10c - c

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Here's a quick construction on why .1r isn't a valid value for c in 10c - c:

When you take .333333333 as c, and you multiply it by 10, you get 3.33333333. That is, one less 3 right of the decimal.

You subtract .333333333 from 3.33333333, and you get 2.999999997. And that really is 9c.

The quick answer is that when you multiply .1r by 10, you don't lose a number of the right side of the decimal. And this alters the mechanics of the operation.
 


That makes a lot of the same points I'm making. And illustrates my point.

(assuming of course that the sequences continue in the same way) and 1.9999999... is the same number as 2. (About this last example, by the way, there can be no argument, since I am giving a definition. I can do this in whatever way I please, and it pleases me to stipulate that 1.999999... =2 and to make similar stipulations whenever I have an infinite string of nines.)

He proceeds from assuming 1.99999.... = 2 without the need to prove it.

I'm not sure if maybe that's your point.

Added: That's definitely my point, by the way.
 
He proceeds from assuming 1.99999.... = 2 without the need to prove it.
Only because the page is just a qualitative explaination. If you want a more rigorous approach see here.
That makes a lot of the same points I'm making. And illustrates my point.
No, it doesn't. You deny the existence of $$\sqrt{2}$$. He explains why it's a real number.
I'm not sure if maybe that's your point.
No, my point is that one of the greatest living mathematicians say you're wrong.

You wouldn't happen to have the first name 'Steven' would you?
 
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.

Who told you that? According to you 1/9 = 0.1, 0.11, 0.111 etc all at the same time.
Using your strange version of maths we can do things like this.

1/9 = 0.1
1/9 = 0.11
1/9 - 1/9 = 0.01
0 = 0.01
 
Only because the page is just a qualitative explaination. If you want a more rigorous approach see here.
No, it doesn't. You deny the existence of $$\sqrt{2}$$. He explains why it's a real number.
No, my point is that one of the greatest living mathematicians say you're wrong.

You wouldn't happen to have the first name 'Steven' would you?


Your contributions to this thread have been funny. That's like 2 strikes. One more and you're out.

I don't deny that the square root of 2 is real. What I deny is an "actual infinity" such that you can express the square root of 2 to an actual infinite number of places. I think any expression must include some indication of how to either determine the next number in the sequence, or otherwise indicate that the sequence is known to be incomplete.

And, again, your link doesn't say I'm wrong. It says exactly what I say. And, in fact, it acknowledges that problems with considering the real numbers as infinite decimal expansions, and making .9r equal to 1 are known deficiencies that mathematicians still struggle with.
 
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.

In other words, you have "progressed" in this thread from misunderstanding fringe mathematics (intuitionism) to misunderstanding even the most basic of mathematics. Have fun in Cuckooland.

BTW, you quoted me out of context here:
No, but it does assume that $$10\times0.\overline{9} = 9.\overline{9}$$ which is why many mathematicians reject this as a proof that $$0.\overline 9 = 1$$
Oh, man, thanks. That's it. I will stop torturing you with this other stuff. This is all I wanted to hear. Thanks.
You cut off the part where I said why mathematicians reject this as a proof. They don't reject it because it is false (it is not false; 0.999...=1); they reject it because it lacks rigor.

1. I take rational to mean "expressed as a ratio", or something like that.
"Expressible as a ratio of integers" is a better way of saying it.

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.
All repeating decimals are expressible as a ratio of integers and are thus by definition rationals.
2. 1/9 expressed as a repeating decimal is .1r.
3. .1r x 9 = .9r
Both statements are true. Your point?

Less plainly obvious:

4. Therefore, .1r is not a valid expression of 1/9.
Ohhhh. That's your point. This is simple, elementary school mathematics that you are rejecting.

Just plain contentious: As a result, mathematicians move reality.
Mathematics is not about reality. That is the realm of science, engineering, finance, etc. Show me a one in the real world -- not one apple or one dog or one dollar or the symbol 1 -- I want you to show me a one. Science, engineering, finance, etc. use mathematics as a tool by relating the concepts developed by mathematicians to the real world.

7. All following proofs that .9r really equals 1 proceed from the assumption that repeating decimals really are valid expressions of rational numbers.
It's not an assumption, its a theorem called the division algorithm.

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.
This time do not quote me out of context. Read and understand this and the next paragraph, please. First off, mathematicians do not define 1/9 to be equal to $$\sum_{n=1}^{\infty}\frac 1{10^n}$$

Defining every possible fraction in this way would be downright silly and utterly fruitless, as there are an infinite number of rational numbers. Fortunately there is no reason to do this because the division algorithm tells us how to translate any rational expressed in the form p/q to a decimal representation (or to a representation in any other base, for that matter).

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.
Yes, it is.
 
