# $e^{x} > x$

1. Dec 25, 2011

### Dschumanji

When you see the graphs of the functions $f(x)=e^{x}$ and $f(x)=x$ it is obvious that the former is always greater than the latter (when x is a real number) for the portion of the graph you are observing. In elementary algebra, this was the "proof" that the following inequality is true for all real numbers x: $e^{x} > x$.

By using calculus, you can prove that that the inequality $e^{x} > x$ is true for all real number x. One method is to employ the properties of exponents to show that $e^{x} > x$ for real numbers less than or equal to 0 and then show that $e^{x} \leq x$ results in a contradiction when you let x increase without bound. Another method, which is much simpler than the first, is to show that the power series of $e^{x}$ necessarily implies $e^{x} > x$ for all real numbers x.

Is it possible to prove $e^{x} > x$ for all real numbers x without resorting to calculus? Can you use elementary algebra to prove it? Is the previous question not possible because the function $e^{x}$ is transcendental?

2. Dec 25, 2011

### mathman

As you noted the exponential function is not algebraic, so algebra can't be used by itself. The simplest was is using the power series for x > 0 and note that for x < 0, the exponential is still positive.