1. 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. 2. Originally Posted by ಠ_ಠ no, it means that ∞ - ∞ = ∞, something your equation doesn't account for
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. Originally Posted by Yarin Is it? Or is 1/3 actually 0.(3) + 0.(0)1. (That is, it simply can't be accurately represented as a single decimal).
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. Originally Posted by Mario F. H
So, it's my turn to play cat and mouse with you: Name me a rational representation of 0.(9) where it becomes equal to 1 and the error margin is 0. Dishonest question, I know. But you made me one too.
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. 3. Originally Posted by EVOEx Calculators show you 0.33333333334.
Whatever calculator you're using is malfunctioning . e.g 3.3 rounds to 3.0 ! 4. Originally Posted by whiteflags It doesn't matter how many places you carry zero out to, it's going to be zero still. Unless I totally misunderstand what you mean by 0.(0)1 -- but I think if you think it means something else, then you are probably wrong.
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. 5. Originally Posted by Sipher Whatever calculator you're using is malfunctioning . e.g 3.3 rounds to 3.0 !
Hmmm you're right. I was thinking (actually I wasn't) of 0.666.... .
Same concept though ;-). 6. Originally Posted by Mario F. 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.
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:
Code:
`lim(x -> infinity) 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x`
I could have used the sigma sign for sums, but I can't write formula's properly here, but you get the point. Originally Posted by Mario F. 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.
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. 7. Originally Posted by Mario F.
This is all to do with the proportions of infinity. Let me give you an example with natural numbers:

Let's look at an infinite series of natural numbers; 1, 2, 3, 4, ....
Let's remove from that series all even numbers; 2, 4, 6, ...

We now have two infinite series. One with all odd numbers and one with all even numbers. It could be said that these two infinities are smaller than the infinity of natural numbers. But all are infinite. We will also have an hard time explaining which of the odd or even infinities is larger. One should be larger than the other because we took each number alternatively from an infinite series.
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. 8. I have read that article before Evoex. I find it curious however that quotes like the following are ignored though:
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
"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.

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. 9. Originally Posted by Mario F. 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.

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.
So do it again with my last proof. The lim x->infinity one. You won't be able to find something wrong there, I bet. 10. Originally Posted by Mario F. 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.
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 11. Originally Posted by EVOEx So do it again with my last proof. The lim x->infinity one. You won't be able to find something wrong there, I bet.
Are you sure?

I'm confident the mistake in your proof is here:
= lim(x -> inf) ( 1*10^-x )
= 0
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?

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?``` Originally Posted by ಠ_ಠ 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 12. Originally Posted by Mario F.
But let's play ball and see what happens if I replace 1 by 2?
You have an error: 2 - 9*10^-1 != 2*10^-1. Rather, you should get:
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``` 13. 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. 14. 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 15. Originally Posted by Mario F. 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?
No that's not what I'm saying. I'm saying THE LIMIT of that number goes to 0 as x goes to infinity. It's the definition of limits, and maybe you should look up limits before telling me I'm wrong with an easy limit ;-). Popular pages Recent additions 