Exponential of Real Number is Strictly Positive/Proof 5
Jump to navigation
Jump to search
Theorem
Let $x$ be a real number.
Let $\exp$ denote the (real) exponential function.
Then:
- $\forall x \in \R : \exp x > 0$
Proof
This proof assumes the definition of $\exp$ as the solution to an initial value problem.
That is, suppose $\exp$ satisfies:
- $ (1): \quad D_x \exp x = \exp x$
- $ (2): \quad \map \exp 0 = 1$
on $\R$.
Lemma
- $\forall x \in \R: \exp x \ne 0$
$\Box$
Aiming for a contradiction, suppose that $\exists \alpha \in \R: \exp \alpha < 0$.
Then $0 \in \openint {\exp \alpha} 1$.
From Intermediate Value Theorem:
- $\exists \zeta \in \openint \alpha 0: \map f \zeta = 0$
This contradicts the lemma.
$\blacksquare$
Sources
- 2011: Robert G. Bartle and Donald R. Sherbert: Introduction to Real Analysis (4th ed.): $\S 8.3$: Theorem $6 \ \text {(iii)}$