Unlocking the Mystery of .999... and .333...: A Math Puzzle

[rajeshmarndi]

I just show how .999... = 1

Let x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x=1

So according to the above .3333... shows as 3.

Let x = .3333...
10x = 3.333...
10x - x = 3.333... - .333...
9x = 3
x=3

But according to me, .333... should equal to .4 , isn't it?

I must be missing something simple, what is it?

Thank you.

 

[Doug Huffman]

The fallacy of truncation.

 

[DrClaude]

9x = 3
x=3

More like x = 1/3, which is pretty much as expected

 

[DrClaude]

But according to me, .333... should equal to .4 , isn't it?

Why? There is an infnity of numbers between .333... and .4, 0.334 for instance. The argument for .999... only works because there can't be any number between .999... and 1.

 

[rajeshmarndi]

My mistake,
9x = 3
so x=1/3.

Why? There is an infnity of numbers between .333... and .4, 0.334 for instance. The argument for .999... only works because there can't be any number between .999... and 1.

Thanks. Next number after .333... can be .333334 or .3333334 or .3333333333334.

 

[Mark44]

My mistake,
9x = 3
so x=1/3.

Thanks. Next number after .333... can be .333334 or .3333334 or .3333333333334.

There is no "next number" after .333... Integers have such a property, but the rationals and real numbers don't.

 

[Tom_K]

I just show how .999... = 1

Let x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x=1

I must be missing something simple, what is it?

Thank you.

That result comes about from truncating the repetend.

Try keeping the repetend intact, like this:
X = .9999…..
X = 9 x .1111….
10X = 10(9 x .1111….)
10X = 90 x .1111….
10X – X = (90 x .1111….) – (9 x .1111….)
9X = 81 x .1111….
X = 9 x .1111….
X = .9999….

 

[Mark44]

That result comes about from truncating the repetend.

Try keeping the repetend intact, like this:
X = .9999…..
X = 9 x .1111….
10X = 10(9 x .1111….)
10X = 90 x .1111….
10X – X = (90 x .1111….) – (9 x .1111….)
9X = 81 x .1111….
X = 9 x .1111….
X = .9999….

So you have proved that if x = .999..., then x = .999...
This is true, but not useful.

 

[Mark44]

I just show how .999... = 1

Let x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x=1

So according to the above .3333... shows as 3.

Let x = .3333...
10x = 3.333...
10x - x = 3.333... - .333...
9x = 3
x=3

But according to me, .333... should equal to .4 , isn't it?

I must be missing something simple, what is it?

Thank you.

The fallacy of truncation.

The fallacy of dividing 3 by 9 and getting 3.

 

[DrewD]

That result comes about from truncating the repetend.

It also comes about because it is true. This is a shaky method of "proving" it, though.

 

[Apogee]

There is no "next number" after .333... Integers have such a property, but the rationals and real numbers don't.

Yeah, isn't the set of all integers a countable infinitely-large set of size aleph-null, whereas rationals are reals are not countable; therefore, they're considered to be of a larger infinity known as aleph-1. And then you can get even larger sets by taking the power set of aleph-1, then aleph-2, etc. I might be getting a bit off topic, but you brought up something interesting, though I'm not sure I explained it correctly.

 

[Mark44]

Yeah, isn't the set of all integers a countable infinitely-large set of size aleph-null, whereas rationals are reals are not countable;

Did you mean, "whereas rationals and reals..."
The rational numbers are countably infinite, just like the integers. The irrational numbers are uncountably infinite.

therefore, they're considered to be of a larger infinity known as aleph-1. And then you can get even larger sets by taking the power set of aleph-1, then aleph-2, etc. I might be getting a bit off topic, but you brought up something interesting, though I'm not sure I explained it correctly.

 

[Tom_K]

It also comes about because it is true. This is a shaky method of "proving" it, though.

Yes, I accept that it is true, (by definition) I just don’t accept that it is proven by the method of truncating the repetend.

This is a little better, but it still won’t stand up as a rigorous proof:
The sum of a geometric series is equal to: a(1-r^n)/(1-r)
Where a is the first term (0.9 in this case), r is the common ratio (0.1)
if r is less than one, as n goes to infinity the sum becomes just a/(1-r)
In this case, Sum = 0.9/(1-0.1) =1, therefore 0.999… = 1

 

[DrewD]

Tom_K, a proof using a geometric series is rigorous. The "by definition" part should be applied to a decimal being defined as a geometric series. The definition of a decimal is (IIRC), $\sum a_i10^{-i}$ where $a_i$ may be equal to zero for all $i>N$ and there are finitely many values of $i<0$. That is a definition from which 0.99...=1 can be proven.

