Positive Real Numbers whose Reciprocals Sum to 1

Theorem
Let $p, q \in \R_{\ge 0}$ be positive real numbers such that:
 * $\dfrac 1 p + \dfrac 1 q = 1$

Then $p > 1$ and $q > 1$.

Proof
From Division by Zero it is immediate that $p > 0$ and $q > 0$.

either $0 < p \le 1$ or $\0 < q \le 1$.

, suppose $0 < p \le 1$.

Note we have that:
 * $\dfrac 1 q = 1 - \dfrac 1 p$

First suppose $p = 1$.

Then:
 * $\dfrac 1 q = 0$

But there exists no $q \in \R$ such that $\dfrac 1 q = 0$.

Hence it cannot be the case that $p = 1$.

Now suppose $0 < p < 1$.

Then:

But this contradicts our assertion that $q > 0$.

Hence by Proof by Contradiction it follows that $p > 1$.

Similarly we reach a contradiction by assuming $0 < q \le 1$.

Hence $p > 1$ and $q > 1$ as required.