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June 19th, 2017, 07:05 PM   #171
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Quote:
 Originally Posted by Mariga many times I will emphasize that the least irrational number is 0.00...001. Such a number exists.
No, it doesn't. If it did, then what is the product of this fictitious number multiplied by $\frac{1}{2}$? Your proposed set of reals would not be closed under multiplication. Also, if I call your fictitious number $x$, then what is the result of:

$$x + x + x + x + x + \dots \text{?}$$

Your version of the reals under this definition lacks the Archimedean property: https://en.wikipedia.org/wiki/Archimedean_property

I suggest you read an introduction to real analysis. It helped me when I was struggling with some of the concepts that you are. In particular, the reals exhibit the Archimedean Property (see page 5 = page 13 of the .pdf file: http://ramanujan.math.trinity.edu/wt...L_ANALYSIS.PDF )

"The property of the real numbers described in the next theorem is called the Archimedean property. Intuitively, it states that it is possible to exceed any positive number, no matter how large, by adding an arbitrary positive number, no matter how small, to itself sufficiently many times."

Quote:
 Originally Posted by Mariga So long as this is mathematically unacceptable, then it is impossible to write down a well ordered set of all positive real numbers.
You won't be able to write it down like you have been attempting to, but a well ordering exists if you adopt the axiom of choice.

"If you work in $L$ (that is, you assume the axiom of constructibility) then a specific formula is known that defines a well ordering of the reals in that context."

Basically, people have pushed these concepts far beyond what you are doing here. You aren't going to get anywhere because you refuse to learn the basics. We can't keep explaining to you that 1 - 0.999... = 0 as opposed to "the least [positive] irrational number." There is no 'least positive irrational number.' It doesn't exist.

Further, a well ordering of the reals does not require a least positive irrational number under the standard $\leq$ ordering. It only requires a least element under some different ordering. For example, 0.5 could be the first element in a well ordering of the reals followed by $\pi$. You still haven't grasped the concept of a well ordering.

 June 19th, 2017, 09:55 PM #172 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Well then. The number 0.00..01 doesn't exist. Even if it exists it is irrelevant in this case. Also operations on such the number doesn't give meaningful results and renders it meaningless. Well ordering is simply about having a pattern from which all elements of the set to be ordered would follow. A standard ordering is an example of such a pattern. That's how I understand the concept and it might be I haven't yet really grasped it. A well order of all positive real numbers exists. It is impossible to write it down.
June 19th, 2017, 10:33 PM   #173
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Quote:
 Originally Posted by Mariga I know the number $0.\dot{9}$ is a number that represents 1, but here am talking about its real value, not as an approximation.
You have this all completely backwards.
• The real numbers are the limits of sequences of rational numbers.
• The decimal representation of a real number is not a number in itself, it's a representation of a sequence of rational numbers.
• In particular the decimal $0.a_1 a_2 a_3 \ldots$ represents the series $$\sum_{n=1}^\infty \frac{a_n}{10^n} = \lim_{m \to \infty} \sum_{n=1}^m \frac{a_n}{10^n}$$ Being a series it represents the limit of the infinite sequence of the finite partial sums as shown above.

Quote:
 Originally Posted by Mariga if the number $0.\dot{3}$ is equal to $\frac{1}{3}$ then $0.\dot{3}$ is not equal to 0.333...
This again is false, it's the same problem. $0.333\ldots$ and $0.\dot3$ are both representations of the same infinite sequence. The limit of that sequence is $\frac13$. It is important to understand that we cannot evaluate the infinite sum $\frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots$. The best that we can do is to evaluate the series - that is the limit of the sequence of finite partial sums. That sequence contains only rational numbers and thus the limit is a real number. And two ways to represent that sequence are $0.333\ldots$ and $0.\dot3$.

Quote:
 Originally Posted by Mariga So long as this is mathematically unacceptable, then it is impossible to write down a well ordered set of all positive real numbers.
You need to understand that a well-ordering of the set of positive real numbers does not use the use the normal ordering relation $\lt$. It uses a different relation under which we say that the first number we pick is the "smallest", the next is the next "smallest", etc.. Then, the only problem is to show that, despite not being listable (countable) we can (in theory) select the real numbers one at a time until they are exhausted. This is completely counter-intuitive, but is facilitated by the axiom of choice. It's kind of barmy but logical at the same time - in fact that is what will send you to the mad house.

Last edited by v8archie; June 19th, 2017 at 10:37 PM.

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