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Math Help - Integral Domain

  1. #1
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    Integral Domain

    I have a doubt. I know C ( [0,1] , R ) ie, continuous functions from [0,1] to the real numbers is not a integral domain. But C ( R , R ), ie, continuous functions the real numbers to the real numbers is a integral domain?

    Regards...
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by orbit View Post
    I have a doubt. I know C ( [0,1] , R ) ie, continuous functions from [0,1] to the real numbers is not a integral domain. But C ( R , R ), ie, continuous functions the real numbers to the real numbers is a integral domain?

    Regards...
    I mean, of course it isn't for the same reason C[0,1] isn't, just construct two nonzero functions which are zero precisely when the other isnt and multiply.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    I mean, of course it isn't for the same reason C[0,1] isn't, just construct two nonzero functions which are zero precisely when the other isnt and multiply.
    Of course is isnīt or of course it is?
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    for example, let

    f(x) = 0, x < 1/3
    f(x) = √[(1/36) - (x-(1/2))^2], 1/3 ≤ x ≤ 2/3
    f(x) = 0, x > 2/3

    g(x) = 0, x < 0
    g(x) = √[(1/36) - (x-(1/6))^2], 0 ≤ x ≤ 1/3
    g(x) = 0, x > 1/3

    then f and g are two nonzero continuous functions on R, but fg = 0, for all x.
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    Mmmm... so how can I prove that if C ( R , R ) is the ring continuous functions the real numbers to the real numbers. And if f is in C ( R , R ) such that f^2=f then f is the constant function f(x)=0 or is the constant function f(x)=1 ?
    Last edited by orbit; May 21st 2011 at 11:16 PM.
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    for any x in R, (f(x))^2 = f(x), and R IS an integral domain, because it is a field. so for any x in R we have: f(x)(f(x) - 1) = 0, hence at any particular x, f(x) = 0, or
    f(x) = 1.

    continuity then forces f to be constant.
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  7. #7
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    Quote Originally Posted by Deveno View Post
    for any x in R, (f(x))^2 = f(x), and R IS an integral domain, because it is a field. so for any x in R we have: f(x)(f(x) - 1) = 0, hence at any particular x, f(x) = 0, or
    f(x) = 1.

    continuity then forces f to be constant.
    HI, but Drextel said C( [R,R]) was not an integral domain....
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  8. #8
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    C([R,R]) isn't an integral domain. but f(x) isn't the FUNCTION, it's the VALUE of the function at the point x. and the values of f are real numbers.
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  9. #9
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    Quote Originally Posted by Deveno View Post
    C([R,R]) isn't an integral domain. but f(x) isn't the FUNCTION, it's the VALUE of the function at the point x. and the values of f are real numbers.
    I understand, but in the first example I gave, ie C([0,1],R), isnīt an integral domain and the values of f(x) also are in R.....
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  10. #10
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    and if f^2 = f on the entire interval [0,1], we have the same result: f must be constant on [0,1] and equal to 0 or 1 on the entire interval.

    yes, if the ring of functions on EITHER domain was an integral domain, then we could conclude that either f or f-1 was the constant 0-function.

    unfortunately, that isn't the case. however, the values of f and f-1, are in an integral domain, R.

    now, that means we have a function f, THAT FOR EACH VALUE OF x IN THE DOMAIN OF THE FUNCTION, is either 0, or 1.

    we can't have f take on both values in any subinterval of the domain, or else (by the intermediate value theorem -since f is CONTINUOUS),

    f would take on EVERY value between 0 and 1, and we know that it ONLY has the values 0 and/or 1.

    so f must be constant.

    this has very little to do with integral domains, and everything to do with continuity.
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