Is 0.999 (repeating) = 1?
I think not. See my sigs.
However I'd like to hear some other opinions, especially if you have a degree in Math.
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Is 0.999 (repeating) = 1?
I think not. See my sigs.
However I'd like to hear some other opinions, especially if you have a degree in Math.
0.9999 rounded is 1, otherwise it is 0.99999
Why did you ever think that it was?Quote:
Originally Posted by capsulecorpjx
Bunch of gamers argued this on a forum (Blizzard gamers specifically).Quote:
Originally Posted by yrwyddfa
Quote:
Originally Posted by capsulecorpjx
On a computer it might be the same because of the rounding in the IEEE standard. But in real life it is not the same....
Quote:
Originally Posted by NoteMe
Exactly.
And it depends on your point of view I suppose but unrounded and not on a computer system 0.99999 is not 1 :)
Cheers,
RyanJ
Hi,
There is no difference between a and b in your sig. Both are 1/0 which is infinite. "Undefined" means nothing.
This, of course, assumes that by 0.999 (repeating) you mean 0.9 recurring, i.e. with an infinite number of 9's on the end.
Here is a proof for you:
Let x = 0.999999999....
Then 10x = 9.99999999999......
Subtracting, 9x = 9
hence x = 0.9999999999....=1
Satisfied?
zaza
OK, Undefined doesn't really mean nothing. That's what a computer will give you. But to all intents and purposes it is infinite. Undefined could, in fact, mean the same thing. But it depends how far into "infinities" you want to go...
Quote:
Originally Posted by zaza
How did you do this step? :confused:
You can't just subtract. And you don't even subrtract the same amount on the same side????
You can always multiply or devide though. To get 9x on the left side you can always multiply with (9/10) then you will have:
9x = 8.99999999....
Can you please elaborate on how you did that?
- ØØ -
Hi,
10x is simply achieved by moving the decimal point one space to the right. That's what multiplying by 10 does. Don't forget, the 9's go on FOREVER, so there's always another 9 on the end. Hence by subtracting x from 10x, all you are doing is removing the part after the decimal point and so you are left with 9x = 9.
HTH
zaza
But you can't simply do that...you are chaning the value of X....
Quote:
Originally Posted by zaza
You're wrong, Undefined <> Infinite.
Sophmore math teaches you that.
Just graph the equations:
y = 1/x
y is undefined if x = 0
y is infinite as x APPROACHES 0
The graph shows the difference.
Quote:
Originally Posted by zaza
To clarify zaza's point
x = 0.999999...
10*x = 9.9999999.....
subtract x from each side
10x - x = 9.99999... - 0.9999....
or
9x = 9
hence
x=1.0
and 1/0 is infinity there is no difference in the graphs
a) 1/(1-0.999...) = postive infinity.
b) 1/(1-1) = undefined.
b) 1/(1-1) = n
1/0 = n
1 = 0*n
n being infinity does not solve the equation so n <> infinity.
At the very least, n (or b) can be pos inf, neg inf or 0.
However a) can only be postive infinity.
Which means a <> b.
http://en.wikipedia.org/wiki/Division_by_zero
There is nothing wrong with zaza's logic.
as far as 1/0, you're right, it is undefined.
Therefore we cannot use it in your proof.
zaza did not divide by zero to reach his conclusion therefore performed no undefined procedure.
You did in your attempt so your result is not valid.
All I'm saying is that to say 1 = 0.999..., you must say undefined = infinity.Quote:
Originally Posted by moeur
Which I contend is not true.
You can have undefined in your proof if you intended to use it.
Anyway, thats my proof.
A math prof has another proof that supports my conclusions 0.9999... < 1 he does this using a new way of representing numbers.
http://www.maths.nott.ac.uk/personal/anw/Research/Hack/
How about this:
How small is the difference between 1 and .999...?
:wave:
sorry, but 0.9999.... = 1
you claim that 1 / (1 - 0.999...) is infinity, but give no proof.
why is this wrong? let x = 1 - 0.99999....
let y be any real number > 0, therefore y > x (as 1 - y < 0.99999....)
so, x must = 0.
another reason:
1 + x + x^2 + x^3 + ... = 1 / (1 - x)
=> 9 + 9x + 9x^2 + 9x^3 + .. = 9 / (1 - x)
let x = 0.1 => 9 + 0.9 + 0.09 + 0.009 + ... = 9.9999... = 9 / (9/10) = 90/9 = 10
another reason (similar to above):
1/9 = 0.111111....
=> 9/9 = 0.99999..... = 1
Note that the 'postponement' multiplication argument that site mentions is not really relevant, as we *know* that all the digits are going to be '1', so it's not a problem of 'postponing' any decision.
