Infinity

[Buffy]

But the proof that it's equal to 1 is by no means unsatisfying Buffy! Remember the geometric series?

 

[math]0.\overline{9}=0.9\sum_{n=0}^{\infty}(0.1)^n=0.9\frac{1}{1-0.1}=0.9\frac{1}{0.9}=1[/math] (and that means exactly!!!)

Sure its unsatisfying because:

 

[math]\sum_{n=0}^{\infty}(0.1)^n=\frac{1}{1-0.1}[/math]

 

is definitional! It's just leaving the [imath]lim[/imath] stuff out! :cheer:

 

Oh you know I'm just playing here dear...

 

If you have an infinite data set where [imath]0.\overline{9999}[/imath] is supposed to describe the probability of a certain property to be exhibited by each element of the data set, does that mean that among any number of elements in the infinite data set you would care to consider that all of them would exhibit this particular property?"

 

(concerning the above) If you were to limit the data set to say [imath]10^{100}[/imath] elements...

Going down the route of analyzing this by bringing in probability is probably going to be less edifying, but there are a few points to be made here.

 

If you are really going to be doing what is called "Discrete Probability"--that is use a limited set of countable elements--then the definition of any particular probability is given by:

 

[math]p = \frac {number\ of\ elements\ possessing\ property\ x} {number\ of elements\ in\ the\ dataset} [/math]

 

Thus if you are going to use a finite data set, [imath]0.\overline{9999}[/imath] really is no longer relevant unless you use the substitution that equates that with [imath]1[/imath]. What you would really have is (simplifying down to a smaller dataset):

 

[math]p = \frac {n-1} {n} = \frac {999999} {1000000} = 0.999999[/math]

 

because there's no "infinite representation" in a discrete probability

 

Conversely if the dataset were infinite, then the substitution [imath]0.\overline{9999}=1[/imath] would say that every element of the infinite set would have the property. But this also presents one of the counter proofs! That is, by saying that the probability is [imath]0.\overline{9999}[/imath] in an infinite set RATHER THAN saying that the probability is [imath]1.0[/imath] you are definitionally saying that exactly one out of the infinite set does not have the property! :cheer:

 

This is basically what you seem to be trying to get at in saying:

...But if the probability for this property to be exhibited per cycle were to be < 1, with any number of 9's after the comma, and you could leave the data set for as long as you wished and the number of elements that did not exhibit the property would start going up.

 

but the "number of elements that did not exhibit the property" would not "go up" because the "inverse probability" implied by [imath]0.\overline{9999}[/imath] is (colloquially) [imath]0.00...infinite\ number\ of\ zeros....0001[/imath] which unfortunately, plugging into the formula for probability is [imath]\frac 1 \infty[/imath]. Mathematically speaking (and this is a corollary to the original argument) these are equivalent to zero, but the 1 in the numerator still *implies* "there is one element without the property x."

 

Note that this is *really* more of an example of the problem of *applying* mathematics to not only "the real world" but to other conceptual--as opposed to concrete--scientific disciplines (don't you dare tell a mathematician that statistics is mathematics or he'll stick his chalk in your eye!).

 

What gets left out is the narrative between the bullets, which would tell us who's going to do what and how we're going to achieve the generic goals on the list, :phones:

Buffy

[KerryK]

Wouldn't that be [math]9x=9.999999'-x?[/math]

 

Tormod,

 

No, it wouldn't. If you take 9.99999' and subtract the 0.99999' from it, you're left with 9. No tricks. But x = 0.99999'.

 

So, 9.999999' - x = 9.

 

And 10x - x = (10-1)x = 9x.

 

So, 9x = 9 and x = 1.

 

(Sorry, I've not had the time to check out LaTex.)

 

Kerry

[Tormod]

I don't get the logic behind that.

 

10x=9.99999'

 

Subtract x:

 

10x-x=9.99999'-x

 

Would yield:

 

9x = 9.99999' - x.

 

Where am I going wrong? Do you divide by nine?

 

-T

[KALSTER]

Tormod:

Let x = 0.999999'

But you could use the same logic to equate anything to 1.:)

[KerryK]

I don't get the logic behind that.

 

10x=9.99999'

 

Subtract x:

 

10x-x=9.99999'-x

 

Would yield:

 

9x = 9.99999' - x.

 

Where am I going wrong? Do you divide by nine?

 

-T

 

Let's take each side of the equation at a time.

 

if x=0.99999' then you surely agree that 10x = 9.9999', right? Multiplying by 10 merely moves the decimal point one position to the right.

 

Now, 10x = x = 9x. Again, we okay with that?

