10c - c

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Don't bother with all this limits and sequence stuff. Try just using common sense, works pretty well for me.

Why do you keep insisting infinity isn't real? That doesn't really matter for this thread.
 
...except in a mathematical sense, whatever that might mean to some people.

To a physicist, real means he has a real value for some measurable, usually.
For a mathematician, it means that numbers must exist, because infinity is only real in a logical sense, as in a value that can't be determined, which means something else to a physicist.
 
Yes but 0.9r and 9.9r are not infinitely large numbers so it doesn't really matter.
 
Why do you keep insisting infinity isn't real? That doesn't really matter for this thread.
Because you cannot construct a field out of just adding infinity. There are such systems, for instance $$\mathbb{C}^{\ast}$$ is the complex plane plus a 'point at infinity' and you can do some great stuff with that (check Youtube for a video on Mobius transformations (maybe spelt Moebius, depends who you ask) but it's not a field and you need a field to do a lot of usual things in maths.
 
Sorry that post went over my head.

To clear up, I never said infinity was real.
From what I understand (I hope) infinity is not real. I just didn't think that really mattered.
 
To be able to work with something 'useful', you need particular rules of algebra. Adding infinity to the reals means you have a system which doesn't have said rules.
 
Makes perfect sense.
Don't see the problem it causes 10c-c though.
 
I think it means infinity must be "outside" something, in order for it (and the something) to be useful.
 
This is a simple argument. (1/3 expressed as a decimal to n places) x 3 is a nice, clean 1.0 for any real number of n.
No, it isn't. 1/3 to n decimal places is 0.3...3, with n 3s. That times 3 gives you 0.9...9, with n 9s. Not 1.0.
My whole point is that infinity isn't a real number,
Nobody said it was...
therefore 1/3 x 3 expressed as .9r isn't real.
No, this doesn't follow. Infinity doesn't need to be a real number for 0.9r to be a real number. Here's another decimal expansion: 1.0r. Is that not a real number? If it is, then why doesn't your logic apply in this case? Even better, try it with the sequence 0.3, 0.33, 0.333,... Are you really going to claim that 1/3 isn't real?
Try it at n = 3. Try it at n = 12341. Try it at n = 2341234823469812634. It is always 1.0.
Except that it's never 1! This is, in fact, the whole point of completion: to also have the limits of such sequences in the set!
My whole point is the way that limits require you to assume that infinity is a number you actually get to.
I think you need to learn more about limits in a formal context, to be honest.
I'm starting to grasp the fundamental disconnect.

<snip>

Real doesn't mean possible to exist.
Ah, see, now I think we're getting to the heart of the matter. Unfortunately, AlphaNumeric (or Guest, I can't remember) was absolutely right: You need to learn what the word "real" means in mathematics. "Real" has a very specific meaning, and your intuitive (not intuitionist) idea of what it means is simply wrong. Furthermore, what you actually seem to be getting at is actually the philosophy of mathematics, where one can ask such questions as: What connection does the mathematical concepts have with reality?

However, that has absolutely no impact on the hard, provable fact that 0.9r = 1.

Oh, and you definitely need to think your examples through a bit more.
Thus, a logically impossible number, such as the largest member of an infinite set is "real" to a classical mathematician and its logical impossibility doesn't make it not real.
This is completely wrong, and shows that you have pretty much no idea what you're talking about.

First of all, it is entirely possible to have a largest member of an infinite set, in a logically possible manner that both intuitionist and "classical" mathematicians will agree on. Consider the set {0,-1,-2,-3,...}. This is an infinite set. However, under the standard ordering of the integers, is has a largest element, 0, but no least element. Or the set $$\{ x \in \mathbb{R} \,|\, 0 \leq x \leq 1\}$$, which has both, under the standard ordering of the reals.

Secondly, if you can show the logical impossibility of a largest element of an infinite set, then no "classical" mathematician would argue that such an element existed. In fact, proof by contradiction is a central tool in almost all branches where mathematics is used.

Third, you have your schools completely reversed: It is the intuistionists, not the "classical", mathematicians that reject the law of the excluded middle, reductio ad absurdam etc. Saying that "classical" mathematians will not accept logical impossibility as proof is simply, and embarrasingly, given the context, wrong.

