Can the Inequality e^{x} > x for All Real Numbers x Be Proven Without Calculus?

[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?

 

[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.