Your contributions to this thread have been funny. That's like 2 strikes. One more and you're out.
It's not my fault you don't understand it.
What I deny is an "actual infinity" such that you can express the square root of 2 to an actual infinite number of places
Why would we need to to know all it's properties?
I think any expression must include some indication of how to either determine the next number in the sequence, or otherwise indicate that the sequence is known to be incomplete..
The link explains how to compute the next term in the expansion. Didn't you read it? For instance, given a random root like $$\sqrt{23}$$ and an length expansion, I can compute the next digit. Can't you?
Ind, again, your link doesn't say I'm wrong. It says exactly what I say.
Yes, it does. It's specifically about thinking of decimals as infinite expansions. He says you can define the reals that way if you want.
and making .9r equal to 1 are known deficiencies that mathematicians still struggle with.
Where does it say that? I went to that lecture course, I remember it well. He never said that and the lecture notes don't say it. He specifically says on his website that 1.9r = 2, makes no mention of maths having problems and you'll find a great many proofs online.

By the way, did you ignore my question about your name because you are called Steven?
 
Well, deconstruct it and tell me what's wrong with it.

One defines .9r. One defines 1/9. Nothing about the first suggests it even claims to represent 1/9.

According to you 1/9 = 0.1, 0.11, 0.111 etc all at the same time.
Using your strange version of maths we can do things like this.

1/9 = 0.1
1/9 = 0.11
1/9 - 1/9 = 0.01
0 = 0.01
 
In other words, you have "progressed" in this thread from misunderstanding fringe mathematics (intuitionism) to misunderstanding even the most basic of mathematics. Have fun in Cuckooland.

Okay, I'll start at the top again. My argument is that it is not possible for classical mathematics to prove .9r = 1, because it assumes .9r = 1. It defines repeating decimals to be rational, even though they equal .9r when multiplied by their reciprocals. There is no "proof" that .9r = 1. It is definitionally so.

You cannot get past my third point without assuming .9r equals 1.

That's it. All I'm saying is that you're not saying anything "deep" when you say .9r = 1. All you're saying is "we define this difference not to exist".
 
I can prove in an easy to understand way that 0.9r = 1

Do you want me to? Or are you just going to ignore the posts you don't like as per the norm?
 
According to you 1/9 = 0.1, 0.11, 0.111 etc all at the same time.
Using your strange version of maths we can do things like this.

1/9 = 0.1
1/9 = 0.11
1/9 - 1/9 = 0.01
0 = 0.01

This is way, way wrong.

The set of numbers in the set the sum of 1/10 + 1/100 ... + 1/10^n as n approaches infininity do not claim to be 1/9 at any n. It does not claim that at n = 1, the result (0.1) equals 1/9, or that at n =2 the result (0.11) equals 1/9.

Which is my problem with it. At any real n, the sum of 1/10 + 1/100 ... + 1/10^n is not 1/9. It is only at n = infinity that it is supposed to equal 1/9.
 
I can prove in an easy to understand way that 0.9r = 1

Do you want me to? Or are you just going to ignore the posts you don't like as per the norm?


Don't this:

Here's a quick construction on why .1r isn't a valid value for c in 10c - c:

When you take .333333333 as c, and you multiply it by 10, you get 3.33333333. That is, one less 3 right of the decimal.

You subtract .333333333 from 3.33333333, and you get 2.999999997. And that really is 9c.

The quick answer is that when you multiply .1r by 10, you don't lose a number of the right side of the decimal. And this alters the mechanics of the operation.

and this:

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.

Show I'm trying not to ignore you?
 
I am quoting YOU here...
This is way, way wrong.

The set of numbers in the set the sum of 1/10 + 1/100 ... + 1/10^n as n approaches infininity do not claim to be 1/9 at any n. It does not claim that at n = 1, the result (0.1) equals 1/9, or that at n =2 the result (0.11) equals 1/9.

Which is my problem with it. At any real n, the sum of 1/10 + 1/100 ... + 1/10^n is not 1/9. It is only at n = infinity that it is supposed to equal 1/9.
1/9 is, by definition, (the sum of 1/10 + 1/100 ... + 1/10^n) + (1/9x10^n) for any n
Stop changing your mind whenever it suits.
 
It's not my fault you don't understand it.

Your two strikes were not knowing what intuitionism was, and linking an article that expressly said it assumed .9r to be 1 by definition and therefore didn't require proof.

Why would we need to to know all it's properties?

Why do you otherwise think I don't think the square root of 2 is real? I think it's real.



By the way, did you ignore my question about your name because you are called Steven?

No. Not Steven.
 
I am quoting YOU here...


Stop changing your mind whenever it suits.

1/9 is, by definition, (the sum of 1/10 + 1/100 ... + 1/10^n) + (1/9x10^n) for any n

At what n is this .1 or .11?

At n = 1 it is .1 and 1/90.

At n = 2 it is .11 and 1/900.
 
Okay, I'll start at the top again. My argument is that it is not possible for classical mathematics to prove .9r = 1, because it assumes .9r = 1. It defines repeating decimals to be rational, even though they equal .9r when multiplied by their reciprocals. There is no "proof" that .9r = 1. It is definitionally so.
One last time. Mathematicians do not define 0.999... to be equal to one. They prove that it is equal to one by a number of means, some more rigorous than others. What mathematicians define is the concept of a convergent series. With this definition, the real numbers are defined as the set of all convergent Cauchy sequences. Nowhere do mathematicians define .999... to be equal to one.

Hey Ben! Isn't it about time to lock this thread? We have a full-fledged crackpot on our hands here. This has deteriorated into yet another thread about $$0.\overline 9 \ne 1\ .$$
 
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