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Proof that irrational numbers do not exist

  1. Mar 1, 2011 #1
    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.
    Last edited: Mar 1, 2011
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  3. Mar 1, 2011 #2

    D H

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    No, it doesn't.

    Even 1/3 doesn't have such a form. 1/3=0.333..., repeating forever. Irrational numbers also go on forever, but they never repeat.
  4. Mar 1, 2011 #3
    yes it does x=0 and all the c's are 3
  5. Mar 1, 2011 #4
    Its just saying that all real numbers have a decimal expansion. Its bad notation, yes I know.
  6. Mar 1, 2011 #5
    It isn't exactly necessary for the argument though. The argument is that the number of decimal places in countable and therefore there better be a countable natural number big enough to multiply and get a natural number.
  7. Mar 1, 2011 #6

    D H

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    Decimal expansion is not bad notation. Yours is, and your bad notation is why your argument is false.

    That certain numbers are irrational such as the square root of 2 was known to the ancients. I also suggest your read up on Cantor's diagonalization argument and Hilbert's hotel.
  8. Mar 1, 2011 #7
    I have read and yes I know that the square root of 2 is considered to be irrational. The argument is that we can consider any number in it's decimal expansion as just having place holders. 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. It seems you are trying to apply inapplicable ideas to a very simple argument. From what I can the argument isn't false it has nothing to do with any specific irrational but it is satisfied by all reals and therefore all irrationals.
  9. Mar 1, 2011 #8
    If you see a flaw then by all means point it out and explain.
  10. Mar 1, 2011 #9
    Your way of argumentation is very unelegant (edit: well, in fact it is quite elegant, but "wrongly worded") and that may a problem why you don't see your error. That is that you assume that limit of rational numbers must be a rational number which of course doesn't hold.
  11. Mar 1, 2011 #10
    (Unconstructive) Proof that irrational numbers does exist can be following:
    Any real number between 0 and 1 in binary notation can be assigned (maped) to exactly one subset of set of natural numbers and vice versa. (for example if on n-th place [in expression like 0.01110...] is 0 then number "n" is not in the assigned subset; if there is "1", it is). Therefore there are as many (concept of "larginess" of infinite sets is nontrivial and I will assume you are familiar with it, because my language abilities are not good enough to explain it more; and I will refer to it simply as "larginess" since I don't know english expression for it) real numbers as there are subsets of natural numbers. Now - there are as many rational number as there are natural numbers, while any set of all subsets of some set is larger than original set itself. This proves the thing. (sry for uncompletness, i could prove both things but it would be needed to define "larginess" of infinite set which I'm not willing to do right now)

    Edit: "larginness" is called cardinality, read about it here if interested http://en.wikipedia.org/wiki/Cardinality
  12. Mar 1, 2011 #11
    > 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

    So 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; but most people won't find your argument very useful, nor will they understand you without a lot of clarification on your part.

    FYI, 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

    It is equal to the geometric series 1 + 10 + 100 + ... = 1 / (1-10) = -1/9. In this sense, -1/9 equal to an infinite integer, and it is larger than any finite integer.

    This kind of thing shouldn't be shocking for anyone familiar with the two's-complement representation uniquitous in computing.
    Last edited: Mar 1, 2011
  13. Mar 1, 2011 #12
    Thank you and I understand what you are saying but even so that step is irrelevant. The property that I used is commonly referred to as scientific notation and it holds in a base 10 number system. What people are failing to realize is that the nth power of 10 is entirely dependent on the nth decimal. This fact makes them equal (the infinities not the power). This is why the Hilbert's Hotel paradox does not apply. It applies to different magnitudes of infinities. The fact that the infinity is countable is purely based on the fact that you can count it which is why the diagonal problem does not apply.
  14. Mar 1, 2011 #13
    Googyae this is exactly what I was saying thank you. I couldn't find the right words. And I need this to prove something very interesting about prime numbers.
  15. Mar 1, 2011 #14
    Other examples:

    pi/4 = 3/4 * 5/4 * 7/8 * 11/12 * 13/12 * ... where each denominator is the prime numerator rounded to a multiple of 4. This clearly shows a quotient of two infinite integers.

    And if you admit regularized values for divergent products, then
    sqrt(2pi) = 1 x 2 x 3 x 4 x 5 x ... x the rest of the positive integers
    4pi^2 = 2 x 3 x 5 x 7 x 11 x ... x all the primes
  16. Mar 1, 2011 #15
    Except there are not infinitely large integers. In fact you can construct them, but such a thing is way beyond present scope of OP, so it's not of much use.

    What you are saying is that any irrational number can be written as a limit of some sequence of rational numbers. That doesn't say so much to OP, i quess.
    Well, i don't really know what you mean, but i would like to point out a fact that property of |q|<1 (q is quocient of geometrical series) is needed to derive formula 1/1-q for it's summation.
  17. Mar 1, 2011 #16
    >Except there are not infinitely large integers.

    They exist if you want them to. Just like 0, -1, i, infinitesimals and things undreamt of which will also be pooh-poohed.

    >This kind of thing shouldn't be shocking for anyone familiar with the two's-complement representation uniquitous in computing.

    Clarification: .....11111 is equal to -1/9 because if you multiply it by 9 and then add 1 you get .....99999 + 1 which "overflows" to produce 0, the expected result. ....111 is also equal to -1/9 because if you multiply by 10 and add 1 you get ....110 + 1 = ...111 and the number stays the same.

    I believe ...11111 might technically be called a "p-adic" number but I never really learned about it through that route.

    >|q|<1 (q is quocient of geometrical series) is needed

    Not 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:
    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 + ...
    Last edited: Mar 1, 2011
  18. Mar 1, 2011 #17
    Thanks for reply.

    Well, then it is quite a different thing (- and interpretation,which is questionable ), I would say. And it is in no way helpfull to OP who has problems with some basic concepts of mathematics (both formally/"exactly" and intuitionaly) .
    I know nothing about it, but it clearly depends on some "non-casual" interpretation. There is no other derivation known to me than this one:

    S_(n+1)=S_(n+1) ; i will write this identity in 2 different ways:
    1+q*S_(n)=S_(n) + q^n
    and use limit to infinity if possible.
  19. Mar 1, 2011 #18
    >There is no other derivation known to me than this one:

    There's this one, too:

    = 1 + x + x^2 + x^3 + x^4 + ...
    -x -x^2 -x^3 - x^4 - ...
    = 1
  20. Mar 1, 2011 #19
    You can't count up to the final "nth decimal" of every irrational number, whether in decimal or scientific notation. Neither can you count up to infinity. Also what do you mean by "the diagonal problem" and why does it not apply?
    Last edited: Mar 1, 2011
  21. Mar 1, 2011 #20
    Sorry, I didn't read your addition to previous post.
    I will need think about what you wrote. I'll (maybe :p) post back later.
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