Author Topic: Why .9̅ ≠ .999...  (Read 3889 times)

Offline Daniel

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Re: Why .9̅ ≠ .999...
« Reply #15 on: December 25, 2016, 02:08:56 PM »
e.g. 1/3 + 2/3
1/3 = .3̅, but its p is 3⅓ (because after all the threes you have more threes)
2/3 = .6̅, but its p is 6⅔ (because after all the sixes you have more sixes)
So 1/3 + 2/3 = .9̅, but 3⅓+6⅔ = 10, so the 1s carry and the 9s are zeroed out, so the overall .9̅ becomes 1.
But wouldn't .3̅ have a p value of 3.3̅? Not 3⅓. So the total would be 9.9̅ which is still less than 10 and doesn't allow for the carry. Unless each of the ps has a p of its own, but then you'd need to somehow get around that infinite regression.

Perhaps the mathematical finitists are right afterall...
 

Offline Kreuzritter

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Re: Why .9̅ ≠ .999...
« Reply #16 on: January 04, 2017, 06:45:58 AM »
I am not a mathematician, but I have read things where matemeticans attempt to show that 1=0.999...  I've never seen what Daniel describing using fractions.

Rather, I have seen something like this:

  • Begin with x = 0.999...
  • Multiply by 10, and you get 10x = 9.999...
  • Take the difference of the two equations, and you get 9x = 9
  • Divide by 9, and you get x = 1
  • Therefore, 1 = 0.999....

As I said, I am not a mathematician, so I have no idea if there is something wrong with this that I am not seeing, but it seems to make sense.

If 1 = 0.999..., then at any place in the proof we should be able to replace x=0.999... with 1 and obtain the same result.

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 1

Therefore

9x = 8.999...

Now divide by 9.  :cheeseheadbeer:

Of course the fundamental problem with these kinds of discussions is that we haven't precisely defined what we mean by the symbol 0.999... in the first place, and we need to know that before we can even carry out such operations as subtraction  and multiplication on it. Generally, it actually stands for this



and the value of such an "infinite series" is defined as the limit at infinity of the n-th partial sum. In this case it's 1. That doesn't mean that if we add the terms of the series together forever we eventually get to 1 and neither does it mean that 0.999... is identical with the natural number 1, although most people will certainly misunderstand it that way. 0.999... REPRESENTS an "infinite series"and the symbolism of an "infinite series" itself just represents a limit at infinity of a finite series. Problems arise when people, mathematicians and laymen alike, draw metaphysical conclusions from these things without paying attention to what they actually mean.




 

Offline Daniel

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Re: Why .9̅ ≠ .999...
« Reply #17 on: January 04, 2017, 10:29:44 AM »
Ok, I've finally figured out what I was doing wrong.

I had falsely assumed the premise that .1̅ is equal to 1/9, and that .3̅ is equal to 1/3. But in reality they're not, since 1/9 and 1/3 are simple constants which are just 1/9 and 1/3 respectively, whereas .1̅ and .3̅ are composite, each consisting of a never-ending series of terms (1/10+1/100+1/1000+... and 3/10+3/100+3/1000+... respectively).

So I take back what I said before. 0.999... is NEVER equal to 1 since it's always just 9/10+9/100+9/1000+... .
« Last Edit: January 04, 2017, 06:52:40 PM by Daniel »
 

Offline Daniel

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Re: Why .9̅ ≠ .999...
« Reply #18 on: January 04, 2017, 01:42:59 PM »
If 1 = 0.999..., then at any place in the proof we should be able to replace x=0.999... with 1 and obtain the same result.

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 1

Therefore

9x = 8.999...

Now divide by 9.  :cheeseheadbeer:
Not so fast. If the purpose of the proof is to prove that x = 1, then 1 cannot be plugged in for x during the proof. That would be begging the question or circular reasoning or one of those fallacies.
And besides, I think anyone who believes that 9.999.../10 = 1 would also believe that 8.999.../9 = 1.

Quote
and the value of such an "infinite series" is defined as the limit at infinity of the n-th partial sum.
My problem with that is, "defining" the sum as the limit seems to violate the principle of non-contradiction. Because a limit and a sum are two separate things. The limit is the value that the partial sum approaches as you make i really big, whereas the sum is the value you would get were you to actually add them all up. Clearly the limit is 1. But the series has no sum, since it's impossible to add up all the terms (there will always be another term to add). And even theoretically if you could somehow add them all up, it would still be less than 1. Because if you move only halfway across the room, and halfway again, and halfway again, you're never going to reach the other side.

