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    • Thread Starter

    f:[1,2,...,n]->[1,2,...,n] is injective if and only if it is surjective

    find a function g:N->N that is

    a) injective but not injective
    b) surjective but not injective
    C) bijective

    done a) but for b) can't figure it out since surely to be surjective range must equal codomain therefore range must be natural numbers but this means to avoid being many-to-one and hence not an actual function, domain must be natural numbers as weel but this forces it to be injective.

    then the next question is

    i) find a bijection from N(subscript 0) to N
    ii) find a bijection from N(subscript 0) to Z
    iii) find a bijection between (0,2pi) and R (hint: project north pole of x^2 + y^2 =1 onto x axis)

    N(subscript 0) means natural numbers including 0

    a) I assume you mean inective but not surjective. g(x) = x^2
    b) g(x)= 1 for x=1,2 and g(x)=x-1 x>2
    c) g(x)=x

    i) f(x) = x+1
    ii) Try and do it in an orderly fashion.
    0->0, 1->1, 2->(-1), 3->2, 4->(-2) ... etc
    So f(x) = (x+1)/2 if x is odd, -x/2 if x is even and 0 if x is 0.
    • Thread Starter

    i dont understand any of what u wrote

    (Original post by latentcorpse)
    i dont understand any of what u wrote
    ssee's post is very clear; it would be very difficult to make it plainer and he's answered the questions in the order you've asked. He even corrected your question part a).

    For (i) which bit of "f(x) = x+1" is proving difficult?
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Updated: October 13, 2006

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