Is 0.999... Truly Equal to 1?

[Integral]

lets see if this works.

$$.99\dot{9}$$

I am not sure what you mean by the reciprocal? The Reciprocal of what?

 

[Adam]

1 is to $$0.\dot{9}$$
As 0 is to... ?

Is there some symbol or way to write that it's an infinitesimally small value? Apart from zero, I mean.

 

[phoenixthoth]

I believe that $$0.\bar{9}=1$$ is what we're after.

As for the reciprocal, the reciprocal of an infinite series is generally difficult. But for our case, let's see...

$$0.\bar{9}=0.9+0.09+0.009+0.0009+...$$
$$0.\bar{9}=9\cdot \frac{1}{10}+9\cdot \left( \frac{1}{10}\right) ^{2}+9\cdot \left( \frac{1}{10}\right) ^{3}+9\cdot \left( \frac{1}{10}\right) ^{4}+...$$
$$0.\bar{9}=\sum_{k=1}^{\infty }9\cdot \left( \frac{1}{10}\right) ^{k}$$

For finite n, we have that
$$ax+ax^{2}+...+ax^{n}=a\frac{x\left( 1-x^{n}\right) }{1-x}$$
Then to do the reiprocal, we can flip both sides:
$$\frac{1}{ax+ax^{2}+...+ax^{n}}=\frac{1-x}{ax\left( 1-x^{n}\right) }$$.

Now take limits of both sides:
$$\lim_{n\rightarrow \infty }\frac{1}{ax+ax^{2}+...+ax^{n}}=\lim_{n\rightarrow \infty }\frac{1-x}{ax\left( 1-x^{n}\right) }$$
$$\frac{1}{\sum_{k=1}^{\infty }ax^{k}}=\lim_{n\rightarrow \infty }\frac{1-x}{ax\left( 1-x^{n}\right) }$$.

Now let a=9 and x=1/10.

$$\frac{1}{0.\bar{9}}=\lim_{n\rightarrow \infty }\frac{1-1/10}{9\left( 1/10\right) \left( 1-\left( 1/10\right) ^{n}\right) }$$. This reduces to
$$\frac{1}{0.\bar{9}}=\frac{1-1/10}{9\left( 1/10\right) \left( 1-\lim_{n\rightarrow \infty }\left( 1/10\right) ^{n}\right) }$$ and further to $$\frac{1}{0.\bar{9}}=\frac{9/10}{9\left( 1/10\right) \left( 1\right) }$$ and therefore $$\frac{1}{0.\bar{9}}=1$$.

There is a way to represent infinitesimal quantities. In fact, in nonstandard analysis there are no limits. We would say that $$\sum_{k=1}^{n }9\cdot \left( \frac{1}{10}\right) ^{k}$$ is infinitesimally close to 1 for all unlimited n.

In standard analysis, real numbers are constructed (or can be) from the rational numbers. Your question is similar to "how does one express or write a real number if all that is known are rational numbers?" And we can usually do only a pathetic attempt at representing real numbers because their decimal expansions rarely repeat (in some sense). The only thing I can think of is if you get into "equivalence classes" (which are just divisions of a set into parts) then you can say that a nonzero infinitesimal is given by, for example,
[{1,1/2,1/3,1/4,...}]. This is nonzero and smaller than every real number x which is represented by [{x,x,x,x,x,x,x,x,...}]. Note the brackets around the sequence. That means really it's the set of all things equivalent to that sequence. But what I mean by equivalent is hard to describe (in fact I barely remember). If you're really interested look up nonstandard analysis--goldblatt is i believe a good book--and get ready for some things call ultrafilters.

 

[Adam]

WTF is an ultrafilter? I'll go look it up...

 

[Adam]

You know, there really should be a simple symbol or something to show it, rather than writing out a set or such.

 

[Adam]

Oh, for those who use the dot above the last digit, as I do, the code is: ( tex ) 0 . \ dot { 9 } ( / tex )
But use the square braces and remove the spaces.

 

[phoenixthoth]

pi stands for [{3,3.1,3.14,3.141,...}]. If you like, you can let any symbol you choose represent the infinitesimal indicated above. But considering what the "long" symbol stands for, it really is compact notation. However, if it's not compact enough, you can let x equal it and then you have just one letter.

