So - in math terms...
The limit .99999 (repeating) = the number 1
The number .99999 (for a whole lot of 999's) is not the number 1
So a limit and a number - some semantics - is at the bottom of this debate?
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So - in math terms...
The limit .99999 (repeating) = the number 1
The number .99999 (for a whole lot of 999's) is not the number 1
So a limit and a number - some semantics - is at the bottom of this debate?
It would appear so, although you don't need "the limit".
I think I've given at least 3 different ways now to show that 0.9 recurring is exactly equal to 1.
And I think that that is enough.
zaza
You got it backwards.
The limit as x approaches 1 <> the number 1
The number .999 (inifinte 9's) is the number 1 (to some people).
Quote:
Originally Posted by szlamany
Anyone who thinks that .999... <> 1 just has to consider the above infinite series. This series is not a limit it is a series and it does equal one.
Also .999... is a number just like .333... is a number. If you say .333... is not a number then is 1/3 a number?
This leads us to another way to look at the problem.
we know that 1/3 = .333... and 3*1/3 = 3*(.333...) = 1
So 1 = 3*(.333...) = 3*3*(.1+.01+.001+...) = 9*(.1+.01+.001+...) = .9999...
So the two are equal again.
After considering these two arguments, I don't know how anyone can say that .999... <> 1
Any flaws in these two arguments?
I posted this link before:
http://www.maths.nott.ac.uk/personal/anw/Research/Hack/
Excerpt from that website claims that the argument you gave is an optical allusion:
Quote:
Conventionally, we multiply two finite strings of digits together from the right, propagating `carries' to the left as necessary. If we work from the left, then we must always be prepared to change the `result-so-far' in the light of later carries. For finite strings, the final results are equivalent, but in working from the left arbitrarily many digits may be only provisionally known at any one time, so that (for example) the computation cannot be performed by a finite-state machine even if the multiplier is specified in advance. So much the worse if either or both strings are infinite. Thus, if we multiply 0.333...3d... by 3, where the next digit, d, is thus far unknown, then all we know is that the answer is no less than 0.999...90 and no more than 1.000...20. If d turns out to be less than 3, then the start of the result is confirmed to be 0.999..., if greater than 3 then 1.000...; but if d is 3, then the decision is postponed. If the 3's persist indefinitely, then so does the postponement of the decision. Of course, if we know that the multiplicand is 1/3, then we know already that the 3's will persist indefinitely, and we can make out a special case for known rational numbers; but the theory required to do this will be entirely equivalent to the formal analytic proof that 0.999... = 1, and there is no didactic gain.
The `result' 0.333... × 3 = 0.999... is thus seen to be an optical delusion. This is confirmed by the observation that most mathematicians are somewhat less happy with the similar `result' 0.14285714... × 7 = 0.999..., where it is clearer that something must be done about the carries. Similarly, the `results' in the third proof that 10 × 0.999... = 9.999... and 9.999... - 0.999... = 9 are visually attractive, but can be justified only by making special cases, which are as hard as the original FAQ. (The Hackenstrings version of this is left to the reader.)
Quote:
Originally Posted by moeur
So what the hacky guy is saying is that 3*(.3+.03+.003+...) = 1 because we already know the answer, but 9*(.1+.01+.001+...) <> 1 because we don't already know the answer.
Sounds kinda like quantum mechanics to me. Where just looking at something changes the result.
I couldn't understand a word of that hack quote but I would say
3(1/3) = 9and
3(0.333...) = 0.999...
It is a matter of the resolution of the representation: (1/3) has an infinite resolution, therefore it is an exact value, however 0.999... has a resolution which approaches infinity, but isn't, no matter how close it gets, a correct representation of (1/3), therefore it is not the same number.
For all real-world purposes, 0.999... = 1, however, purely theoretically, 0.999... < 1 by definition, therefore 0.999... != 1.
What does this mean?Quote:
0.999... has a resolution which approaches infinity
Both .333... and .999... can be represented by an infinite series
.333... = 3*(.1 + .01 + .001 +...)
.999... = 9*(.1 + .01 + .001 +...)
Do you disagree?
No, I agree.
By resolution I mean the the number of decimal places, something most people would probably call "precision".
I believe (1/3) has an "infinite" resolution because it is a concise representation of the figure which fits exactly three times into 1; however 0.333..., no matter how many extra .3's you give it, is never exactly that figure. It would be if it had an infinite number of .3's, however you can't have an infinite number, therefore it is approaching infinite resolution.
In other words, 0.999... is constantly tending toward 1, and 0.333... tends toward (1/3) (seem to be going back to limits...). You can't mathematically prove that a number which is infinitely approaching x, is x, however for all intents and purposes it can be assumed that it is.
0.333... represents a number with an infinite nuber of 3's (not approaching) and exactly equals 1/3. They are just two different representations of the same number, I don't think people are questioning that.
0.999... has an infinite number of 9's it is not tending toward any thing it is a representation of an exact number. The question is, is 1 also a representation of that number.
If you have an infinite number of decimal places, then it can't represent an exact number; since infinity is not an exact number, it is a concept.
