Prove that the exponential function exp: R to R(positive) defined by :?

Question

for all X in R exp(X)=e^X , is a one to one correspondence ?

Answer 1

We know that exp(x) >0 for every real x and that (exp(x))' = exp(x). Therefore, (exp(x))' >0 for every x, which implies exp is strictly increasing over R. In addition, the differentiability of exp implies continuity.

Since exp(x) -> 0 when x -> -oo and exp(x) -> oo when x -> oo, iof folows exp takes on every value in (0, oo). Since it's stritly increasing, it follows it's one to one and is a surjection for R to (0, oo), that is, it's a bijection from R to (0, oo).

Answer 2

Let f: R -> R+; f(x):= exp(x)

For a function to have a one-to-one correspondence. It must be both one-to-one (injective) and onto (surjective).

onto simply means the codomain (R+ in this case) is equal to the range (R+ in this case!)

so f is onto.

one-to-one means that one value in the domain of f takes one value in the range.

I agree with mathemagician in that

'Since exp(x) = exp(y) implies x = y, then the outputs of x are equal only when the inputs are equal'.

This means that the function is not many-to-one. It is worth pointing out that the function is not one-to many. Leaving the only other possibility that f(x) is one-to-one.

We have show that f(x) is both onto and one-to-one, therefore it is a one-to-one correspondence.

Answer 3

Onto (surjective): Let y be positive. Then exp(ln y) = y. Since for any positive y we can find something that maps to it (namely, ln y), then exp is onto.

One-to-one (injective): Suppose exp(x) = exp(y). Taking the natural logarithm of both sides gives x = y. Since exp(x) = exp(y) implies x = y, then the outputs of x are equal only when the inputs are equal, so exp is one-to-one.

Since exp is onto (surjective) and one-to-one (injective), then it is a one-to-one correspondence.