 

Now let's take the RHS (right hand side):

 

9.99999' - x = 9.99999' - 0.99999' = 9.0 (all the nines to the right of the decimal point "cancel" out by the substration). We okay?

 

So 10x-x = 9x = 9. Divide by 9 and x = 1.

 

There's no trick. It's just very simple arithmetic.

 

Kerry

[KerryK]

Tormod: But you could use the same logic to equate anything to 1.

 

Really and truly, NO, you can't. This is all simple arithmetic. No dividing by 0 anywhere. No trick anywhere. Just straight simple arithmetic.

 

Now, I felt the same way when I first saw the proof. (No, it isn't mine; I'd love to take credit for it; the reality is, I had it in algebra a million years ago. I was fascinated by the proof BECAUSE it is so delightfully simple.)

 

Take care,

 

Kerry

[Tormod]

Now, 10x = x = 9x. Again, we okay with that?

 

I must be absolutely blown. The only solution I know to that equation is 0. What am I not understanding?

 

If x=1 then 10x=10, x=1, 9x=9.

 

(Edit: I suspect this is a typo!)

 

9.99999' - x = 9.99999' - 0.99999' = 9.0 (all the nines to the right of the decimal point "cancel" out by the substration). We okay?

 

This one I get. But I don't understand the step inbetween.

 

There's no trick. It's just very simple arithmetic.

 

Now I'm embarrassed. That's what I get for sitting up at night.

[KALSTER]

Kerry:

 

When you say: "Consider 10x = 9.999999' (right?)", don't you already assume [imath]0.\overline{9999}=1[/imath]? I mean, if you were to say that [imath]0.75 = 1 = x[/imath], then your equation would also look like that, i.e.

 

[imath]10x = 9.75[/imath]

 

[imath]10x - x = 9[/imath]

etc., which would not make much sense. Although, I am about the farthest thing from a mathematician, so I am probably talking nonsense.:shrug:

No?

 

Buffy:

 

Thus if you are going to use a finite data set, [imath]0.\overline{9999}[/imath] really is no longer relevant unless you use the substitution that equates that with [imath]1[/imath]. What you would really have is (simplifying down to a smaller dataset):

 

[math]p = \frac {n-1} {n} = \frac {999999} {1000000} = 0.999999[/math]

 

because there's no "infinite representation" in a discrete probability

Agreed. What I was trying to do, was to work with the structure within infinity (if that makes sense). That is, if you were to consider a finite data set, with a finite, [imath]p < 1[/imath] probability for property [imath]x[/imath] to be exhibited, then there would not be any problems. But if you were to consider an infinite number of finite data sets, how would that affect the overall probabilities?

 

By

"But if the probability for this property to be exhibited per cycle were to be < 1, with any number of 9's after the comma, and you could leave the data set for as long as you wished and the number of elements that did not exhibit the property would start going up."

, I meant that if you had a counter that showed the number of elements that exhibited the property after each iteration in a finite data set with a limited probability, then the elements that exhibited the inverse property would start to add up, which would not happen if the probability was [imath]0.\overline{9999}[/imath].

 

Basically, I am wondering what the conditions need be for the finite, [imath]p <1[/imath] probability of an element in a finite data set to exhibit property [imath]x[/imath], to translate into [imath]0.\overline{9999}[/imath] probability for any element in an infinite data set to exhibit the property, and with both co-existing.

 

I hope I stated that clearly.

[nichaleh]

ow my brain hurts.

[KerryK]

I must be absolutely blown. The only solution I know to that equation is 0. What am I not understanding?

 

If x=1 then 10x=10, x=1, 9x=9.

 

(Edit: I suspect this is a typo!)

 

 

 

This one I get. But I don't understand the step inbetween.

 

 

 

Now I'm embarrassed. That's what I get for sitting up at night. ;)

 

You are correct that the line had a typo. That is,

 

Now, 10x = x = 9x. Again, we okay with that?

 

Should have read

 

Now, 10x - x = 9x. Again, are we okay with that.

 

I'm embarassed, too, because I'd already caught myself in the -/= type and was trying to be very careful, but obviously wasn't.

[KALSTER]

KerryK:

Nevermind my last post concerning your proof. I was being an idiot.;)

 

I still can't help feeling that something must be wrong with it though.....

[KerryK]

Kerry:

 

When you say: "Consider 10x = 9.999999' (right?)", don't you already assume [imath]0.\overline{9999}=1[/imath]? I mean, if you were to say that [imath]0.75 = 1 = x[/imath], then your equation would also look like that, i.e.

