Find a function such that where is the image of f at E and is the pre-image of f at E.

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- Oct 15th 2008, 12:06 PMArythFunctions
Find a function such that where is the image of f at E and is the pre-image of f at E.

- Oct 15th 2008, 03:34 PMPlato
Suppose that .

Define a function

- Oct 17th 2008, 08:22 AMpoutsos.B
The theorem covering the above is:

For all functions f : ,if f is**not**onto then f[ ( R(f))]=/= R(f) ..........where R(f) is the range of f.

But due to contrapositive law the above theorem is equivalent to:

For all functions f : if f[ ( R(f))]= R(f) then f is onto.

So**any function which is not onto**will satisfy f[ ( R(f))]=/= R(f).

Now to prove the above .

But to prove that f is onto if f[ ( R(f))]= R(f),we must prove that:

For all yεR(f) then there exists an xεD(f) such that y=f(x),where D(f)is the domain of f .

The above in symbols is:

[ yεR(f) -------> ( xεD(f) & y=f(x))].

Let yεR(f),since f[ ( R(f))]= R(f),then yεf[ ( R(f))].

But yεf[ ( R(f))] <=====> y= f(x) & xε[ ( R(f))]..

But xε[ ( R(f))] <====> xεD(f) & f(x)εR(f).

Hence; yεf[ ( R(f))]<=======> y=f(x) & xεD(f) & f(x)εR(f).

Thus we see that there an xεD(f) such that y= f(x), so f is**onto**