=== Forward Implication === Proved in Exponential Function Inequality. {{qed|lemma}} === Reverse Implication === Consider $f_a \left({x}\right) = a^x - x - 1$. Then we need to prove: :$a = e \impliedby \forall x \in \R: f_a \left({x}\right) \ge 0$ By Linear Combination of Derivatives, Derivative of Power of Constant, Derivative of Identity Function, and Derivative of Constant, we have: :$f_a' \left({x}\right) = a^x \ln a - 1$ By Linear Combination of Derivatives, Derivative of Power of Constant, Derivative of Constant Multiple, and Derivative of Constant, we have: :$f_a'' \left({x}\right) = a^x \left({\ln a}\right)^2$ Now, we divide the theorem into two cases: ==== Case 1: $0 < a \le 1$ ==== Consider $x = 1$. Then: {{begin-eqn}} {{eqn | l = f_a \left({1}\right)       | r = a^1 - 1 - 1 }} {{eqn | r = a - 2 }} {{eqn | o = \le       | r = 1 - 2 }} {{eqn | r = -1 }} {{eqn | o = <       | r = 0 }} {{end-eqn}} Thus the right hand side is not true. {{qed|lemma}} ==== Case 2: $a > 1$ ==== We have: {{begin-eqn}} {{eqn | l = \ln a       | o = >       | r = 0       | c = Logarithm of 1 is 0 and Logarithm is Strictly Increasing }} {{eqn | n = 1       | ll = \implies       | l = \left({\ln a}\right)^2       | o = >       | r = 0       | c = Square of Non-Zero Real Number is Strictly Positive }} {{eqn | n = 2       | l = a^x       | o = >       | r = 0       | c = Power of Positive Real Number is Positive over Reals }} {{eqn | ll = \vdash       | l = a^x \left({\ln a}\right)^2       | o = >       | r = 0       | c = Real Number Axioms: $\R O2$: compatibility with multiplication, using $(1)$ and $(2)$ }} {{eqn | l = f_a'' \left({x}\right)       | o = >       | r = 0 }} {{end-eqn}} Hence, by Second Derivative of Strictly Convex Real Function is Strictly Positive, we have that $f_a$ is strictly convex. Trying to find its minimum, we solve: :$f_a' \left({x}\right) = 0$ Therefore: {{begin-eqn}} {{eqn | l = a^x \ln a - 1       | r = 0 }} {{eqn | l = a^x \ln a       | r = 1 }} {{eqn | l = a^x       | r = \frac 1 {\ln a} }} {{eqn | l = x       | r = \log_a \left({\frac 1 {\ln a} }\right) }} {{eqn | l = x       | r = - \log_a \left({\ln a}\right) }} {{end-eqn}} At this point: {{begin-eqn}} {{eqn | l = f_a \left({x}\right)       | r = f_a \left({- \log_a \left({\ln a}\right)}\right)       | c =  }} {{eqn | r = \frac 1 {\ln a} + \log_a \left({\ln a}\right) - 1       | c =  }} {{eqn | r = \frac 1 {\ln a} + \frac {\ln \left({\ln a}\right)} {\ln a} - 1       | c = Change of Base of Logarithm }} {{eqn | r = \frac {1 + \ln \left({\ln a}\right)} {\ln a} - 1       | c =  }} {{end-eqn}} It is to be noted that when $a = e$, the minimum is: : $\dfrac {1 + \ln \left({\ln e}\right)} {\ln e} - 1 = 0$ meaning that the only solution to $f_e \left({x}\right) = 0$ is $x = 0$. Also, from Exponential Function Inequality, we have: :$\forall x \in \R: e^x \ge x + 1$ Substituting $x = \ln \left({\ln a}\right)$: :$\ln a \ge \ln \left({\ln a}\right) + 1$ From Logarithm is Strictly Increasing and Logarithm of 1 is 0: :$\ln a > 0$ Thus: :$\dfrac {\ln a} {\ln a} \ge \dfrac {\ln \left({\ln a}\right) + 1} {\ln a}$ Therefore: :$\dfrac {1 + \ln \left({\ln a}\right)} {\ln a} - 1 \le 0$ Then, we solve the case of equality: {{begin-eqn}} {{eqn | l = \frac {1 + \ln \left({\ln a}\right)} {\ln a} - 1       | r = 0 }} {{eqn | l = \frac {1 + \ln \left({\ln a}\right)} {\ln a}       | r = 1 }} {{eqn | l = 1 + \ln \left({\ln a}\right)       | r = \ln a }} {{end-eqn}} From earlier, we have that the only solution is $\ln \left({\ln a}\right) = 0$, when $a = e$. Therefore, for other values of $a$, the minimum is negative. {{qed}} Category:Euler's Number 3yo1vwh450sokmb6cg2rhbk0uv4xjbe