TITLE: Prove that $\exp(x)>0$ using only formal definition of exp
QUESTION [8 upvotes]: This problem would be easy if I could use the fact that $\exp(x)=e^x$, but I have to use the following definition:
$$\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
I can also use the fact that
$$\exp(x+y)=\exp(x)\exp(y)$$
So how do I prove, using those two equations, that
$$\forall x\in \mathbb{R}:\exp(x)>0 $$
I mean, I can't just use the definition, because if $x<0$ then it isn't so obvious that $\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}>0$. Can someone give me a hint or two?
Thanks!

REPLY [17 votes]: $\exp(x) = \exp(x/2+x/2)=\exp(x/2)^2 \ge 0$ for all $x$.
$
\exp(x)=\exp(x_0)\exp(x-x_0)
$ implies that if exp is zero for some $x_0$ then it is zero for all $x$.
$\exp(0)=1$ shows that exp is not the zero function.

REPLY [10 votes]: It's quite simple:  We can show that $\exp(x) > 0$ for all $x > 0$ using the series definition.  Then note that $\exp(0) = 1$, again using the series definition.  Then for $x < 0$, consider the second definition $1 = \exp(0) = \exp(x+(-x)) = \exp(x) \exp(-x)$, from which it easily follows that $\exp(x) = 1/\exp(-x) > 0$.