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There exists one number N

  1. Mar 27, 2008 #1
    The set N of natural numbers = {1, 2, 3, 4, ...}.

    But there exists one (1) number N, such that

    N = 12345678910111213... (where the unit's place is at infinity).

    A good example of an irrational number then would be

    Last edited: Mar 27, 2008
  2. jcsd
  3. Mar 27, 2008 #2
    Doesn't that make N = infinity?
  4. Mar 27, 2008 #3


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    so, put a decimal in front of it.

    oh he did that.
  5. Mar 27, 2008 #4


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    No, there is no such number. All integers have only a finite number of digits. By the way, it is not at all a good idea by using "N" to represent the set of natural numbers and then say that "N" is a number.

    Now THAT is a perfectly good irrataional number.
  6. Mar 27, 2008 #5


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    Are you sure that decimal string actually denotes a number? How can the unit's place be 'at infinity'? What digit is in the unit's place?
  7. Mar 27, 2008 #6


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    That's 10 times Champernowne constant.
  8. Mar 28, 2008 #7


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    If it were possible to construct any irrational number by putting a decimal into some positive integer, that would imply that the set of irrational numbers is countable.
  9. Mar 28, 2008 #8

    good question
  10. Mar 28, 2008 #9
    It is necessary that N is not an interger, but it is one number.
  11. Mar 28, 2008 #10
    you can call it anything you want
  12. Mar 28, 2008 #11
    lim f(x) (as x approaches infinty) is infinity, but N is a single number (not a variable).
    Last edited: Mar 28, 2008
  13. Mar 28, 2008 #12
    a definition of an irrational number is a number that cannot be expressed in the form m/n, where m and n are intergers and n not equal to zero. such numbers are infinite to right of the decimal point and do not repeat. for example,

  14. Mar 28, 2008 #13


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    Yes, so together with Hallsofivy's statement you know that 123456789101112... is not an integer.
  15. Mar 28, 2008 #14
    All Numbers must have a meaning such that a rational number can be found to approximate the number within a chosen value, a expression that is an infinite string of numbers without any fixed decimal point does not have any meaning and is not a number.
    Last edited: Mar 29, 2008
  16. Mar 30, 2008 #15


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    Yes it was. Since it was about your post, do you have a good answer?

    Okay, what do you mean by "number". And my criticism was simply about using the same symbol, N, with two different meanings.

    Thank you. But I do prefer to use standard terminology. If you did that, it might be easier to understand what you are trying to say.

    ??This is the first time you mentioned "f(x)". Where did that come from. Once again, the N you posit is NOT a "number" by any standard definition.

    Yes, we know that- it is not necessary to state the obvious.
  17. Mar 31, 2008 #16
    Yes. I think it does.

    If f(x) = x, then

    lim (of f(x) as x approaches infinity) = infinity = N. (but the unit's place of N is at infinity.)

    Last edited: Mar 31, 2008
  18. Mar 31, 2008 #17


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    But since "infinity" is not an integer, you know that N isn't an integer.
  19. Apr 2, 2008 #18
    1234567891011... is not a conventional way of representing real numbers, so unless you introduce your own convention, it doesn't mean anything. Whereas if you put a disimal point somewhere, it represents a real number in a conventional sense. Because, by convention, 1.234567... represents some real number to which the sequence, 1, 1.2, 1.23, 1.234, ... converges. This is what we call the completeness of R. If we agree to say that 1234567891011... represents where the sequence 1, 12, 123, 1234, ... go, then we may call it infinity, or more precisely, we introduce the concept of infinity.
  20. Apr 2, 2008 #19


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    Nitpicking: that's "a concept of infinity", not "the concept of infinity".
  21. Apr 2, 2008 #20
    No need to stay confused. Let go of your mindset which says it should be possible to put a decimal at a point of infinity of a string of numbers and have something meaningful. There can only be a finite string of numbers prior a decimal point to have anything resembling a number.
  22. Apr 2, 2008 #21
    1. N = infinity = 1234567891011121314... (a single number, where the unit's place is at N).

    2. If f(x) = x, then

    lim (as x approaches N) of f(x) = N.

    3. There must be a field, a set of elements having two operations, designated addition and multiplication, satisfying the conditions that multiplication is distributive over addition, that the set is a group under addition (where N is the unit of all the other elements in the set), and that the elements with the exception of an additive identity form a group under multiplication.

    a. If X, Y, and Z are elements in the said set, then

    1.) X + Y = Y + X.

    2.) X*Y = Y*X.

    3.) (X + Y) + Z = X + (Y + Z).

    4.) (X*Y)*Z = X*(Y*Z).

    5.) X*(Y + Z) = X*Y + X*Z.

    6.) 0 + X = X.

    7.) N*N^(-1) = 1.

    4. (0.123456789101112...)*10^N = N.
  23. Apr 2, 2008 #22
    This is nonsense [tex]\frac{\infty}{\infty}[/tex] is undefined and [tex]\infty[/tex] is not a number.
  24. Apr 3, 2008 #23


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    Once again, that is nonsense. You have not defined any number space in which such a thing can exist.

    Since N is undefined, this is also nonsense.

    Why must there exist such a field? Because you say so? You may be attempting to do what was suggested before and trying to define a number space in which N can exist. Unfortunately, you cannot define N first and then define the number space!

    What do you mean by "1" here? I thought you had said that N was the "unit" (multiplicative identity) for this field so "1" makes no sense.
  25. Apr 3, 2008 #24


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    I don't understand your point #1, the 'definition' of N. When you're defining your own field you can't just use ellipses that vaguely! But I can address your third point, assuming that N is some distinguished element of a set S over which your field lies:

    There does exist at least one such field: GF(2) suffices, for example. Here's the correspondence:
    0 := 0
    1 := 1
    N := 1
    N[itex]^{-1}[/itex] := 1

    GF(3) also suffices. Here's one correspondence:
    0 := 0
    1 := 1
    N := 2
    N[itex]^{-1}[/itex] := 2

    Heck, any field suffices, since there has to exist some invertible element, which is all 7 requires. (Of course all nonzero elements are invertible in a field.)

    Now in the context of a field, let's examine the 'definition' for N:
    "N = infinity = 1234567891011121314..."

    Now "infinity" has no meaning in abstract algebra, so we ignore that part: clearly, that's just an alternate name for N. But "1234567891011121314..." seems to have meaning: let's examine that.

    Definitions (over a generic field with additive identity 0 and multiplicative identity 1):
    0 := 0
    1 := 1
    2 := 1 + 1
    3 := 1 + 1 + 1
    4 := 1 + 1 + 1 + 1
    . . .
    9 := 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
    f(n) := n * (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1)

    Further, define g(n) to be the digit in the nth decimal place of the Champernowne constant, defined additively as above.

    Let [tex]N_0 = 0[/tex] and [tex]N_k=f(N_{k-1})+g(k).[/tex]

    Now N is naturally defined as the limit of the [tex]N_k[/tex] if such limit is defined. But I can't think of any field in which it is defined, unless you consider the degenerate 0=1 (which convention does not consider a field). It's not defined for any Galois field, and it's not defined for the real or rational numbers.
  26. Apr 3, 2008 #25
    I have newbie question.
    I have recently discovered that X^0 = 1
    What happens if X = 0
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