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Math Help - Injective / surjective function

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    Injective / surjective function

    a. let f:R^2-->R^2 be given by f(x,y)=(x,xy) . If the mapping injective? is it surjective?

    b. Represent F(a) by a picture when A - {(x,y) ∈ R^2 | x ∈ [1,2], y ∈ [0,2]}
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Hint for part (a).

    What can you say about f(0,0) and f(0,y) where y\in\mathbb R,\,y\ne0\,? Also, can you find some (a,b)\in\mathbb R^2 such that f(a,b)=(0,y)\quad(y\ne0)\,?
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  3. #3
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    Quote Originally Posted by champrock View Post
    a. let f:R^2-->R^2 be given by f(x,y)=(x,xy) . If the mapping injective? is it surjective?
    i) f(0,1)=?~\&~f(0,2)=?

    ii) If f(a,b)=(0,2)\text{ then }a=?~\&~b=?
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    Quote Originally Posted by Plato View Post
    i) f(0,1)=?~\&~f(0,2)=?

    ii) If f(a,b)=(0,2)\text{ then }a=?~\&~b=?
    Please let me know if this is correct:
    1. the function is not Injective since f(0,1) = (0,0) and f(0,2) = (0,0). So, there exist differnt values in the domain which have the same image.

    2. The function is surjective since if f(a,b) = (0,2) then it is always possible to find atleast one value of a & b which give that image?
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  5. #5
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    Quote Originally Posted by champrock View Post
    Please let me know if this is correct:
    1. the function is not Injective since f(0,1) = (0,0) and f(0,2) = (0,0). So, there exist differnt values in the domain which have the same image.
    Yes, that is correct and was Plato's and TheAbstractionist's point.

    2. The function is surjective since if f(a,b) = (0,2) then it is always possible to find atleast one value of a & b which give that image?
    NO. Showing that there exist a, b such that f(a,b)= (0,2) does not show it is surjective. You would need to show that for any numbers x, y, there exist a, b such that f(a,b)= (x, y).

    And you did not show that there must be a, b such that f(a,b)= (0,2), you simply asserted it! If f(a,b)= (0, 2), HOW would you find a and b? Exactly what must they be?
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    And you did not show that there must be a, b such that f(a,b)= (0,2), you simply asserted it! If f(a,b)= (0, 2), HOW would you find a and b? Exactly what must they be?
    OK ya, got it. There cannot be any values of a&b which give the image as (0,2). [Since if x=0 then x*y=0 always). So the fuction is not surjective. There does not exist any value in the domain which is giving this particular value in the image.
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