1/9 (which applies directly to a certain proof) but for practical purposes you can assume it does. [snip]

.999999.... never equals one. [snip] as the decimal points increase onto infinity the space between 1 and .9999... becomes smaller, but the two never do connect to equal the exact same number. there is an infinite number of spaces to which decimals can extend (as noted by pie), so there is no reason for .999999999.... ever to change to one.

you can argue that .9999999 REPRESENTS one in the language system you happen to be using, but .99999... will never actually equal one in real life.

.
You people who assert that .999... and .111... can never reach 1 or 1/9 th etc. are simply stuck with the idea that you can consider the term with out the ellipses. You can't it is the ellipses that make the term complete and it is impossible to speak of the term apart from the ellipses. It is trival to speak of the term without the ellipses and say that you will never have the complete number. There is no argument about that. It also is true that the ellipses mean that the term goes ont into infinity and has more decimal places than the total number of particles in the universe (no matter how small the particle), but that of course doesn't mean that the term can not equal a real or rational number. We mathematicians have already proven that it does. See for instance the post of Char Limit. Anyone who says that there must be an error in such proofs are simply not serious mathematicians.
 
  • #20
724
0
Here's an interesting proof, based on knowledge from 4th grade:

I'll pose it as questions, so as to allow the reader to reach their own conclusion.

Are there real numbers between any two unequal real numbers?
What is a number between .999... and 1?

You learn in 4th grade, in simpler terms, that the reals are dense. In any dense set, there are infinitely many members between any two members. All real numbers have a decimal expansion. There is no decimal expansion between .999... and 1, therefore there is no number between them. There is no number between them, therefore they are equal.
 
  • #21
841
0
could you explain the above in layman's terms (i haven't seen anything like that in several years)?

and does it account for the possibility of an infinity of something in a finite space?
Yes with the ellipses, i.e. considering the number of terms as infinite, for instance the 1 by 2 rectangle posted by jhar2.718 and repeatedly dividing the rectangles in half. The sum of the infinite parts is indeed 2 and not something less and this can be proven mathematically. Also, try googling Zeno's Paradox.
 
Last edited:
  • #22
151
0
Curd, the ellipses imply the limit of SUM (.9/10^n) as n --> infinity, so if you are arguing that .9 repeating does not equal 1, then you are arguing, in principle, against limits, and suggesting that all of Calculus is "wrong."

Would you also, for instance, argue that 1/999.972 (999.972 kg. being the mass of water contained inside a 1 meter cube box at ~3.98 degrees Centigrade) is not equal to lim n --> infinity .001*SUM (28^(1*n)/10^(6*n))?

In the real physical world, if we are talking about the mass of water (at maximum density under one atmospheric pressure), you could make a very rational case that those numbers up above are approximations (even as scientists toy with the idea of making the kilogram an SI derived unit of measurement...). After all, the original ratio back in the late 1700's was 1/1000 (units: m/kg).

But in the idealized mathematical world, which is what we are dealing with here, then the fraction representing the ratio of the edge length of a 1 meter cube containing water to the mass of the water contained in that cube, 1/999.972 is precisely equal to lim n --> infinity .001*SUM (28^(n)/10^(6*n)). That which the fraction represents may be inaccurate, but the infinite sum representation of that fraction is 100% accurate.

Seems to me that you have a problem with infinite sums in general and are only willing to accept partial sums as mathematically valid. If so, then that should be the focus of your argument (good luck), not does .9 recurring equal 1. Because .9 recurring DOES equal one (as the sum of an infinite series), even if the way it "gets there" and the constituent (infinite) parts from which it is constructed are different from, say, n - (n-1) = 1

P.S. Or, just think of it this way: If you draw a circle of diameter 1 meter inside a square of edge length 1 meter, then is there an exact area inside the circle and also an exact area outside the circle but inside the square? If so, is that area any less exact simply because one could never, in an infinite amount of time, write out that area in decimal notation form?
 
Last edited:
  • #23
47
0
well Curd, I think I have a way you can understand this plainly. First I'll give a simple formal proof and then a more informal one which seems to follow your thinking strategy against the idea that 0.99999... = 1.

Formal but simple proof:

(1/3) = (0.333333333...)
(1/3)*3 = (0.999999999...)
3/3 = 0.99999999...
1 = 0.999999999

Then a more informal proof which seems to be following the way you are thinking against the 0.999... = 1 idea.

You seem to think that because adding another 0.09 and a 0.009 and a 0.0009 and so in is never enough for the expression to finally equal 1. Once we get up to 0.99999, it is still 0.000001 short of equaling 1. Remember that we are saying 0.999999... with an INFINITE number of 9s is equal to 1. If you still think that there will never come a nine which is exactly enough to make the expression equal 1, you are right. Because if a nine does come to make it equal, it truly isn't an infinite decimal and another 0.00000000000000001 difference still exists.

We are saying that the number of nines is infinite. So, the difference between 0.999... and 1 is equal to:

0.000000000000...1

with an infinite number of zeros, the number is never quite ready to add on the extra one. The one essentially doesn't exist, the number of zeroes will carry on forever and ever and the one will never show up. I hope you see the connection between the idea that there will never be a 9 in the infinite decimal 0.999999... which will make it equal to 1 and the idea that there isn't ever going to be enough zeroes to make the number ready to add the 0.000000...1



I hope this helps ease your confusion.


Zac
 
  • #24
828
2
I don't know if this argument has been given on this particular thread, but this is how I like to show that .9999...=1.

Let n = .999....
multiply both sides by 10:

10n = 9.999....

Subtract n from both sides:

9n = 9

Divide both sides by 9:

n=1

but by assumption, .99...=n. Thus, .99...=n=1

EDIT: My calc prof. taught us this one just before we started with series.
 
Last edited:
  • #25
78
1
OK, the 0.999... question... again

Question for the moderators: isn't it a good idea to put an FAQ in the math forums where such things are explained? So that people who want to post on the issue, have at least heard what we think of it? If you want, I'm willing to write such an FAQ, containing basic questions like 0.999... and division by zero.




Then the question obviosuly becomes: what is 0.9999... in real life? Can you give me an example what it is?

Here's an easy proof that 1=0.999...
Let x=0.999...
Then 10x=9.999...
Then 10x-x=9.999... - 0.999...=9
Then 9x=9
Then x=1

Of course, this isn't really a proof, it's merely an indication why this should be true. The real proof that 1=0.999... can only be given with the explicit construction of the real numbers, i.e. when working with Dedekind sets or Cauchy fundamental sequences.

The truth is actually that we've CHOSEN 1 to be equal to 0.999... If you want, you can choose it another way, but then there's a lot of arithmetic that won't hold. So in order to keep the nice laws of arithmetic, we have to define 1=0.999... You may not like it, but it's much more beautiful this way.
This paragraph is actually personal opinion, you may find many mathematicians who disagree. But you won't find any mathematician who says that 1=0.999... isn't true.

doesn't matter if it's beautiful or not. if it is not reflective of reality then it's not accurate. what do physicists say? that's what will really count.
 

Related Threads on I would like to argue about .999

  • Last Post
Replies
4
Views
2K
Replies
6
Views
1K
  • Last Post
Replies
13
Views
4K
  • Last Post
2
Replies
25
Views
8K
  • Last Post
Replies
8
Views
1K
  • Last Post
2
Replies
45
Views
8K
Replies
11
Views
2K
Top