Well then write it that way. But 0.333... means 1/3 carried to infinite places, so the infinitesimal is taken into account. Plus, lim x -> infinity ( 1 + 1/x ) = 1. There is no difference here.
Well then write it that way. But 0.333... means 1/3 carried to infinite places, so the infinitesimal is taken into account. Plus, lim x -> infinity ( 1 + 1/x ) = 1. There is no difference here.
I know. It was a joke. I mean, my "proof" "showed" that 1 = 0 (you think I didn't realize that wasn't true?) and if 1 is indeed 0 then one can proof AND disproof ANYTHING. That the world is square, round, but not square, and not round, and that I exist but do not.
Yes, EVERYTHING becomes true.
Tell me what's wrong about my proof then. You can use exactly the same proof for 1/3 = 0.3333
Calculators show you 0.33333333334. That's because they simply don't have enough 3's to represent it, because calculators, unfortunately, can not display an infinite amount of 3's.
In fact, if a 4 ever does appear, then 3*(1/3)rd is GREATER than 1.
Let me show you: 0.3333333334... Multiply that by 3. The 4 becomes 2 and generates a carry: everything to the left of it becomes 0 and generates a carry.
Hence EVER adding the 0.(0)1 would mean that 1*(1/3) > 1, which is obviously false.
It's an informal proof. Still not convinced? I can make it formal if you didn't understand it.
Read my last proof, and tell me what you think of it. Tell me where you think I was wrong.
I propose this to anyone who does not believe that 0.9999... = 1. Read my proof and point me where my "error" is. You can't, because there isn't any.
It may be though that you misunderstand "...". It doesn't mean, repeated a lot of times, it means repeated infinitely, so it is about limits.
Yeah, an infinitesimal must be named. Can't be represented by something like 0.(0)1. In fact our numeric system capabilities to handle any sort of infinite numerations (uncountable sets) is perhaps at the basis of the misconceptions here. The attempt to represent 0.999... or 0.333... and even perform arithmetic operations on them is the type of error that allows one to reach conclusions such as 0.999... equal 1.
There's an error introduced on such calculations that is simply left unexplained in the eagerness to show the radical idea that irrational numbers can amount in fact to integers by some mathematical operation. A proposition that -- and forgive me my rudeness -- I find strange coming from folks dealing every day with the problems of floating point arithmetic.
But if nothing else, at least the thought that 0.999... represents a number that is close to, but never reaches, 1 is an universally accepted idea. That this is being disputed is a complete surprise to me. Maybe someone is playing devil's advocate. I can accept that since I don't have the math skills many of you do in here and haven't be able to formalize a proof. But I'm not dumb either. And can read proofs. I just haven't seen any yet that shows me 0.999... equals 1. And that's the type of claim that requires proof.
Originally Posted by brewbuck:
Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.
Of course we can represent 0.9999..., as you and I both just did. "0.999..." is the representation mathematicians agreed upon, just as "1" is what we agreed upon for using the number 1.
Another representation, according to the definition, would be in a formula:
I could have used the sigma sign for sums, but I can't write formula's properly here, but you get the point.Code:lim(x -> infinity) 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x
So as I asked you; what's wrong with MY proof. And is it really "universally accepted"?
Not according to wikipedia:
0.999... - Wikipedia, the free encyclopedia...
And any other actual mathematical resource.
The textbook answer for non-mathematicians like me is that those infinite sets are of the same size since there is a one-one correspondence between the set of natural numbers and that set of odd numbers, and a one-one correspondence between the set of natural numbers and that set of even numbers, and of course there is a one-one correspondence between the set of natural numbers and the set of natural numbers.Originally Posted by Mario F.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
I have read that article before Evoex. I find it curious however that quotes like the following are ignored though:
"Varying degrees of mathematical rigour" being the key expression here. It is ok to attribute the equality between those two numbers (or two ways to represent one, whatever you wish) within varying contexts. But not as an universal truth. The very nature of infinity stops math on its tracks and starts producing non-numbers as final results.Proofs of this equality have been formulated with varying degrees of mathematical rigour, taking into account preferred development of the real numbers, background assumptions, historical context, and target audience
You may claim all sorts of proofs. But for every single one of them, there's a counterpoint that destroys them. I did that already, with no math skills of my own to speak off, to two of so-called "proofs" in here. Both also shown on that article. You cannot name one single proof in that set of "proofs" that remains true for an infinite requirement for mathematical rigour. And that is the requirement when you are dealing with infinity.
Originally Posted by brewbuck:
Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.
no, the universally accepted idea is that 0.(9) is 1
if you want to do some reading, 0.999... - Wikipedia, the free encyclopedia... has quite a few equations that show this
edit- looks like I'm a bit late with this one
Last edited by ಠ_ಠ; 03-02-2011 at 11:37 AM.
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Are you sure?
I'm confident the mistake in your proof is here:
Now you are saying that an infinitesimal equals 0 without first proving it. You ostensively reduce accuarcy to an integer without even flinching But let's play ball and see what happens if I replace 1 by 2?= lim(x -> inf) ( 1*10^-x )
= 0
Please read the posts above before replying. I already answered that 30 minutes before you posted.Code:0.9999... is, per definition: 0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) (that's the definition of ..., agreed?) If so: 2 - lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = lim(x -> inf) ( 2 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = lim(x -> inf) ( 2*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = lim(x -> inf) ( 2*10^-2 + 9*10^-3 + ... + 9*10^-x ) ... = lim(x -> inf) ( 2*10^-x ) = still 0? Really useful it would be 1. But it isn't, is it?
Last edited by Mario F.; 03-02-2011 at 11:41 AM.
Originally Posted by brewbuck:
Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.
You have an error: 2 - 9*10^-1 != 2*10^-1. Rather, you should get:Originally Posted by Mario F.
Code:2 - lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = lim(x -> inf) ( 2 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = lim(x -> inf) ( 1 + 1*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x ) = lim(x -> inf) ( 1 + 1*10^-2 + 9*10^-3 + ... + 9*10^-x ) ... = lim(x -> inf) ( 1 + 1*10^-x ) = 1
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Ah thanks for the correction laserlight.
We still have the problem on how we get from lim(x -> inf) ( 1*10^-x ) to 0. Certainly this requires evidence of its own. We are supporting the proof of one idea by using itself. That can't do.
Originally Posted by brewbuck:
Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.
But we can do the limit and then by induction the proof would be true. Notice that the limit you mentioned involves an infinitesimal. I'd say that lim x->infinity (1/x) = 0 on the basis that for any x as x grows the quotient will be closer to zero. The limit is zero. Now you can evaluate 10^-x as -x gets smaller and get answers close to zero. That limit is also zero
Last edited by whiteflags; 03-02-2011 at 01:09 PM.