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Math Help - Set Theroy

  1. #1
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    Set Theroy

    Let D={2,3,4} and E={1,2,3} define the Funcation f: D->E by the rule
    f(1)=1, f(2)=3, f(3)=1 and f(4)=2



    a)What is the image of 3
    b) what is the codonmain of f
    c)draw the digraph representing f
    d) is the funcation injective or subjectiv enad why

    so i have done question similiar to this but i am totally stumped by this one do if any help would be much appricated and the more detailed the more i can learn form it

    many many thanks

    Tom
    Last edited by Tom101; April 8th 2008 at 12:14 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Tom101 View Post
    Let D={2,3,4} and E={1,2,3} define the Funcation f: D->E by the rule
    f(1)=1, f(2)=3, f(3)=1 and f(4)=2



    a)What is the image of 3
    another way to phrase this question is: what does f(3) = ?

    can you answer now?

    b) what is the codonmain of f
    if g:A -> B is a function. the domain is A and the codomain is B

    you should look up the definitions of "domain" and "codomain"


    c)draw the disgraph representing f
    what's a "disgraph"?


    d) is the funcation injective or subjectiv enad why
    injective means one to one.

    one to one means. If f(a) = f(b), then a = b, for all elements in the domain. So basically, if two images are the same, then the number in the domain that maps to them must be the same.

    Hint: find two images that are the same, do they come from the same thing?


    surjective means onto. this means that EVERY element in the codomain is mapped to, that is, every element in the codomain is the image of some element in the domain
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  3. #3
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    thanks, i have done most of it now, a disgraph was a typo and i meant a digraph

    many thanks
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Tom101 View Post
    thanks, i have done most of it now, a disgraph was a typo and i meant a digraph

    many thanks
    still don't know what a digraph is. i will look it up if i have time, but you probably solved the problem already
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