 

[imath]10x = 9.75[/imath]

 

[imath]10x - x = 9[/imath]

etc., which would not make much sense. Although, I am about the farthest thing from a mathematician, so I am probably talking nonsense.:shrug:

No?

 

Buffy:

 

Agreed. What I was trying to do, was to work with the structure within infinity (if that makes sense). That is, if you were to consider a finite data set, with a finite, [imath]p < 1[/imath] probability for property [imath]x[/imath] to be exhibited, then there would not be any problems. But if you were to consider an infinite number of finite data sets, how would that affect the overall probabilities?

 

By , I meant that if you had a counter that showed the number of elements that exhibited the property after each iteration in a finite data set with a limited probability, then the elements that exhibited the inverse property would start to add up, which would not happen if the probability was [imath]0.\overline{9999}[/imath].

 

Basically, I am wondering what the conditions need be for the finite, [imath]p <1[/imath] probability of an element in a finite data set to exhibit property [imath]x[/imath], to translate into [imath]0.\overline{9999}[/imath] probability for any element in an infinite data set to exhibit the property, and with both co-existing.

 

I hope I stated that clearly.

 

Kalster,

 

I'm really doing no such thing (that is, assuming [imath]0.\overline{9999} = 1[/imath]). Nowhere in my derivation do I say this.

 

Honestly, with all humility, you're all making this too hard. It's a very simple algebraic/arithmetic derivation.

 

Let [imath]x = 0.\overline{9999}[/imath]

 

Then

 

[imath] 10x = 10*\overline{9999} = 9.\overline{9999}[/imath]

 

because you're just moving the decimal point to the right one place...that's what multiplying by 10 does

 

Now, subtract x from both sides.

 

[imath] 10x - x = 9.\overline{9999} - \overline{9999}[/imath]

 

Then, just doing the math on EACH side (that is 10x - x = 9x AND [imath] 9.\overline{9999} - 0.\overline{9999} = 9[/imath]):

 

9x = 9

 

Divide both sides by 9 and you get

 

x = 1.

[KerryK]

KerryK:

Nevermind my last post concerning your proof. I was being an idiot.;)

 

I still can't help feeling that something must be wrong with it though.....

 

Kalster,

 

I was in the process of writing my last reply to you when you were submitting this, so I didn't see it soon enough.

 

Nevertheless, I thank you for maybe now understanding it.

 

And, yes, I agree with your last sentence, there DOES seem to be something wrong with it. But, there isn't. Really.

[Buffy]

Agreed. What I was trying to do, was to work with the structure within infinity (if that makes sense). That is, if you were to consider a finite data set, with a finite, [imath]p < 1[/imath] probability for property [imath]x[/imath] to be exhibited, then there would not be any problems. But if you were to consider an infinite number of finite data sets, how would that affect the overall probabilities?

It wouldn't! It doesn't matter if you divide up the elements in this way, the overall probability is going to be the same as [imath]p[/imath]!

 

Let's take a look at this as you proposed it:

...if you had a counter that showed the number of elements that exhibited the property after each iteration in a finite data set with a limited probability, then the elements that exhibited the inverse property would start to add up,

Now any particular "set" could have any number of elements on either side of the probability--just as if you do a coin-flip, [imath]p=0.5[/imath] but any particular set could have any number of heads, or all heads for that matter--but as you increase the number of data sets, the *average* number of "tails" (your inverse probability)--*per set* would converge to your probability [imath]p[/imath]. If you make this number of sets *infinite* then you will have exactly [imath]p[/imath] by definition!

 

So yes, the counter would go up, but never--over the long run--outside of a well defined range (a function of the number of data sets!) from [imath]p[/imath].

 

...which would not happen if the probability was [imath]0.\overline{9999}[/imath]

Right! Because its the same as 1.0, which means that your inverse probability counter would always be zero!

 

Events in the past may be roughly divided into those which probably never happened and those which do not matter, :phones:

Buffy

[CraigD]

I still can't help feeling that something must be wrong with it though.....

And, yes, I agree with your last sentence, there DOES seem to be something wrong with it. But, there isn't. Really.

We seem to have reached the consensus that, despite seeming somehow wrong, 0.[9] = 1 really is true.

 

The “[]” notation means that the part within the bracket repeats. It’s a bit easier to write in text, especially in computer programs, than notations like [math]0.\overline{9}[/math], and more precise than using …s, which can get you into trouble with numbers like 0.123[456], where not all of the preceding digits repeat.

 

Note that every rational number (a number that can be written as a fraction using integers, eg: 1/2, but not [math]\sqrt2[/math] or [math]\pi[/math]) can be considered to have a repeating part of at least one (eg: 1/2 = 0.5[0]).