If, using your logic, you say
.999... = 9*(.1 + .01 + .001 +...)I agree completely with that. As you have shown, for every .1, .01, .00....1 you add to .999, you are increasing .999 by a decreasing amount. But no matter how many times you do this, you will never reach 1, not even if you carry on an infinite number of times; and so .999... does not equal 1.
BTW congrats on 1,000 posts :)
By your argument, the infinite series (.3 +.03 +.003+...) does not equal 1/3 and that would be wrong since the series clearly converges to that value.
There is a different discussion going on here. Does the series (.9+.09+.009+...) converge to that value of 1?
Is the original question: does the infinite series .999... converge to 1?
Or is the question: is the sum of a convergent series equal to the number it converges to?
I think we've shown the first to be true, the second then would just be a definition you'd have to accept or deny with no real proof either way.
Hi,
Re. 0.9 + 0.09 + 0.009 ....
I think there may be some confusion between the Limit and the Evaluation of this series. In the limit of this series for any finite number of terms, the sum of the series gets ever closer to 1. However, you can always get closer by including an extra term.
Lets think about it in these terms: Suppose we sum an infinite number of terms and the answer is NOT 1. Then we must be able to get closer to 1 by adding another term. But by doing so, we have assumed that our current number of terms is finite, because you cannot add another term to an infinite number of terms and arrive at a different result. Hence our original assumption must be false.
I think that some people will never be convinced of this, particularly if they can dig up a maths prof who appears to back their argument. I can't really be bothered to go into the details of what he's doing and argue it, but I bet I can find more than one math prof who agrees with the majority of us on this. Good call with the "permanently unresolved" tag.
zaza
EDIT:
I've now read what this prof is going on about, and my summary would be as follows.
If you multiply two decimals of indefinite length number by number from the left, then you have to wait until you reach the end of the decimal before you get a final decision as to what the result is. This, of course, is true for COMPUTATIONALLY evaluating 3 x 0.333333d...., and is what you let yourself in for by doing a multiplication from the left. However, I give you the following quote:
i.e. if rather than doing the multiplication of 3 x 0.33333d digit by digit, in which case we cannot be sure that the next digit is 3, we already know that the 3's persist indefinitely, then we have our answer which is of course the same as the proof being discussed here.Quote:
Of course, if we know that the multiplicand is 1/3, then we know already that the 3's will persist indefinitely, and we can make out a special case for known rational numbers; but the theory required to do this will be entirely equivalent to the formal analytic proof that 0.999... = 1, and there is no didactic gain.
I very much agree that if we consider the decimal 0.99999999d, where d is the next digit, and go from the left, then we can never be sure that the decimal we have is equal to 1, because we are considering our decimal on a digit-by-digit basis and we do not know whether it is infinite or even whether the next digit is a 9. However, given that we are considering 0.9 recurring in which we know that the next digit is indeed 9 and that there are indeed infinitely many of them, then this question never arises. We aren't trying to count infinitely many digits one by one, which would indeed take forever and would not be a sensible way of trying to prove that 0.99999.. does equal 1.
You lot have got me confused.
0.999 rec is a mathematical symbol
1.000 is a mathematical symbol
The symbols are not identical. Are they numerically equivalent?
I don't think so. If you add 0.00...1 to 0.9999999 rec you will get 1.00000
0.999 = 1-0.0...1
Simple algebra therefore shows that the assertion that 1=0.99rec is incorrect.
That's what I've been trying to say. No matter how many 0.000...1's you add to 0.999... you always need to add one more to get 1 and so you will never reach it.Quote:
Originally Posted by yrwyddfa
My example was just that.Quote:
Originally Posted by penagate
The fact of the matter is that there is a number (i) - however difficult it is for mathematics to descibe - that we can add to 0.9999rec (j) to equal 1 (x)
Therefore x = i+j.
As far as I am aware there is no wayyou can rearrange this equation to make j=x
I order to destroy this you will need to show that there is absolutely no number that can be added to 0.9999999 in order to make it 1.
No I don't think that you have at all.Quote:
Originally Posted by moeur
I think there are two good example that show that .999... = 1 in this discussion. If you disagree with any step in these examples then point it out.
For both examples, put the number into an infinite series
0.999... = (.9+.09+.009+...)
since each term in the series is divisible by nine we can pull it out front
9(.1+.01+.001+...)
The series is of the form of an infinite geometric series
(1+.1+.12+.13+...)-1
which we know is equal to 1/9
http://www.maths.abdn.ac.uk/~igc/tch...es/node48.html
Therefore .999... = 9(1/9) = 1
A second example relies on the fact that we know 1/3 = 0.333... = 3(.1+.01+.001+...)
and
.999... = 3*3(.1+.01+.001+...)=3*1/3 = 1
Should this be the case then please demonstrate why the simple algebra I posted above is incorrect.Quote:
Originally Posted by moeur
On reflection . . .
I assume you believe that your 'proof' is valid.
I believe that my 'proof' is valid.
We then have a problem that 0.999r=1 and 0.999<>1 cannot both be simultaneously true.
