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Math Help - Functions Proof

  1. #1
    Junior Member
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    Functions Proof

    Hi

    Could someone please help me with the functions problem.


    Functions Proof-untitled.jpg

    I know by definitionlecture notes)


    injective, or one-to-one, if different elements of X have different images under F: x1 6= x2 implies F(x1) 6= F(x2) (or equivalently, F(x1) = F(x2) implies x1 = x2).


    surjective, or onto, if its range is equal to Y: that is, for every y 2 Y, there is some x 2 X such that F(x) = y.



    bijective, or a one-to-one correspondence, if it is both injective and surjective.



    But i find it hard to determine wether a function is onto or one-to-one, so can someone please help me.
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  2. #2
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    Quote Originally Posted by 1234567 View Post
    Hi

    Could someone please help me with the functions problem.


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    I know by definitionlecture notes)


    injective, or one-to-one, if different elements of X have different images under F: x1 6= x2 implies F(x1) 6= F(x2) (or equivalently, F(x1) = F(x2) implies x1 = x2).


    surjective, or onto, if its range is equal to Y: that is, for every y 2 Y, there is some x 2 X such that F(x) = y.



    bijective, or a one-to-one correspondence, if it is both injective and surjective.



    But i find it hard to determine wether a function is onto or one-to-one, so can someone please help me.
    there is only finitely many to check ie only 7 and 11 to see if 1-1 and onto...so don't be discouraged, this is a quick excercise.

    hint part (a), 2^3=8=1=1^3 \in \mathbb{Z}_7 and therefore not 1-1.
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