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]}
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?
Yes, that is correct and was Plato's and TheAbstractionist's point.
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).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?
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.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?