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 $q > 1$.

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

Aiming for a contradiction, suppose either $0 < p \le 1$ or $0 < q \le 1$.

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:

$\ds \dfrac 1 p$

$>$

$\ds 1$

$\ds 1 - \dfrac 1 p$

$<$

$\ds 0$

$\ds \leadsto \ \ $

$\ds \dfrac 1 q$

$<$

$\ds 0$

$\ds \leadsto \ \ $

$\ds q$

$<$

$\ds 0$

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.

$\blacksquare$