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January 10th, 2016, 05:22 PM   #1
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Which infinity for irrational numbers?

The value of a decimal number depends on the number of its digits. For irrational numbers that have infinity of digits, their values seem to be definitive. However, the meaning of infinity is ambiguous because there exist several kinds of infinities. If the infinity used to define the number of digits is not clear, the values of irrational numbers will not be well defined. This is why we have to answer the question of the title.
Please read the article at

PDF Which infinity for irrational numbers? PengKuan on Maths: Which infinity for irrational numbers?
or Word https://www.academia.edu/20147272/Wh...tional_numbers
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January 11th, 2016, 04:38 AM   #2
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Since you don't understand infinity, I suggest you stop writing nonsense about it. The irrationals are non-terminating decimals, as are many rationals. It's quite straightforward to see that the number of decimal places is countable.
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Last edited by skipjack; February 29th, 2016 at 02:13 AM.
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January 11th, 2016, 03:36 PM   #3
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How do you show the number of decimal places is countable?
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January 11th, 2016, 03:57 PM   #4
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Quote:
Originally Posted by zylo View Post
How do you show the number of decimal places is countable?
It's actually part of the very definition of the decimal representation of a real number. If $\displaystyle x \in \mathbb{R}$ then the portion to the right of the decimal point has a representation as a convergent infinite series

$\displaystyle \sum_{i = 1}^\infty d_i 10^{-i}$

where each $\displaystyle d_i$ is a decimal digit, and this is exactly what we mean when we write $\displaystyle .d_1 d_2 d_3 \dots$ (To be accurate, some real numbers have two such representations, such as .999... = 1).

It doesn't really make sense to add up uncountably many real numbers, because it can be shown that if you have some convergent uncountable sum of real numbers, all but countably many of the numbers must be 0. So countable sums are the only infinite sums that make sense to define.
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Last edited by Maschke; January 11th, 2016 at 04:15 PM.
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January 11th, 2016, 05:10 PM   #5
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Quote:
Originally Posted by Maschke View Post
It's actually part of the very definition of the decimal representation of a real number. If $\displaystyle x \in \mathbb{R}$ then the portion to the right of the decimal point has a representation as a convergent infinite series

$\displaystyle \sum_{i = 1}^\infty d_i 10^{-i}$

where each $\displaystyle d_i$ is a decimal digit, and this is exactly what we mean when we write $\displaystyle .d_1 d_2 d_3 \dots$ (To be accurate, some real numbers have two such representations, such as .999... = 1).

It doesn't really make sense to add up uncountably many real numbers, because it can be shown that if you have some convergent uncountable sum of real numbers, all but countably many of the numbers must be 0. So countable sums are the only infinite sums that make sense to define.
If "decimal places" of v8archie is places of digits in the same decimal expansion, then cardinality is not the question.

If he means decimal numbers on the real line, then the places of non-terminating decimal numbers are not countable.

Last edited by skipjack; February 29th, 2016 at 02:44 AM.
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January 11th, 2016, 06:13 PM   #6
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Quote:
Originally Posted by Pengkuan View Post
If he means decimal numbers on the real line, then the places of non-terminating decimal numbers are not countable.
But that's not what I mean a) because it's not the meaning of the phrase "decimal place"; and b) because your apparent meaning of "position in a list of all real numbers (ordered according to magnitude)" is nonsensical because there is no such list of all real numbers".

I used the standard meaning for "decimal place" because you were talking about "number of digits". Clearly, the "number of digits" in the decimal representation of a real number is countable, because each digit occurs in a decimal place.

Thus, there is no problem such as you describe: we know how many digits an irrational has - it must have countably many.

Last edited by skipjack; February 29th, 2016 at 02:46 AM.
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January 12th, 2016, 02:03 PM   #7
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Any definition containing a summation from 1 to infinity requires further clarification, because it is not at all clear how we can count from 1 to infinity.

Attempts at defining real numbers include declaring them as equivalence classes of Cauchy sequences of rational numbers, Dedekind cuts, or certain infinite "decimal representations". These definitions all require the acceptance of 'infinitely many' iterations or occurrences.

When we say "for n = 1 to infinity" or “infinitely many” what number type are we referring to?

It can't be natural numbers or integers because these cannot "go to infinity". By definition they can only take finite values. It can't even be ‘reals’ because these do not allow infinity as a value.

So presumably we must be talking about something like the Hyperreal numbers or the Surreal numbers? Note that these are extensions to the reals, which they both take as already being well-defined.

So all definitions of ‘reals’ are based on the assumption that ‘reals’ are already well-defined. In short, real numbers are not well-defined at all.
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January 12th, 2016, 02:29 PM   #8
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I suggest that, since you don't understand the notation $\displaystyle \sum_{n=0}^{\infty} a_n$, you refrain from trying to tell people about what it means.

The definition is that $\displaystyle \sum_{n=0}^{\infty} a_n =\lim_{m \to \infty} \sum_{n=0}^m a_n $, which in turn is the limit of the finite partial sums as $\displaystyle m$ grows without bound.

Note that there is no mention of infinity, so there is no confusion amongst those that know what they are talking about.

Last edited by v8archie; January 12th, 2016 at 02:31 PM.
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January 12th, 2016, 05:13 PM   #9
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Quote:
Originally Posted by v8archie View Post
I suggest that, since you don't understand the notation $\displaystyle \sum_{n=0}^{\infty} a_n$, you refrain from trying to tell people about what it means.

The definition is that $\displaystyle \sum_{n=0}^{\infty} a_n =\lim_{m \to \infty} \sum_{n=0}^m a_n $, which in turn is the limit of the finite partial sums as $\displaystyle m$ grows without bound.

Note that there is no mention of infinity, so there is no confusion amongst those that know what they are talking about.
Here you claim to avoid the issue of n going to infinity by defining the summation as the limit as m tends to infinity, then you try to avoid this use of infinity by saying this is the limit of the partial sums. But the partial sum can not equal the limit however big you make m. You are just saying it somehow can because you are defining it to.

You are simply defining the summation to be equal to the limit. You claim there is no mention of infinity but you appear to stand by the definition which is a summation of infinitely many terms.

You are simply defining the summation to be what you want it to be. I cover all this and much more in my article (http://www.extremefinitism.com/blog/...on-of-0-999-1/).

Last edited by Karma Peny; January 12th, 2016 at 05:30 PM.
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January 12th, 2016, 05:40 PM   #10
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Ummm...

I think maybe you should read up on infinite series and, in particular, limits. Once you have a thorough understanding of those then you can argue it out with archie all you like.
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