The funny thing about that site is that it seems to think recurring decimals in base 10 are nasty, imprecise limits, but recurring decimals in base 2 (which is what hackenstrings effectively are) allow exact representation... :ehh:
That article is about enumerating games, and the theory can't be applied simply. For one, game enumeration allows concepts such as 'infinity plus one', 'infinity plus two' etc... which are always 'greater than' infinity. In fact, it's not seen as infinity, but is denoted by 'omega'. Real numbers do not have this concept.
Disclaimer: I am *not* a gamer, I am a programmer from a strong maths background.
And don't just take it from me:
http://www.cs.uu.nl/wais/html/na-dir.../0.999999.html
http://forums.topcoder.com/?module=T...start=0&mc=142
Quote:
Originally Posted by Something Else
This is actualy a very interesting question. Since it is here the kernel of the dispute is...:)....one side here means that it can become ZERO/NIL/NOTHING because they are to lazy to write an infinte amount of 9999999999, the other side says that it goes towards 0, but never reaches it since they are still sitting on the 9 key on their keybarod, and never reach 1.0, at least not before they are out of memory on their computer...:)
- ØØ -
I guess you can, but you can't operate on it; therefore you can't doQuote:
You can have undefined in your proof if you intended to use it.
0*1/0 = 1
by defenition 0*1/0 is undefined.
You guys don't understand limits.Quote:
Originally Posted by NoteMe
0.999... is not a number more than a limit.
And approaching a number IS NOT that number.
Anyway, my proof seems logical to me, but maybe it has some flaws.
What are you thoughts about the proof provided at this university site?
http://www.maths.nott.ac.uk/personal/anw/Research/Hack/
Okay, I don't get this..maybe I am an idiot...but I believe what I see
0.9999(repeated) doesnt even involve the number 1...how the heck can you say it = 1? Unless you change the number..it will always be 0.9999(repeated)....
The difference isQuote:
Originally Posted by Something Else
lim x
x-> 0
In other word, the difference is x as x approaches 0.
Repeated means forever means infinity - we are talking about a pretty abstract concept here.Quote:
Originally Posted by kfcSmitty
Note the only math term used here was "infinity" ;)
In floating point (since this is a computer language forum - at it's roots) even 4.9 = 5 - that's just the way the value gets stored.
I've not yet figured out if this thread is about floating point representations of values or about what appears to be a math-class teaser question.
Quote:
Originally Posted by capsulecorpjx
Ehhhh...did I ever say that it was?
capsulecorpjx,
You are arguing two points.
Your argument is flawed so abandon it.Quote:
my proof seems logical to me, but maybe it has some flaws.
This is a better way to approach the problem.Quote:
How small is the difference between 1 and .999...?
What you'll find out is the whole argument depends on how you define the difference: is it zero or infinitesimal; a value arbitrarily close to but greater than zero.
And so the question, "does .999... = 1", cannot be proved but depends on how you define that difference. Those that subscribe to the Arcimedean Axiom say there is no such thing as an infinitesimal number so .999… = 1. Others that accept the idea of an infinitesimal will say that .999 < 1.
I’m not a mathematician, but a Physicist; I work with real world things. Another definition of infinitesimal is: immeasurably or incalculably small. So if it is immeasurable it is indistinguishable from 0.
Hence in the world where we most work, as well as the mathematician's world where the Arcimedean Axiom is accepted; .999… = 1.
I was trying very hard to keep from getting in this debate (help me - please - don't let me hit "Submit Reply"!Quote:
Originally Posted by NoteMe
NoteMe - you know very well that it's not "possible" to write an infinite amount of 999999999's.
You might be able to spin them around very fast and crash them into other particles ;)
Yeah I know....;)...and your point is....;)Quote:
Originally Posted by szlamany
And now you are in...;)
- ØØ -
EDit:
you should come to CERN...;)Quote:
Originally Posted by moeur
Are you saying those particles are real?Quote:
Originally Posted by NoteMe
I'd love to.Quote:
you should come to CERN...
Hear, Hear!!!Quote:
Originally Posted by moeur
And, also, if it is incalculably small, it also is indistinquishable from zeroQuote:
This is a better way to approach the problem.
What you'll find out is the whole argument depends on how you define the difference: is it zero or infinitesimal; a value arbitrarily close to but greater than zero.
And so the question, "does .999... = 1", cannot be proved but depends on how you define that difference. Those that subscribe to the Arcimedean Axiom say there is no such thing as an infinitesimal number so .999… = 1. Others that accept the idea of an infinitesimal will say that .999[edit]...[/edit] < 1.
I’m not a mathematician, but a Physicist; I work with real world things. Another definition of infinitesimal is: immeasurably or incalculably small. So if it is immeasurable it is indistinguishable from 0.