Anyway, I have come across a video that makes this all easy to understand. (And part 2.)

Basically, the reason the proof doesn't work is because the multiplication followed by subsequent subtraction is illegal.

Begin with x = 0.999...
Multiply by 10, and you get 10x = 9.999...
Take the difference of the two equations, and you get 9x = 9 (<-- the problem)
Divide by 9, and you get x = 1
Therefore, 1 = 0.999....

I'll first mention, "0.999..." is the same thing as the sum of the infinite series "9/10, 9/100, 9/100, ...". And as I've mentioned before, there is no actual sum. There is, however, a partial sum of the first n terms in the series, since you can potentially add up all terms 9/10^n from n=1 to whatsoever value of n you choose. This partial sum can be calculated according to the formula 1 - 1/10^n.

So there's one problem right off the bat: If this "0.999..." number in any way comes from the formula 1 - 1/10^n, then it's automatically not the same thing as 1. Because 1 would be 1 - 0. But no matter what value you make n, 1/10^n will never equal 0.

But as for why the proof doesn't work, it's because when you multiply by 10, the "9.999..." then signifies 10 - 1/10^(n-1), not 10 - 1/10^n. And (10 - 1/10^(n-1)) - (1 - 1/10^n) is NOT 9.
« Last Edit: January 04, 2017, 06:47:38 PM by Daniel »
 

Offline Akavit

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Re: Why .9̅ ≠ .999...
« Reply #19 on: January 05, 2017, 11:33:28 PM »
That's what you get for letting the alphabet into math.

R

Or maybe someone let the math into grammar?

Offline cgraye

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Re: Why .9̅ ≠ .999...
« Reply #20 on: January 06, 2017, 01:17:52 AM »
Hello Daniel,

I think you are looking at some of this the wrong way.

First of all, there is a matter of notation.  You start out with the idea that the ellipsis and the vinculum are different.  But these are just different notations for a repeated decimal.  Whatever 0.999... and 0.9̅  mean, they mean the same thing.

You also start out with the idea that 0.999... means something in itself, and we must discover what it means.  But this isn't true.  0.999... is a sequence of symbols, and we define it to mean something.  And what we define it to mean is the value of the series 9/10 + 9/100 + 9/1000 + ...

So the question is, what is the value of this series?  It is the limit of the partial sums, as you yourself say here:

There is no denying that the limit of .999... (as n approaches infinity) is 1. In fact it is, as proven above.

As Kreuzritter points out, this value of a series is defined to be this limit.

Quote
My problem with that is, "defining" the sum as the limit seems to violate the principle of non-contradiction. Because a limit and a sum are two separate things. The limit is the value that the partial sum approaches as you make i really big, whereas the sum is the value you would get were you to actually add them all up. Clearly the limit is 1. But the series has no sum, since it's impossible to add up all the terms (there will always be another term to add).

What is 2 x 3?  6, obviously.  But why?  Because multiplication is repeated addition, and 2 added to itself 3 times is 6.  Now, can you multiply 2.5 and 3.5?  It would seem not, because you cannot add 2.5 to itself 3.5 times.  You cannot do anything 3.5 times.  And yet, you know this problem has an answer.  You learned to do problems like this in elementary school.  And the way to do it is so easy that you didn't even notice they changed the definition of multiplication on you to make it possible.  They changed the definition of multiplication from repeated addition to a kind of scaling.  But the great thing about it is that when one of your factors in a natural number, this scaling corresponds exactly to repeated addition.  So it wasn't really that the definition of multiplication was changed so much as it was generalized in a useful way, with the special case of it you knew at first still perfectly valid.

So here it is not that the definition of a sum has been changed to something contradictory; it has been generalized to something useful that still includes the old understanding of a sum.  You are right to point out that this is not a "sum" in the simple sense of repeated counting - it is something more general that includes that simple sense.

Quote
And even theoretically if you could somehow add them all up, it would still be less than 1. Because if you move only halfway across the room, and halfway again, and halfway again, you're never going to reach the other side.

And yet you do reach the other side.  You can get up and do it right now.  You really can cross the room, even though at some point during your trip you were half way across the room.  And then another quarter across the room.  And then another eighth across the room.  And every other fraction in that series.

Quote
It does seem to make sense. But I am thinking there must be a trick to it. Looks a lot like this similar (yet completely unrelated) proof I once saw:

Start with 1. Subtract 1 from it. Then add 1. Then subtract 1. Keep doing that forever. What do you end up with?
In reality, it all depends on whether the series ends on +1 or on -1. Because suppose the series was not infinite. We start with 1 which is 1. Then 1-1, which is 0. But 1-1+1 is 1. And 1-1+1-1 is 0 again. And 1-1+1-1+1 is 1. So it basically keeps bouncing back and forth between 0 and 1.
But the series is infinite. Which means it has no answer. I mean, really, how can there be an answer? No answer could possibly make any sense, because no matter how many iterations you have, the series does not approach anything.