You might want to check out
http://en.wikipedia.org/wiki/Nonstandard_analysis
and
http://en.wikipedia.org/wiki/Hyperreal_number
and
http://en.wikipedia.org/wiki/Infinitesimal

 

[Adam]

Thanks very much for the work you put into the explanations.

 

[krusty the clown]

Nobody in my family but me believes that 1/3 is exactly equal to 0.333...
When I asked what it does equal they told me that you can't write 1/3 as a decimal, even after I showed them a paragraph in a mathematics "encyclopedia" that says all common fractions can be represented by either a finite or an infinite repeating decimal.

I looked for the definition of "common fraction" but I couldn't find one; except possibly that a common fraction is a fraction whose numerator has a smaller magnitude than the magnitude of the denominator.

1)Is 1/3 a common fraction?
2)What is the definition of a common fraction?
3)Are there any definitions that show 1/3=.333...?


-Thanks, Erik

 

[phoenixthoth]

You're welcome.

Nobody in my family but me believes that 1/3 is exactly equal to 0.333...
When I asked what it does equal they told me that you can't write 1/3 as a decimal, even after I showed them a paragraph in a mathematics "encyclopedia" that says all common fractions can be represented by either a finite or an infinite repeating decimal.

I looked for the definition of "common fraction" but I couldn't find one; except possibly that a common fraction is a fraction whose numerator has a smaller magnitude than the magnitude of the denominator.

1)Is 1/3 a common fraction?
2)What is the definition of a common fraction?

I've never heard of a "common" fraction. Maybe they mean common as in a fraction that typically appears on a wrench or in the stock market or a recepie. Truth is any fraction (of integers) has a repeating decimal, common or not.

3)Are there any definitions that show 1/3=.333...?


-Thanks, Erik

Let me give you two ways, one that a third grader would believe (which is not to say that's a bad or good thing) and one that requires some second semester calculus.

1.
Divide 3 into 1 using pencil and paper. You always get the same remainder and there's no reason why the remainder should suddenly change because you're working with the same numbers.

2.
$$0.\bar{3}=0.3+0.03+0.003+0.003+...$$. Then turn that into an infinite series and sum it. You get 1/3 thus showing that $$0.\bar{3}=1/3$$.

 

[krusty the clown]

Thanks phoenixthoth, the problem with "common fraction" might be that I got it out of a really old book, but nevertheless all fractions with integers can be written exactly with decimals, correct.

I have already tried explanation one, and they continue to insist that 1/3 can't be written as a decimal.

As I am currently taking calc II and haven't gotten to infinite series I don't fully understand the second way. I see the general idea behind it, but hopefully I can spend some time on this tonight and figure out how to do it myself; and also how to explain it to my parents.

-Thanks, Erik

 

[phoenixthoth]

Check out specifically geometric series in your book. A geometric sequence has the form a, ax, ax^2, ax^3, ... and a geometric series is the sum of a geometric sequence. All repeating decimals are really geometric series (don't know what the plural of series is). In your book, it will discuss how limits are involved in series and the main point is that an infinite series is the limit of a sequence of partial sums. In other words, if S_1, S_2, S_3,... is a sequence where the subscript denotes how many terms have been added, then the limit of this sequence is the sum of the infinite series.

In the case of 1/3=0.3(bar), you have the sequence
0.3, 0.33, 0.333, 0.3333, ... .

This is the sequence of partial sums of the infinite series
$$\sum_{k=1}^{\infty }3\cdot \left( \frac{1}{10}\right) ^{k}$$.

In your book, they'll prove that
$$\sum_{k=1}^{\infty }a\cdot r^{k}=\frac{ar}{1-r}$$ or something equivalent to that.

In this case, a=3 and r=1/10:
$$\frac{3\left( 1/10\right) }{1-\left( 1/10\right) }=\frac{3/10}{9/10}=1/3$$.

Geometric series are cool for a few reasons, not the least of which is that all repeating decimals are really ones. Another thing is that their sum is actually findable; often infinite series have sums that can only be approximated but not found exactly in a simple way.

 

[Alkatran]

Do it by the difference.