:thumb:Quote:
Hence in the world where we most work, as well as the mathematician's world where the Arcimedean Axiom is accepted; .999… = 1.
a difference that makes no difference is no difference.
.00000000000000000000000000...=0
:wave:
Really just a question of semantics. I say 0.999... approaches 1, not 1 itself.
You're saying 0.999... = 1.
Its a different story if you say something that approaches a number is that number, that is wrong.
Quote:
Originally Posted by Something Else
Hi,
I think you seem to have led yourself up the garden path with your signature. You have assumed that a <> b, and then "discovered" that in fact they are not the same! I would say that your a and b options are in fact exactly the same, by dint of the fact that 0.9 recurring is indeed 1. Thus both a and b are what you term "undefined". You've just decided to call one of them "infinity" and the other "undefined". The semantics, I'm afraid, are in your argument. The lesson, I'm afraid, is that dividing by zero is never a good thing to do.
zaza
Incidentally [carefully puts cat amongst pigeons], what do you call 1/infinity?
i still dont see how 0.99999 = 1
its like that einstein thing...if you move towards something, moving half the distance of the previous each time, you will never reach your destination...same goes for this
Yeah - Einstein had a lot of those things - saw a great exhibit at the NY Museam of Natural History on Einstein a couple of years ago...Quote:
Originally Posted by kfcSmitty
The keyword you used is never - kind of like the repeated used in many of the posts here...
Those abstract facts (is that an oxymoron?)
those abstract facts...what? It kinda seems like you stopped in the middle of the sentence there :sick:
I thought to myself how funny the "phrase" abstract fact is - that seems like an oxymoron - in that those two words are polar opposites.
I thought that was enough said..
1 is a number.
.99999 (repeating) is a limit - or so people have said in this thread.
Apples and oranges?
Seems like a game of semantics at this point.
I've seen posts here using incalcuably small to describe the difference - but that's not true, since the .9999 repeats for infinity, then in reality this is not a number - so subtracting it from 1 (a solid good old number) against something that is not a number (we obviosly cannot have a number with an infinite amount of digits - infinity is like a NULL VALUE in a database - rather a lonely thing) makes no sense.
But, then again maybe it's about thinking outside the box here...
Since we seem to have taken a physics turn...
You know how a gas is a loosely organized pile of molecules - a solid is a bit more organized. That desk your PC on is really rather porous (much more vacuum space then "molecule" space) - but gives the illusion of being a flat hard solid object. Those darn sub-nuclear forces at work giving it the appearance of something we cannot pass through. Maybe it's that our own particles interfere/interact with those desk particles.
Maybe if this poor girl was in the "proper frame of mind" she would have avoided that "crash"...
http://www.vbforums.com/showthread.php?t=351029
Which is closer to 1:
0.9 or 0.99?
What about:
0.999 or 0.9999?
How do you make 0.9 closer to 1? Add a 9 on the end = 0.99.
What about 0.99? Add a 9 on the end = 0.999.
Now imagine 256 9's on the end. How do you make it closer to 1? Add another 9 on the end, so you have 257.
Every time you put a 9 on the end, you get closer to 1. However, with any finite number of 9's you would never be exactly at 1, even if you had a hundred million billion of them. There would always be an end point for you to add another 9 onto. But an infinite number of 9's, there is no end point, therefore it is impossible to make it any closer to 1 by adding a 9 on the end. Hence if it can't be any closer to 1, it must BE 1.
BTW, when we say 0.999..., those dots actually mean "going on forever". Nobody is trying to claim that 0.999 is, in fact, 1.
zaza
Great argument - and I have no problem thinking both ways in this thread...Quote:
Originally Posted by zaza
But you cannot say a hundred million billion - give a fact...
Then say infinite number of 9's and give a fact...
As soon as you say infinite you are on a totally new playing field.
My opinion anyway ;)
All you're saying is that it's infintely close to 1.
You're not saying it is 1.
There are many limits and graphs in math that are infinitely close to something, but never quite there.
Quote:
Originally Posted by zaza
Hi,
You seem rather fixed on this idea of the difference between limits and actual values. However, you haven't actually said what your limit actually refers to. I would say that it refers to the following series:
0.9 + 0.09 + 0.009 + 0.0009 + ....
This is a geometric progression. The sum to infinity of this is defined (check your maths textbook) to be:
S = lim Sn = 0.9 / 1-(1/10) = 0.9 / 0.9 = 1
n->inf.
Just because there is a limit does not mean that it must always just APPROACH a particular value although, as you say, just because there is a limit does not mean that the limit IS that particular value.
In this case, however, it is.
zaza
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