OKQuote:
please demonstrate why the simple algebra I posted above is incorrect.
There is no basis for this assumption. It is not a fact unless of course you mean i=0.Quote:
The fact of the matter is that there is a number (i) - however difficult it is for mathematics to descibe - that we can add to 0.9999rec (j) to equal 1 (x)
Hold on for a little while.
I am working on a proof that denies x+x+x=1 can be a valid proposition where x is a non-terminating decimal sequence.
Might take a while.
Right OK I wasn't too descriptive by what I meant by this.Quote:
Originally Posted by moeur
(i) 0.9+0.1=1 (n=0)
(ii) 0.09+0.01=1 (n=1)
if steps (i) and (ii) are true (which they are) it follows, by induction, for the whole series that for 0.n9 there must be a value 0.n1 which can be added where the total becomes equal to 1 (where n is the number of successive digits to create the number.This holds for the entire infinite set.
Therefore x=i+j is valid by induction.
0.999r is therefore not equal to 1
Again you have a false assumption.
you can see that the number you are adding to the series is getting smaller.
so, as the series approaches an infinite series, that number approaches zero.
So, the number you add to an infinite sseries is equal to zero.
I've just realized that this thread is a self-fulfilling prophecy!
Each post is not quite right - as I'm sure the others would agree to.
So each post is .9 correct.
If we keep this up for an indefinite (infinite, long, long time) will the answer be 1?
I'm guessing 42 myself :D
I see where you are coming from, now.Quote:
Originally Posted by moeur
I believe that your definition of infinity is flawed. If you firstly have a set of all numbers you have infinity If you subtract the infinite set of all numbers >4 the numbers left in the set are {1,2,3,4} Infinity-Infinity<>0
The opposite of infinity is not necessarily zero. In my induction the infinite set is for all numbers where 0.n9 holds true. In this set - which I have to add is required for the problem domain - my induction holds true for any number in the set.
You're assumption that if a series approaches infinity then the opposite series will approach zero is incorrect.
Not quite true. By induction I am correct.Quote:
Originally Posted by szlamany
In this situation, it would be one minus infinity. Is that zero?Quote:
Originally Posted by yrwyddfa
If you create a set of numbers of which it's cardinality is infinity then infinity can be (I suppose) anything you need it to be. The only logical way to think of infinity is in terms of set theory.
Therefore assigning infinity an absolute value at implied, or prescribed limits or anywhere else, is simply mathematical nonsense.
I don't know much about geometric series, I have to admit, so I don't see how 1/3=0.3333r
Furthermore:
9(.1+.01+.001) (etc) gives a figure of 0.9999r
if 0.9999r = 1/9 how can it also be 1. By it's own definition the geometric series is dividing 1 into 9 equal parts. It cannot be 1
0.111r = 1/9Quote:
if 0.9999r = 1/9 how can it also be 1.
this can be obtained by applying know mathematical methods to solve an infinite series.
I don't think that it does - I'm not sure, though.
Nevertheless my argument by induction stands.
You can prove this to yourself by just doing the longhand division.Quote:
I don't see how 1/3=0.3333r
1 divided by 3 = .3 remainder 1
.1 divided by 3 = .03 remainder 1
you will always get remainder 1 so there are an infinite number of 3's in the solution.
Yup my mistake thinking of the non-infinite regression. Doh! :eek:Quote:
Originally Posted by moeur
Hi,
I quote:
Quote:
Right OK I wasn't too descriptive by what I meant by this.
(i) 0.9+0.1=1 (n=0)
(ii) 0.09+0.01=1 (n=1)
if steps (i) and (ii) are true (which they are) it follows, by induction, for the whole series that for 0.n9 there must be a value 0.n1 which can be added where the total becomes equal to 1 (where n is the number of successive digits to create the number.This holds for the entire infinite set.
Therefore x=i+j is valid by induction.
0.999r is therefore not equal to 1
First of all, I would point out that step ii) is not, in fact, true. Secondly, I would point out that you haven't actually done any induction. You have shown that the first step (0.9+0.1=1) works, but you have not shown how the truth of the nth step requires the truth of the (n+1)th step, nor have you shown how this is true even if n is infinite.
The problem lies with your definition of what you are adding, 0.n1. This is a finite number! However, the 9's go on for infinity. Thus the decimal place to which you assign the 1 is, in fact, the infinite'th dp. i.e. never.
So, as moeur says, what you need to add to 0.9 recurring to get 1 is 0.
I'm sure I already said this...
zaza
Nope you're wrong.Quote:
Originally Posted by zaza
You cannot use induction to prove infinites...
you've proven it true for every n in the set of positive integers. This is obvious though, because 0.99..9 is definitely not equal to 1 if there is a finite number of 9s. However, 0.99... has an infinite number of 9s.
I just noticed the [Perm Unresolved] tag :p
I was waiting for someone to notice that you can't substitute a philisophical proof for a mathematical one.Quote:
Originally Posted by sql_lall
I don't really know why you can't - it appears to me to be convention rather than logic, to me . . . .
;)