Proof that the exponential function is injective using its definition as a power series

Question

If we define $\exp (x) : \mathbb{R} \to ] 0, \infty [$ by its power series $\exp (x) = \sum_{i = 0}^{\infty} \frac{x^i}{i!}$ then I want to prove that $\exp (x)$ is injective. I do not want to use the $\log$ function (as I want to define $\log$ as the inverse of $\exp$). To do this I want to prove that $| \exp (x) - \exp (y) | > | x - y |$ if $x > y$, I managed to prove this if $x > y > 0$, but fails to do this for the cases $x \geqslant 0 > y$ and $0 > x > y$ . Can someone help me with the two last cases or another proof of injectivity.

Answer 1

You can prove from the definition as power series that$$(\forall x,y\in\mathbb R):\exp(x+y)=\exp(x)\exp(y)$$and that $x>0\implies\exp(x)>1$. But then$$y>x\implies\exp(y)=\exp\bigl((y-x)+x\bigr)=\exp(y-x)\exp(x)>\exp(x).$$

Answer 2

First note that $\exp(a)\exp(b)=\exp(a+b)$ from the power series definition; the $a^kb^l$ coefficients are $\frac{1}{k!l!}=\frac{\binom{k+l}{k}}{(k+l)!}$. Hence $\exp(x)=(\exp(1))^x$ for all $x\in\Bbb R$. (Prove the case $x\in\Bbb Z$ by induction in two directions, and get to $x\in\Bbb Q$ viz. $((\exp(1))^{p/q})^q=(\exp(1))^p$; irrational $x$ follow by continuity.) If $\exp(x)=\exp(y)$, $\exp(y)=0$ or $\exp(x-y)=1$; but the former lacks real roots, and the latter gives $x-y=0$ as required.

Answer 3

We have $\exp(x)$ mapping $\mathbb R$ into $(0,\infty),$ and $\exp'(x) = \exp(x)$ (the latter following easily from the power series). Thus if $x<y,$ the MVT shows $\exp(y)-\exp(x)= \exp(c)(y-x)$ for some $c$ between $x$ and $y.$ Since $\exp(c)>0,$ we see $\exp(y)-\exp(x)>0.$ This proves injectivity.