 

[Apogee]

Did you mean, "whereas rationals and reals..."
The rational numbers are countably infinite, just like the integers. The irrational numbers are uncountably infinite.

Ahhh that makes sense. Rationals and integers are countably infinite, whereas reals are not. Therefore, rationals and integers are of size aleph-null and reals are of size aleph-1. And thank you for the grammatical correction :)

 

[jbriggs444]

Ahhh that makes sense. Rationals and integers are countably infinite, whereas reals are not. Therefore, rationals and integers are of size aleph-null and reals are of size aleph-1. And thank you for the grammatical correction :)

Technically, it is undecidable whether the cardinality of the reals is aleph-1. That is the continuum hypothesis. The cardinality of the reals could be greater than aleph-1. It cannot be smaller.

 

[Apogee]

Technically, it is undecidable whether the cardinality of the reals is aleph-1. That is the continuum hypothesis. The cardinality of the reals could be greater than aleph-1. It cannot be smaller.

Oh, yeah. Basically, it stipulates that no set can have its size between integers and real numbers, but we don't actually know if it's true, right?

 

[nuuskur]

As there is an infinite sequence of 9-s in the decimals, there logically can not be any other real between 0.9(9) and 1 - the density property of the reals states that there is always another real between any 2 reals, conclusion - the 2 reals are equal.
As to what 0.3(3) is exactly equal to is answered by the axiom of continuity - it's equal to 1/3, since it's the least upper bound for the sequence and likewise there can not be another real between 0.3(3) and 1/3.

( I'm curious what the cardinality of reals has to do with the problem at hand )

 

[Tom_K]

What about a number like 1/9999…… where the decimal place is infinitely far to the Right.
Could that theoretically fit between 0.9999…… and 1.0 ?
My only point is that just saying there can be no real number between 0.9999… and 1.0 is not a proof so much as it is a definition.
I do accept the geometric-series proof I posted earlier that they are the same number.

 

[nuuskur]

1/999.. most certainly does not fit between 0.9(9) and 1.
As 9 < 99 < 999 < 9999 ... etc, then their reciprocals 1/9 > 1/99 > 1/999 ..., where 1/9 = 0.1(1) < 9*0.1(1) = 0.9(9).
If you demand a rigorous proof, the problem should be more abstract and we should not be allowed to deal in real numbers, for real numbers have a handful of useful properties to be exploited and neither can you deny the validity of an argument that is based on the axioms and properties of real numbers.

 

[jbriggs444]

What about a number like 1/9999…… where the decimal place is infinitely far to the Right.
Could that theoretically fit between 0.9999…… and 1.0 ?

One of the principles of mathematics is that one is not allowed to use an expression in an argument unless you have a guarantee that the expression is meaningful -- that it denotes something that actually exists. Just because you can string some symbols together in a particular fashion is no guarantee that they are meaningful. In particular, the decimal notation for a real number demands that there be at most a finite number of digits to the left of the decimal point. The meaning for digit strings that do not fit this requirement is left undefined.

If you try to go ahead and define a meaning for such a syntax, you will be departing from the system of real numbers and will be (perhaps) working in the system of p-adics, for instance.

 

[Tom_K]

That’s well said, thank you! I’m not a numbers theorist by any stretch, so the number I posted ( 1/9999…… may well lie outside of the accepted norm for real numbers. My only point was I don’t believe a statement that no real number can lie between 0.9999… and 1.0 is in itself a proof. It is only a statement; a definition. Not that I disagree with it…….I simply question the validity of stating it.
Would the geometric-series-proof I posted earlier be sufficient to validate the statement?

 

[jbriggs444]

That’s well said, thank you! I’m not a numbers theorist by any stretch, so the number I posted ( 1/9999…… may well lie outside of the accepted norm for real numbers. My only point was I don’t believe a statement that no real number can lie between 0.9999… and 1.0 is in itself a proof. It is only a statement; a definition.

Agreed. Perhaps not a definition per se. But a simple consequence of the definition of 0.9999... The hard part is nailing down the structure of the real numbers. Once that is done, the definition for 0.9999... is easy -- there is essentially only one reasonable possibility.

