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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- Jun 10th 2009, 07:20 AMchamprockInjective / 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]} - Jun 10th 2009, 07:33 AMTheAbstractionist
Hint for part (a).

What can you say about $\displaystyle f(0,0)$ and $\displaystyle f(0,y)$ where $\displaystyle y\in\mathbb R,\,y\ne0\,?$ Also, can you find some $\displaystyle (a,b)\in\mathbb R^2$ such that $\displaystyle f(a,b)=(0,y)\quad(y\ne0)\,?$ - Jun 10th 2009, 07:42 AMPlato
- Jun 12th 2009, 10:26 PMchamprock
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? - Jun 13th 2009, 02:30 AMHallsofIvy
Yes, that is correct and was Plato's and TheAbstractionist's point.

Quote:

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?

**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? - Jun 13th 2009, 03:33 AMchamprockQuote:

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?

**not**surjective. There does not exist any value in the domain which is giving this particular value in the image.