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Math Help - Advanced Calculus Help

  1. #1
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    Advanced Calculus Help

    A) For each part, find a function f: R -> R that has the desired properties: neither onto nor one-to-one

    B)Under what conditions does A\(A\B) = B?

    C)Define f:J -> N(natural numbers) where f(n) = 2n-1 for each n element of N(natural numbers).

    D) Given A = {1,2,3,4,5}, B = {2,3,4,5,6,7} and C = {a,b,c,d,e} state an example of f: A ->B, g: B-> C, such that g(f)(the composition of g onto f) is 1-1 but g is not 1-1.
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  2. #2
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    Talking

    Quote Originally Posted by algebrapro18 View Post
    B)Under what conditions does A\(A\B) = B?
    See #2 of http://www.mathhelpforum.com/math-he...ts-proofs.html by kathrynmath.
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    I read that thread and still can't see the answer.
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  4. #4
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    Question Solution for B

    As we have msntioned previously, A\backslash B=A\cap B^{c}, where B^{c} is the compliment set of B.
    Noting that (B^{c})^{c}=B, and simplifying the given expression, we get
    A\backslash(A\backslash B)=A\backslash(A\cap B^{c})
    .............. =A\cap(A\cap B^{c})^{c}
    .............. =A\cap(A^{c}\cup B)
    .............. =\underset{\emptyset}{\underbrace{(A\cap A^{c})}}\cup(A\cap B)
    .............. =A\cap B.\qquad(*)
    Since (*) is equal to B, we infer that B\subset A is true.
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  5. #5
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    Thanks, that helps me with part B now all I need help with is C because I got A and D done by my self.
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  6. #6
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by algebrapro18 View Post
    Thanks, that helps me with part B now all I need help with is C because I got A and D done by my self.
    Can you please explain C a little bit more?
    What is J or should we find what it is?
    Also do we still want the function f not to be onto and into again?
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    C)Define f:J -> N(natural numbers) where f(n) = 2n-1 for each n which is an element of N.

    I need to use what was given(the line above is all that is given) to find: the image of f, if f is 1-1, if f is onto, and if f is onto I need to find its inverse, the domain of the inverse, and the range of the inverse.

    here is what I have so far:

    The Im(f) has to be all odd numbers because that is what you get when you plug numbers into f(n)=2n-1.

    from there I get stumped.
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  8. #8
    Senior Member bkarpuz's Avatar
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    Lightbulb Solution for C

    Quote Originally Posted by algebrapro18 View Post
    C)Define f:J -> N(natural numbers) where f(n) = 2n-1 for each n element of N(natural numbers).
    Although the following remark is not applicable for this exercise, I would like to tell it.

    Remark. Let f:A\to B be a function and A,B be finite sets if f is onto, then it is one-to-one; vice versa, if f is one-to-one, then it is onto.

    if J\not\subset K:=\Big\{1,\frac{3}{2},2,\frac{5}{2},\ldots\Big\}, then f can not be a function with an image which is a subset of \mathbb{N} (pick an element which is not in the set K, and see that it is mapped into the set \mathbb{R}\backslash\mathbb{N}).
    Therefore, we must have J\subset K.
    If J=K, then f is one-to-one and onto.
    If J\neq K, then f is only one-to-one.

    Just, try to figure it out by yourself by letting f(n)=2n-1=m, where n\in K and m\in\mathbb{N}. Note that f is strictly increasing, which indicates that it is one-to-one.
    Then obtain the set K by isolating n...
    Last edited by bkarpuz; September 11th 2008 at 10:41 AM. Reason: Title added.
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  9. #9
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    In english please...I understood about 1 persent of what you said...
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  10. #10
    Senior Member bkarpuz's Avatar
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    Exclamation In English

    Quote Originally Posted by algebrapro18 View Post
    In english please...I understood about 1 persent of what you said...
    If f is a function from J to \mathbb{N} defined by f(n)=2n-1, we see that 2n-1\in\mathbb{N} for all n\in J.
    This means that for every n\in J, there exists m\in\mathbb{N} such that 2n-1=m holds.
    Note that for every m\in\mathbb{N}, we may not find n\in J.
    Therefore, we see that
    <br />
n=\frac{m+1}{2}\text{ for }m\in\mathbb{N}<br />
    holds.
    Since m\in\mathbb{N}, we this indicates that
    <br />
n\in K:=\Big\{1,\frac{3}{2},2,\frac{5}{2},3,\frac{7}{2}  ,\ldots\Big\}.<br />
    Hence, the domain of f can be picked to be any subset of K, i.e. J\subset K.
    Now consider the following possible cases.
    If J=K, then f is one-to-one and onto.
    If J\neq K, then f is only one-to-one.

    Hint. f is strictly icreasing, and it is hence one-to-one.

    I guess it is more clear now?
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