But moderns don't like that.

And some of them think that all religions are equal, so they need to come up with a compromise. "How about 1/2?", they say. Because 1/2 is half way between 0 and 1.

1/2 is not the value of this series because someone shrugged his shoulders and came up with a compromise between 0 and 1.  Certainly not a modernist attempting to do the same the thing with religions.  This is Grandi's series, and Grandi was an Italian monk and priest.

This is again a matter of context, like with the multiplication before.  Does 2.5 x 3.5 exist?  It depends on the context.  In the context of multiplication as repeated addition, then absolutely not.  And that's a perfectly fine answer.  But in the context of multiplication as a more general kind of scaling, then it does exist, and that's a perfectly fine answer too.  Does the sum of an infinite series exist?  In the context of sums as repeated counting, absolutely not.  And again, that's a perfectly fine answer.  But there is also a more general context, with sums as the limit of the partial sums, where the sum of an infinite series does exist.

So, can we find an even more general context than that, where we can assign a value to a divergent series like Grandi's series, where there is no limit of the partial sums?  Again, yes.  One example of this is Cesàro summation, where the value of a series is the limit of the average of the partial sums.  Let's work that out for Grandi's series:

The series:
1 -1 + 1 -1 + 1 - ...

The partial sums:
1, 0, 1, 0, 1...

The averages of the partial sums:
1, 1/2, 2/3, 2/4, 3/5...

And the limit of these is 1/2.  And like before, the more general case also captures the special cases we had before.  In the case of a convergent series, Cesàro summation just gives the limit of the partial sums.

Quote
And then they go using this proof to further prove that the sum of all positive integers is -1/12. Because they are completely out of their mind. I mean really, why would it be -1/12?

Remember, this is not the sum in the simple context of repeated counting.  In that context, there is no sum.  But is there a context in which we can assign a value to this divergent series?  Yes.  Cesàro summation will not help us here, but there are other methods that will.  Analytic continuation and Ramanujan summation, for example, allow us give this divergent series a value of -1/12.

But is this actually useful for anything except playing mathematical games?  It is.  In fact, assigning values to divergent series is done all the time in physics.

Quote
But they don't care. They say it's -1/12. Because that's the answer they get when they do the math. And they then go using that -1/12 number to get their string theory to work, which they then pass off as "science" even though all these hypothetical dimensions have never been observed...

It's true that -1/12 is where the 26 dimensions of string theory comes from, but this isn't just used in theories as speculative as string theory.  This is used all over the place in quantum field theories, such as quantum electrodynamics, which is not only testable via experiment, it is arguably the most precisely tested theory in the history of science.  These kinds of divergent sums come up in that theory, and when we actually run the experiments in real life, we don't get infinite results, we get results like -1/12, as the math predicts.

To finish up, I want to return to the original question of 0.999... being equal to 1.  I think a lot of people have trouble with this because they have an intuition that 0.999... should mean something like "infinitely close to 1 but not 1."  Is something like that possible?  Again, this depends on the context.  In the context of the real numbers, absolutely not, because there are no numbers that are infinitely close to other numbers.  But is it possible to extend the real numbers to include infinitesimal numbers like this?  Actually, yes, there are hyperreal systems that do this.  And in those contexts you can have numbers that are infinitely close to 1 (though you have to specify which infinity).  But even in these hyperreal contexts, 0.999... still normally just stands for the same limit we defined above, so it is still equal to 1.  But hey, you are perfectly free to define your own hyperreal system of numbers where 0.999... is defined to be something else, one of those numbers infinitely close to 1.  If it turns out to be useful, maybe they'll even name it after you someday!
« Last Edit: January 06, 2017, 02:31:37 PM by cgraye »
 
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Offline Jayne

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Re: Why .9̅ ≠ .999...
« Reply #21 on: January 06, 2017, 01:21:59 PM »
Hello, cgraye!!  I have not seen you for ages.  I am grateful that Daniel came up with a topic that could lure you over here in spite of your caution regarding traps.

Now I must post something mathy so I won't be completely off topic:

Jesus, meek and humble of heart, make my heart like unto Thine.
 

