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- #27

mathwonk

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this belongs in the circle squaring or angle trisecting thread.

- #28

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Even tough there is an infinite number of integers, all of them are finite.

- #29

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Infinity is not an integer. That is the problem. 10^infinity makes no sense in the way that we wants to use it. You can't say infinity + infinity and get anything other than infinity. And furthermore, if lim f(n) = x as n -> a this does not mean that when f(a) = x. Consider, for example, 1/x. As x -> 0 1/x -> infinity but 1/0 isn't defined.

Even tough there is an infinite number of integers, all of them are finite.

Again, what is wrong with the proof that sqrt(2) is irrational. If his conclusion is correct, this proof has an error. Find it, please.

- #30

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Any number c in the real numbers has the form [tex]x.{c_1}{c_2}...{c_n}[/tex], in which x is an integer and [tex]0 \le {c_n} \le 9[/tex] is a natural number. From the way that we have enumerated the decimal places, clearly number of decimal places is countable. Then there is a bijection from the indexes of the decimal places onto the set of the first n natural numbers. Consider the irrational numbers, namely [tex]\mathop {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})[/tex]. The infinity in the enumeration of decimal places must then be the same infinity as that of the natural numbers. Then, [tex]\mathop {\lim }\limits_{n \to \infty } [\frac{{{{10}^n}}}{{{{10}^n}}}(x.{c_1}{c_2}...{c_n})] = \frac{{\mathop {\lim }\limits_{n \to \infty } ({{10}^n} \times x.{c_1}{c_2}...{c_n})}}{{\mathop {\lim }\limits_{n \to \infty } {{10}^n}}} = {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})[/tex]. Is a rational expression in which the numerator and denominator are both integers.

If we consider Euclid's argument and acknowledge that the factor in the numerator is not unique in yielding an integer multiple then we can understand Euclid's flaw. The numerators and denominators will have infinite prime factorizations and therefore it would be impossible to simplify the fraction.

This is even worse than I thought. You can't evaluate this limit:

[tex]

\mathop {\lim }\limits_{n \to \infty } [\frac{{{{10}^n}}}{{{{10}^n}}}(x.{c_1}{c_2}...{c_n})]

[/tex]

like you think you can.

- #31

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Any number c in the real numbers has the form [tex]x.{c_1}{c_2}...{c_n}[/tex], in which x is an integer and [tex]0 \le {c_n} \le 9[/tex] is a natural number. From the way that we have enumerated the decimal places, clearly number of decimal places is countable. Then there is a bijection from the indexes of the decimal places onto the set of the first n natural numbers. Consider the irrational numbers, namely [tex]\mathop {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})[/tex]. The infinity in the enumeration of decimal places must then be the same infinity as that of the natural numbers. Then, [tex]\mathop {\lim }\limits_{n \to \infty } [\frac{{{{10}^n}}}{{{{10}^n}}}(x.{c_1}{c_2}...{c_n})] = \frac{{\mathop {\lim }\limits_{n \to \infty } ({{10}^n} \times x.{c_1}{c_2}...{c_n})}}{{\mathop {\lim }\limits_{n \to \infty } {{10}^n}}} = {\lim }\limits_{n \to \infty } (x.{c_1}{c_2}...{c_n})[/tex]. Is a rational expression in which the numerator and denominator are both integers.

If we consider Euclid's argument and acknowledge that the factor in the numerator is not unique in yielding an integer multiple then we can understand Euclid's flaw. The numerators and denominators will have infinite prime factorizations and therefore it would be impossible to simplify the fraction.

As an additional response to this already completely debunked, completely asinine post, I'll point out that it is absolutely no surprise that it is possible to express irrational numbers as an infinite sum, which is exactly what you have done.

As a side note, assuming that what you have written actually makes any sense, I would really like to know what you have allegedly proven about prime numbers

- #32

disregardthat

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-1/12 is not the infinite sum 1 + 2 + 3 + ... The formula will simply not apply if x = -1. What you are doing is extending the domain of 1/(1-x)^2 to x = -1, but it willNot according to Euler. It's a fact that using it with |q| >= 1 is a tremendous aid to calculation, e.g. for computing values of the Riemann zeta function (pretty badly named, since of course it was Euler who discovered its reflection forumla). Quick example:

1/(1-x)=1+x+x^2+x^3+...

d/dx 1/(1-x) = 1/(1-x)^2 = 1+2x+3x^2+...

Now substitute -1 to obtain

1/4 = 1 - 2 + 3 - 4 + ... = (1 - 4) (1 + 2+ 3 + 4 +...)

-1/12 = 1 + 2 + 3 + 4 + ...

- #33

Hurkyl

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You are confusing two different things.Then we can number the place holders, i.e. tenths -->1, hundredths -->2 and so on which is clearly a countable set. Since it is countable then there must be a countable power of 10 to multiply it by and give an integer.

- For each decimal place, there exists a power of 10 to multiply it by to move it to the left of the decimal place.
- There exists a power of 10 to multiply a number by that moves all decimal places to the left of the decimal place

I'm doubt that itSo you're saying that irrational numbers aren't really irrational. They're just ratios between infinitely large integers. This makes some sense, and if this helps you think of them, go for it;

e.g. this numeral doesn't name an integer, and it isFYI, you do not always even need the quotients of infinitely large integers to produce irrational numbers. Sometimes a single integer will do.

Consider the infinitely large integer

...111111

p-adic numbers are quite useful; that's why we've defined them and even have whole fields of study devoted to them -- but it is not useful to pretend that p-adic integers are integers; that is a sure path to get all sorts of wrong ideas stuck in your head.

Yes -- but in doing so you are not computing what is commonly known as "the sum of an infinite series". In this case, you are probably referring to computing a "zeta regularized sum".But I was able to calculate in a few lines that zeta(-1) = -1/12. Divergent series, when used carefully, can be extremely practical.

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- #35

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The fact that there are reals that cannot be written in terms of integer ratios is, in fact, an important distinction if only because of the fact that they are countable whereas the entire real line isn't. There is nothing trivial about that.

Putting this aside, I grant that what you are saying is, in some sense, correct (at least what you wrote in the OP is "correct", even if it is trivial). So, what is the result you have proven with respect to prime numbers?

- #36

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I dare say that what the OP purports to prove is false because all rational numbers are either finite decimal numbers or infinite repeating decimals. That is only some infinite decimal representations, e.g. 12.66123123123..., where the ending part, i.e. 123, repeats forever are rational numbers while those infinite decimal numbers which do not have an repeating ending are irrational numbers.

The fact that there are reals that cannot be written in terms of integer ratios is, in fact, an important distinction if only because of the fact that they are countable whereas the entire real line isn't. There is nothing trivial about that.

Putting this aside, I grant that what you are saying is, in some sense, correct (at least what you wrote in the OP is "correct", even if it is trivial). So, what is the result you have proven with respect to prime numbers?

- #37

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When I said that his OP was "correct" I only meant that he was correct in asserting that irrational numbers are the limits of sequences of rational numbers. But, as I said, that is not new to anyone who was taken a course in Real Analysis.I dare say that what the OP purports to prove is false because all rational numbers are either finite decimal numbers or infinite repeating decimals. That is only some infinite decimal representations, e.g. 12.66123123123..., where the ending part, i.e. 123, repeats forever are rational numbers while those infinite decimal numbers which do not have an repeating ending are irrational numbers.

I am just interested in what he has allegedly proven about prime numbers.