Proof 8.3.3,4,5
Consider the functions, and .
We can show that is not an onto function because there are points in the plane that cannot be obtained as outputs of . Counterexample: The point (1,0) exists in the target but there does not exist a corresponding (x,y) in the domain. That is, there is no point (x,y) such that f(x,y)=(1,0).
We can show that is an onto function. Let . Then and . Both a and b are defined for all values of x and y. Thus, for any (a,b) in the target, there exists an (x,y) in the domain such that f(x,y)=(a,b).
We can easily show that is one-to-one. Consider the element such that . We see that the x-coordinate always maps to itself, thus there is a one-to-one relationship for the x-coordinates. By rearranging, we see that . Taking the reciprocal of any real number gives another unique real number and thus the mapping between the y-coordinates is also one-to-one. That is, is one-to-one.
We can easily that is also one-to-one. Consider the element such that , The y-coordinate always maps to itself, thus there is a one-to-one relationship for the y-coordinates. By rearranging, we see that . This equation gives a unique real. Thus the mapping between the x-coordinates is one-to-one. Therefore, is one-to-one.
We can see that neither function is distance preserving. Consider the points A=(a,b) and B=(c,d). Then . For , .
For , .