First, ask him if he agrees that:
1/infinity = 0
x^-y = 1/x^y
x*0 = 0
x^infinity = infinity

Then do this:
1 - 0.9 = 1 - 9*10^-1 = .1 = 10^-1
1 - 0.09 = 1 - 9*10^-2 = .01 = 10^-2
skip ahead...
0.999~ = 1 - 9 * 10^-infinity = 1 - 9 * 1/10^infinity = 1 - 1/infinity = 1 - 0 = 1

The difference is 0. The have the same value but a different representation.

 

[Adam]

You know I asked about representing the reciprocal of the $$0.\dot{9}$$, earlier? Would that be the limit $$i\rightarrow0$$ ?

 

[Alkatran]

You know I asked about representing the reciprocal of the $$0.\dot{9}$$, earlier? Would that be the limit $$i\rightarrow0$$ ?

Latex isn't working, please print out your equation

 

[Adam]

You know I asked about representing the reciprocal of the 0.9..., earlier? Would that be the limit i->0 ?

 

[phoenixthoth]

Division is an operation that is defined for real numbers and infinity is not a real number. Therefore, 1/infinity is not equal to 0.

Similarly, other operations are not defined for non real numbers such as exponentiation. You might as well say 1/rabbit=0 or + =1/ .

That's the standard comeback but now let's turn to nonstandard analysis to fix what you have.

Let x and y be hyperreal numbers. Denote their infinitesimal closeness by writing x==y. x==y means |x-y|<r for all positive real numbers r.

The main fact your assuming by writing 1/infinity=0 is that IF x and y are real numbers then x==y if and only if x=y in which case |x-y|=0<r for all positive real numbers r.

The second thing you're doing is dividing non real numbers. While in standard real analysis, this makes about as much sense as dividing apples into oranges (both of which are nonreal numbers), you can kind of do this in nonstandard analysis.

Let x be a hyperreal number. x is called unlimited if |y|<|x| for all real numbers y. Then instead of saying 1/infinity=0 here's the correct statement:

1/x==0 for all unlimited x.

If x is unlimited then 1/x is an infinitesimal. An infinitesimal y is a hyperreal number such that y==0.

x^-y = 1/x^y
x*0 = 0
x^infinity = infinity

The second result holds if x is limited and we can say that a noninfinitesimal x raised to an unlimited power is unlimited.

Then do this:
1 - 0.9 = 1 - 9*10^-1 = .1 = 10^-1
1 - 0.09 = 1 - 9*10^-2 = .01 = 10^-2
skip ahead...
0.999~ = 1 - 9 * 10^-infinity = 1 - 9 * 1/10^infinity = 1 - 1/infinity = 1 - 0 = 1

Technically, 0.9999...==1 because 1/x==0 for unlimited x, not =0. But this is ok since for two real numbers x and y, x==y if and only if x=y. Therefore, 0.9...=1.

0.9... is a real number because it is the least upper bound of the set {0.9,0.99,0.999,0.9999,...} which is bounded above by 1. This is part of the definition of real number.

 

[Aphex_Twin]

2) If 1 is greater than .999..., then by how much, exactly, is it greater? After all, 1 is greater than .9 by .1. It is greater than .99 by .01. It is greater than .99999999999999999999 by .00000000000000000001. And so on, and so on. However, if the number of nines is infinite, then the number of zeros preceding the 1 will be infinite, and 0.000... is obviously equal to 0, which means that 1 is greater than .999... by exactly 0.

If you're taking the discussion to the "Surreal number set" then there can be a "one" after an infinity of "zeros".

 

[Hurkyl]

First off, there are "too many" surreal numbers to fit into a set.


It's very easy to describe sequences that have infinitely many of one thing, and then another; you just need to index your sequence by an infinite ordinal that is not a limit ordinal. For example, the ordinal $\omega + 1$ consists of a copy of the natural numbers adjoined with a single element. ($\omega$) that comes after the natural numbers. You could then write the sequence $\{a_n\}$ where $a_n = 0$ for finite n, and $a_{\omega} = 1$.

This sequence, however, is not a decimal. By definition, decimals are indexed by the limit ordinal $\omega$, the natural numbers, all of whose elements are finite. There is no place in a decimal that has infinitely many predecessors.

End broken record.

 

[hypnagogue]

Thanks for the response Hurkyl.

I believe this thread has run its course. 0.999... = 1 is simply a mathematical fact, and much effort has gone into this thread to provide various rigorous and intuitive support of this fact. There is no need to continue recycling old debates on what is certainly a completely settled issue.