Offline MundaCorMeum

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Re: Why .9̅ ≠ .999...
« Reply #22 on: January 06, 2017, 01:24:17 PM »
Hello, cgraye!!  I have not seen you for ages.  I am grateful that Daniel came up with a topic that could lure you over here in spite of your caution regarding traps.

Now I must post something mathy so I won't be completely off topic:




I'm totally going to use this during my 7th grader's next math lesson.  How much do you want to be that she rolls her eyes?  :lol:
 
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Offline cgraye

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Re: Why .9̅ ≠ .999...
« Reply #23 on: January 06, 2017, 09:14:59 PM »
Hello, cgraye!!  I have not seen you for ages.  I am grateful that Daniel came up with a topic that could lure you over here in spite of your caution regarding traps.

Thank you, Jayne.  I have occasionally come by to read the forum and see what my old friends are up to.  I almost joined a physics discussion once or twice, but there are some very knowledgeable people here, and I didn't really have anything to add.  But I didn't see a good answer to this thread, and I didn't want any aspiring young mathematicians to read this and miss out on a lifelong friendship with divergent series.
 
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Offline Rosarium

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Re: Why .9̅ ≠ .999...
« Reply #24 on: February 08, 2017, 07:56:56 PM »


And then they go using this proof to further prove that the sum of all positive integers is -1/12. Because they are completely out of their mind. I mean really, why would it be -1/12? It clearly has no answer, since you can never add all positive integers together.

A Ramanujan summation is not meant to be presented in the way you have stated.

Quote
I mean really, why would it be -1/12?

If you understood in what context that number is assigned to that divergent series, you would not think it strange. It is your ignorance which resulted in your rant about enemies in mathematics.

Be careful about declaring things you do not understand to be nonsense or wrong. Anything which is not understood is nonsensical.

In fact, it is the prime method of attacking sacred doctrine.




« Last Edit: February 08, 2017, 08:38:27 PM by Rosarium »
In like manner these men also defile the flesh, and despise dominion, and blaspheme majesty.  When Michael the archangel, disputing with the devil, contended about the body of Moses, he durst not bring against him the judgment of railing speech, but said: The Lord command thee.  But these men blaspheme whatever things they know not: and what things soever they naturally know, like dumb beasts, in these they are corrupted.
 

Offline Heinrich

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Re: Why .9̅ ≠ .999...
« Reply #25 on: February 08, 2017, 08:42:19 PM »
You haven't lost a step.
Schaff Recht mir Gott und führe meine Sache gegen ein unheiliges Volk . . .   .                          
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Offline Daniel

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Re: Why .9̅ ≠ .999...
« Reply #26 on: February 09, 2017, 08:50:22 AM »
If you understood in what context that number is assigned to that divergent series, you would not think it strange. It is your ignorance which resulted in your rant about enemies in mathematics.

Be careful about declaring things you do not understand to be nonsense or wrong. Anything which is not understood is nonsensical.
Well my post was intended to be somewhat satirical. I did not mean to imply that the -1/12 number is entirely arbitrary. All I was saying was, the sum of all positive integers is NOT -1/12 as they claim.

I do not entirely understand their math, but I do understand it well enough to know why they claim the sum to be -1/12: They use "proofs" such as these, or else they get those values graphically from plotting the Riemann zeta function. But again, none of this proves anything. Because those so-called proofs are based on the unverified assumption that you are allowed to add and subtract infinite series from one another in that manner. (Finitists say you are not.) And the Riemann zeta function uses analytic continuation, so (while the graph looks nice and symmetrical and everything,) it really doesn't count (because that value comes from the left side, yet the entire left half of the graph is plotted just by mirroring the right half).
« Last Edit: February 09, 2017, 08:58:55 AM by Daniel »
 

Offline Rosarium

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Re: Why .9̅ ≠ .999...
« Reply #27 on: February 09, 2017, 09:14:09 AM »
All I was saying was, the sum of all positive integers is NOT -1/12 as they claim.

But the thing is, nobody claims that. If people do think that is meant to be a sum of a divergent series, then it is a severe mistake.

Quote
I do not entirely understand their math, but I do understand it well enough to know why they claim the sum to be -1/12:

Poorly presented Youtube videos are not really significant in the context of mathematics.

Clickbait ad laden videos are not significant for anything at all in fact.



In like manner these men also defile the flesh, and despise dominion, and blaspheme majesty.  When Michael the archangel, disputing with the devil, contended about the body of Moses, he durst not bring against him the judgment of railing speech, but said: The Lord command thee.  But these men blaspheme whatever things they know not: and what things soever they naturally know, like dumb beasts, in these they are corrupted.
 
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