Prove that $e^x \gt 0$ for $x \in \mathbb{R}$

Question

This is a consequence of the exponential rule, but how do I actually prove it to be true?

Answer 1

How do you define $e^x$?

The most common way to define it is

$$e^x = \sum_{n = 0}^{\infty} \frac{x^n}{n!}$$

If you take this definition, then if $x > 0$, $e^x$ is the sum of an infinite number of positive terms.

One should show that the series converges, but if you take that for granted, then of course if converges to a positive number.

If $x < 0$ then $e^{-x} > 0$ ; and since $$ e^x \cdot e^{-x} = 1 > 0 \Rightarrow e^x > 0$$ since $e^{-x} > 0$

Answer 2

Every definition of $\exp(x)$ leads to the property, $\exp(a+b)= \exp(a)\exp(b)$. Furthermore every definition also tells us that $\exp$ is defined on the entire real line and takes real values when given a real number as an argument.

So we will consider the exponential function evaluated at $x$ and notice that$x=x/2+x/2$.

$$ \exp(x) = \exp(x/2)\exp(x/2) = (\exp(x/2))^2 > 0 $$

Since the square of every real number is positive we have our result.

Answer 3

We have that $e>0$. Then for every $n\in \Bbb N$ we have that $e^n>0$.Because $\sqrt[m] .$ is increasing then $\sqrt [m] {e^n}>0$. Thus $$e^q>0$$ for every $q\in \Bbb Q$. Now $\Bbb Q$ is dense in $\Bbb R$ thus $e^x>0$ for every $x\in \Bbb R$.