10c - c

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.

 

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

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.

 

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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?

 

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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

 

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.

 

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.

 

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:

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.

"Expressible as a ratio of integers" is a better way of saying it.

All repeating decimals are expressible as a ratio of integers and are thus by definition rationals.

Both statements are true. Your point?

Ohhhh. That's your point. This is simple, elementary school mathematics that you are rejecting.

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.

It's not an assumption, its a theorem called the division algorithm.

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).

Yes, it is.

 

It's not my fault you don't understand it.

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

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?

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.

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?

 

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

 

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?

 

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.

 

Don't this:

and this:

Show I'm trying not to ignore you?

 

What is wrong with that sum to infinity? Have you heard of Zeno's paradox?

 

I am quoting YOU here...

Stop changing your mind whenever it suits.

 

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

No. Not Steven.

 

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.

 

Just let it lye for fuck sake. 0.9 does equal 1 mathematically.

 

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 \(0.\overline 9 \ne 1\ .\)

 

It does? This comes to me as a very great surprise. Would you care to explain your reasoning here?