2782.11 – Minding Your p's and q's

Since $q = 1 - p$, we can work exclusively with $p$. We want to know if the quantity $p q = p (1 - p) = p - p^2$ is less than $\dfrac{1}{2}$. In other words, we want to solve the following inequality:

$$\begin{aligned}p - p^2 &< \dfrac{1}{2} \\2 p - 2 p^2 &< 1 \\f(p) = 2 p^2 - 2 p + 1 &> 0\end{aligned}$$

Our task is now to prove that $f(p)$ is always positive. Well, $f(p)$ is a quadratic function. Its graph is a parabola. In this case the parabola opens upwards. Such a parabola is always positive if it is never zero. Thus we ask for the solution of the equation:

$$f(p) = 2 p^2 - 2 p + 1 = 0$$

According to the quadratic formula the solution is:

$$p = \dfrac{2 \pm \sqrt{4 - 8}}{4} = \dfrac{2 \pm \sqrt{-4}}{4}$$

But since there is no (real) square root of $-4$, it follows that $f(p)$ is never zero. This proves $p q < \dfrac{1}{2}$, as desired.

By trying some values, you might guess that in fact, $p q < \dfrac{1}{4}$. Applying the quadratic formula as we just did, you can show that this is true, except if $p = \dfrac{1}{2}$, when $p q = 0$. Therefore, $p q \geq \dfrac{1}{4}$.

If you have ever been told to "mind your p's and q's" you may wonder where the expression comes from. The OED has a great article about it – you can google it.