Just a gentle (or not-so-gentle) reminder that a bit of modesty is a good idea before you go putting other, far more knowledgable, members down. Especially in fields you admit to being a novice in.
 
Someone please tell me

What we're discussing here is 0.9r which has infinitely many digits in it's decimal. It is not infinity which is larger than any real number. The sum to infinity of a geometric series is something that's well defined in certain circumstances, that is, you can add together infinitely many terms and get a finite answer.

Consider the sum

$$\frac{9}{10}+\frac{9}{100}+ \frac{9}{1000}+\ldots$$

You can see that each subsequent number a tenth of the previous one. Consider adding up the first n terms - you'll get 0.999... with n 9's in it. If you sum this series to infinity you get 1 but it is also clear that this is a representation of 0.9r.

There is a formula for the sum to infinity of a geometric series: $$S = \frac{a}{1-r}$$ where a is the first term and r is the common ratio (which must be less than 1) Plug the numbers in and you will get 1.
 
What we're discussing here is 0.9r which has infinitely many digits in it's decimal. It is not infinity which is larger than any real number. The sum to infinity of a geometric series is something that's well defined in certain circumstances, that is, you can add together infinitely many terms and get a finite answer.

Consider the sum

$$\frac{9}{10}+\frac{9}{100}+ \frac{9}{1000}+\ldots$$

You can see that each subsequent number a tenth of the previous one. Consider adding up the first n terms - you'll get 0.999... with n 9's in it. If you sum this series to infinity you get 1 but it is also clear that this is a representation of 0.9r.

There is a formula for the sum to infinity of a geometric series: $$S = \frac{a}{1-r}$$ where a is the first term and r is the common ratio (which must be less than 1) Plug the numbers in and you will get 1.

Yes I know.
What I wanted telling is why infinity not being real has anything to do with the self inconsistency of the 10c - c proof of 0.9r = 1.
 
It doesn't have anything to do with it as far as I can tell. The number 0.9r has a representation that is a recurring decimal, but the number is still a number despite our awkward representation of it.
 
It's funny that this started as an effort to understand 10c - c = 9c, and has spun into me no longer believing in rational numbers. Anyway, this argument to me couldn't be any simpler. It is plainly obvious.

1. I take rational to mean "expressed as a ratio", or something like that. I don't take it that repeating decimals are, by definition, rational. It is the ability to be expressed as a fraction that makes them rational.
2. 1/9 expressed as a repeating decimal is .1r.
3. .1r x 9 = .9r


Less plainly obvious:

4. Therefore, .1r is not a valid expression of 1/9.
5. What this really shows is that you can not have a fraction expressed in infinitely repeating decimals. You need a finite number of places to express the remainder, so that the result multiplied by its reciprocal equals 1.


Just plain contentious:

6. As a result, mathematicians move reality. They say that this doesn't show that .1r isn't 1/9, it means that .9r really is 1.
7. All following proofs that .9r really equals 1 proceed from the assumption that repeating decimals really are valid expressions of rational numbers. As an example,

1/9 is not, by definition, the limit of the sum of 1/10 + 1/100 ... + 1/10^n as n approaches infinity.
1/9 is, by definition, (the sum of 1/10 + 1/100 ... + 1/10^n) + (1/9x10^n) for any n. This literally describes the process of long division in a way the above doesn't.

Math argues that the two are equal because the limit of 1/9 x 10^n as n approaches infinity is zero. But it's not a given that the limit is the expression's actual value. It says "the limit is zero", but it remains to be proven that the limit means anything. Math "proves" the limit is the actual value based on a proof that assumes repeating decimals are valid expressions of rational numbers. Which brings us back to step 1, above.

Cliffs:
8. Once you accept that repeating decimals are valid expressions of rational numbers, .9r = 1 is built in as an assumption and the resulting proofs aren't really proofs, they're demonstrations of consistency. The process has to start at proving repeating decimals are valid expressions of rational numbers. I won't say it's impossible, but I don't know how you go about that given points 1, 2 and 3, above.

9. Completely unrelated, but just to address the point, 10c - c = 9c doesn't work for .1r because .1r isn't a valid value for c. We can argue that, but it can be shown that invalid values for c exist. You cannot perform 10c - c on infinity, and my argument for .1r would be similar.
 
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