Would the geometric-series-proof I posted earlier be sufficient to validate the statement?

To make it rigorous, one would need to first define what is meant by the sum of a geometric series, proove that convergent geometric series have well-defined sums, that multiplication distributes over convergent infinite series and that the series used in the proof converge.

The key part that is easy to gloss over is the distribution of multiplication over series sums. The sum of an infinite series is not, strictly speaking, a sum. It is a limit of a sequence of partial sums. The fact that multiplication distributes over addition does not automatically apply. It has to be proven.

 

[Mark44]

What about a number like 1/9999…… where the decimal place is infinitely far to the Right.
Could that theoretically fit between 0.9999…… and 1.0 ?

I don't know how you can think that. What you have here is a cumbersome way to write 0.

My only point is that just saying there can be no real number between 0.9999… and 1.0 is not a proof so much as it is a definition.
I do accept the geometric-series proof I posted earlier that they are the same number.

 

[jbriggs444]

I don't know how you can think that. What you have here is a cumbersome way to write 0.

One assumes that the suggestion is that 1 - 1/9999... would be between 0.999... and 1. That would amount to naive reasoning about infinitesimals rather than something that is obviously wrong.

 

[Mark44]

One assumes that the suggestion is that 1 - 1/9999... would be between 0.999... and 1.

Yes, but as posed, Tom_K's question was about whether 1/999... is between .999... and 1.0

 

[Tom_K]

I don't know how you can think that. What you have here is a cumbersome way to write 0.

Of course it is! Just as 0.9999... is a cumbersome way to write 1

 

[micromass]

Of course it is! Just as 0.9999... is a cumbersome way to write 1

With the exception that $0.999...$ is a valid notation, but $999...$ is not.

 

[Mark44]

Of course it is! Just as 0.9999... is a cumbersome way to write 1

My comment was more about what you wrote earlier (quoted below), about whether 1/9999... was between 0.999.. and 1.

I didn't know if you really believed that, or it was a typo on your part. In answer to your question, no, 1/9999... is NOT between .999.. and 1. Furthermore, there are no numbers between .999... and 1, since these are merely two different ways to write the same number.

Here is your quote that I commented on:

What about a number like 1/9999…… where the decimal place is infinitely far to the Right.
Could that theoretically fit between 0.9999…… and 1.0 ?

 

[arydberg]

You said,
Let x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x=1

But you can't arbitrary use varying numbers of decimal points.
change line 2 so 10X = 9.99 that is make it correct
then 10X -X = 9.99 - .999 = 8.991
or 9X = 8.991
or X .999 back where we started.

 

[micromass]

You said,
Let x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x=1

But you can't arbitrary use varying numbers of decimal points.
change line 2 so 10X = 9.99 that is make it correct
then 10X -X = 9.99 - .999 = 8.991
or 9X = 8.991
or X .999 back where we started.

No, you're missing that there are infinitely many numbers after the decimal point, not finitely many. The proof is completely correct.

 

[Mark44]

You said,
Let x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x=1

But you can't arbitrary use varying numbers of decimal points.
change line 2 so 10X = 9.99 that is make it correct

You are mistaken. There are not "varying numbers of decimal points" anywhere in the work above. Evidently you are not familiar with the ellipsis notation, ... , which means "and continuing in the same fashion. In the 2nd line above, the notation 9.999... means that the 9's repeat endlessly.

Similarly, n = 1, 2, 3, ... is shorthand for saying that n can be any positive integer.

then 10X -X = 9.99 - .999 = 8.991
or 9X = 8.991
or X .999 back where we started.

No. .999 is very different from .999...

 

[rajeshmarndi]

Is .999... tends to 1(as it is always less than 1) or equal to 1.

 

[micromass]

Is .999... tends to 1

That is as meaningless as saying that 3 tends to 4.

 

[WWGD]

A standard nonzero Real number can not be indefinitely small, by the Archimedean property. If it can be made indefinitely small, it must equal zero ( this is not so for the non-standard Reals, where you can have infinitesimals ). The expression 1-0.99999... can be made as small as you want , so it must equal 0. Then x-y =0 implies x=y. More formally, you have an argument using the Reals as a metric space, so that d(x,y)=0 iff x=y . Then apply Archimedean principle.