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0.99999... != 1
Posted: Fri Mar 09, 2007 6:45 pm
by Android Mouse
I've seen this debate elsewhere on the internet, and am curious what others think here.
I actually wonder how many people actually believe 0.9999... == 1 and how many are just saying they do for the sake of trolling.
Posted: Fri Mar 09, 2007 7:13 pm
by nick8325
But it's true that 0.9recurring and 1 are equal.
If two real numbers are unequal, you can find a number that comes between them (for example, the average of the two numbers). But there's no number between 0.9recurring and 1. So they're equal.
Posted: Fri Mar 09, 2007 7:20 pm
by Alboin
Edit: Ah, okay. .999... does equal 1....Yet, I still don't see how this is so; .999999999999999999999999999999999999999 is still not one. It's very close, but not 1.
Posted: Fri Mar 09, 2007 7:24 pm
by GLneo
if .9... and 1 were == the .9... would be written as 1, but the 0 at the beginning means its less no mater how far down you go. Of course you could write 1 as (0 10/10) hmmmm.... Let the debit begin!
Posted: Fri Mar 09, 2007 7:25 pm
by nick8325
Alboin wrote:For all practical purposes, I think .9999... can just as well represent 1. However, technically speaking, it does not equal 1. So, I take both sides.
But technically speaking, it
does equal one
(for example, see
http://www.nrich.maths.org/askedNRICH/edited/1347.html)
GLneo wrote:if .9... and 1 were == the .9... would be written as 1, but the 0 at the beginning means its less no mater how far down you go.
Ah, but that only works for finitely long decimals. All the decimals 0.9, 0.99, 0.999, 0.9999, ... are less than 1, but in the limit the sequence converges to 1.
Posted: Fri Mar 09, 2007 7:49 pm
by GLneo
it would be an infinitively small amount less, but still less...
Posted: Fri Mar 09, 2007 7:55 pm
by ~
At least for machine's integer math, even if there were something nearer to 1 than 0.999999999999999999999999999999999999999999999999999999..., such number is not equal and would be in the form of an integer in the remainder.
At least it could be rounded up wheter using emulated or true floating point math, but the machine would differentiate them and would round them up, but certainly I don't see how it would consider them the same thing.
Posted: Fri Mar 09, 2007 8:24 pm
by pcmattman
Well, if you put both into an integer type, 0.9999 becomes 0 and 1 becomes 1.
Posted: Fri Mar 09, 2007 8:31 pm
by ~
I was thinking in something like multimedia math arithmetic (SSE, MMX), or something like dividing RDX:RAX by RBX.
In the first case, I am sure that a binary compare of the actual data between the representation of 0.999999999 and 1 will be different if there is no rounding taking place. In the second case, the remainder will be in RDX and the integer part will be in RAX.
Posted: Fri Mar 09, 2007 8:48 pm
by frank
All repeating decimals can be written as a fraction. To do the conversion you put the repeating part over the number of 9s that is equal to the number of repeating digits.
Example 0.7 repeating is 7/9. or 0.23 repeating is 23/99. Check it out and you will see.
When you do the math to convert .9 repeating to a fraction you end up with 9/9, which is 1.
Posted: Fri Mar 09, 2007 9:17 pm
by ~
It means that for 0.4193 it would be 4193/9999. It seems to me like a heavily worked out concept that is beyond a simple division without rounding from the machine.
As a mathematical concept it may be correct, but for a short division of something like 23/41, for example:
RDX=0 high dividend half
RAX=23 low dividend half
RBX=41 divisor
[RDX:RAX]/[RBX]
[RDX]=23 remainder
[RAX]=0 quotient
This is what I'm up to.
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Now let's assume that the infinitely recurrent number 0.99999999999 cannot be reproduced naturally (does anybody know some division in which X/Y produces 0.9999999999?).
Now, let's also note the division of the following numbers:
1/9=0.1111111111111.........
2/9=0.5
3/9=0.3333333333333...........
4/9=0.4444444444444...........
5/9=0.5555555555555...........
6/9=0.6666666666666...........
7/9=0.7777777777777...........
8/9=0.8888888888888...........
9/9=1
OK, so let's assume that 0.999999999999......... is a hypotetical number, (if somebody knows of some division like X/Y=0.999999999..... please clarify!), that could only be obtained hypotetically produced by the division of 9/9.
But since it is the same number, it can only produce 1.
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Now, we could say the following both for integer and decimal numbers and its combinations:
* If 0.999999999999........ is the hypotetical result of 9/9, then there is no controversy over a nonexistent number and we can say indeed that it leads us to 1 when we try to reconstruct both dividend and divisor.
* If 0.99999999999....... can be produced by some division other than 9/9, then we can certainly say that:
- For 9/9, or for any other division of the exact nature, the quotient number 0.999999999...... is hypotetical and REALLY is a quotient of 1
- For any other division not such aligned mathematically, the quotient number 0.999999 is different from a quotient of 1
Posted: Fri Mar 09, 2007 10:18 pm
by iammisc
This has to do with the way computers store floating point numbers and the whole inaccuracies involved in that.
Posted: Fri Mar 09, 2007 10:29 pm
by ~
Sure; the integers are the only ones that would not lose because they keep being integers.
But for the 0.99999999... thing, it's not really something that we can refer to as the way computer stores numbers. It's not the case; it has absolutely nothing to do with the fundamental binary representation of numbers. Forget about it being a computational issue; it's not the intention of the actual debate.
It's a debate of a seemingly hypotetical unique result from dividing 9/9 that somebody came up to think because of the sequence I tried to explain in my previous post in this thread.
Posted: Fri Mar 09, 2007 10:32 pm
by m
Even on some scientific calculator.For example on a CASIO one 2x2^0.5-8^0.5=10^-15 but not 0.
Floating numbers expand the range at the cost of accuracy because they still use as many bits as same-length integers do.
As for 0.999999...,err,:
Code: Select all
1 / 9 =0.111111...
2 / 9 = 0.222222...
3 / 9 = 0.333333...
...
9 / 9 = ?
Posted: Fri Mar 09, 2007 10:47 pm
by ~
Hmmm.......
The actual point of discussion, as I get to understand it, is not the accuracy of any machine. No; it just is talking about the saying that "any floating point number that repeats can be represented as a fraction".
Hence, if we talk about a number like 0.23232323, we can see that the decimal cipher string that is repeating is 23, so, for the "2" we will place a 9, and for the "3" another 9, like this:
23/99
That's the very point where we clash with 0.9999999. Since the cipher string that repeats is "9", we place a 9 for that 9, so we end up with:
9/9
Did you see that? It looks like sort of a trolling, and, assuming that 9/9 could be the only division that could produce a recurring 0.999999, and that's not the case and that's where they say that the hypotetical 0.99999... is the same than 1.
Anyway, try to re-read the first 10 posts and you will see that it has nothing to do with binary representations, nothing to do with scientific math, only has to do with a curiosity and with a hypotetic value tought to try to contradict the self logic and for trying of not making possible to reach a conclussion.
I already tried to explain clearly how this apparent discrepancy can be conciliated and understood. It has nothing special. Pure trolling, if you like, a bad yet well constructed, time and brain-consuming joke.
At least, are there some other clarifications from somebody, some other division other than 9/9 that produces the 0.99999999999..., just to feed the solution with examples?