A function o:R-->R that is not onto but is 1-1
e^(x) is a function o:R-->R so that f is not onto but is 1-1.
We know from calculus that e^x is a function. But, it is not so easy to rigorously prove - t ry it without assuming what you are trying to prove ! Let's assume that we have proved that e^x is a function.
Recall that the derivative of e^x is e^x itself. We know that (e^x) is always positive for all x by definition of a positive number(2.71828...) to a power x. Since the derivative is always positive, we know that the function o(x) is increasing. Thus if a<b then e^(a) < e^(b).
Proof that o(x) is 1-1 assuming that we know that o(x) is increasing.
So, assume for contradiction that o(x) is not 1-1.
Then, we can find x1 and x2 that are not equal so that o(x1)=o(x2). But, if x1 and x2 are not equal,
then one must be smaller than the other - without loss of generality, assume that x1<x2. Since o(x)
is increasing, we know that o(x1)<o(x2). We have arrived at a contradiction to the fact that
o(x) is not 1-1. Hence o(x) is 1-1.
Proof that o(x) is not onto .
To show that o(x) is not onto, we'll produce a y in R so that o(x) is not equal to y for all x in R.
Take y=-1. Assume for contradiction that o(x)=-1 for some x in R.
Then e^x=-1 for some x in R.
But then x=Pi*I, which is complex, and so we have arrived at a contradiction to the fact that x is in R.
Hence o(x) is not equal to -1 for all x in R, and so o(x) is not onto.