 

Some people find the sense of wrongness reduced by examples like:

1/3 + 1/3 + 1/3 = 0.[3] + 0.[3] + 0.[3] = 0.[9] = 1.

 

The false sense of wrongness, and confusion, stems I think from a failure to appreciate that when you write something like 0.[3] or [math]0.\overline{3}[/math] or 0.333…, you’re not really writing the full representation of a rational number, but an algorithm for generating the full representation. This is a really deep “map is not the territory” kind of cognitive paradigm thing-y leading down deep philosophical and psychological rabbit holes, which I’ll now stop mentioning. :)

 

Practically, for any rational number, you can sidestep the need for repeating parts after the radix point (a more general term for the decimal point in base 10) by choosing a base other than 10 (eg: [math]1/3 = 0.\overline{3}_{10} = 0.1_3[/math]), but no matter what base you chose, the 0.999… conundrum will always be there, so it’s best, IMHO, to come to grips with it.

 

It’s worth noting that the derivation

A = 0.[9]

10A = 9.[9]

10A - A = 9.[9] - 0.[9]

9A = 9

A = 1

is a special case of a general technique for finding the numerator and denominator of a repeating fraction.

 

For the case there are no non-repeating digits (eg:

12/34 = 6/17 = 0.[3529411764705882],

as opposed to

123/456 = 41/152 = 0.269[736842105263157894],

where “269” are not part of the repeating sequence of digits), the technique is to take the repeating sequence of digits as the numerator, and [math]B^{L_R}-1[/math] as the denominator, where [math]B[/math] is the base of the numeral system (usually 10), and [math]L_R[/math] is the length of the sequence. For example,

0.[123] = 123/999 = 41/333.

 

For case where there are repeating digits, you need just use [math]\frac{N ( B^{L_R} -1) +R}{( B^{L_R} -1) B^{L_N}}[/math]

where [math]N[/math] and [math]R[/math] are the non-repeating and the repeating parts, and [math]L_N[/math] is the length of the non-repeating part. For example,

[math]0.123\overline{456} = \frac{123 \cdot 999+ 456}{999000} = \frac{123333}{999000} = \frac{41111}{333000}[/math]

 

Repeating digits are an interesting and well-studied subject. They’ve been discussed in several threads, including 4639 and 13369.

[Tormod]

Okay. So I started thinking (bad idea) and began wondering.

 

When exactly does

 

[math]0.999999999......=1[/math]

 

become more true than

 

[math]0.999999999.............8=1[/math]

 

?

 

Is it simply because we define 0.[9] as an infinite set?

[CraigD]

When exactly does

 

[math]0.999999999......=1[/math]

 

become more true than

 

[math]0.999999999.............8=1[/math]

 

Is it simply because we define 0.[9] as an infinite set?

This is one of the rabbit holes I warned of in my previous post! All brace yourselves for philosophically ladened writing!

 

0.999999999.............8, which could be written [math]\left ( \sum_{n=1}^{x} \frac9{10^n} \right ) + \frac8{10^{x+1}} [/math]. where [math]x[/math] is an integer greater than 9, is not equal to 1. It’s equal to exactly [math]1 - \frac2{10^{x+1}}[/math].

 

It’s not meaningful to set [math]x=\infty[/math], because its not meaningful to perform arithmetic operations like “+1” on infinite numbers. In other words, you can’t put anything after the “…” without changing its meaning from “an infinitely number of repeats of the preceding numeral” to “a finite number of repeats of the preceding numerals”, and what you’ve written from a repeating decimal to a terminating one – or, if you prefer, a repeating decimal where the repeating part is 0, ie: 0.999…98[0].

 

Reusing the “1/3” example in my previous post, 0.999999999.............8 = 1/3 + 1/3 + ”something a little less than 1/3”.

 

0.[9], which can also be written [math]\sum_{n=1}^{\infty} \frac9{10^n}[/math] is the same number as is written 1. It’s a single number, not the set containing an infinite number of members {0.9, 0.99, 0.999, …}. The important idea here is that a number is a single, ideal thing not the description of an algorithm to approximate or exactly specify the number, or the sequence of numbers generated by such an algorithm.

 

Strictly speaking, a number is not the same thing as any glyph or vocal utterance used to describe the number. The number we call “one” is not the same as the string “one”, the numeral “1”, or the sound of someone saying the word for one in any language.

 

These are the sort of "the ideal is not the real"/"the map is not the territory"–type philosophical points that mathematicians both delight in and strive to avoid. You want to “get” them, but as soon as you do, imagine they never existed, until you're to confused to work, and need to "get" something in some way similar. :)