# 2782.11 – Minding Your p's and q's

Given that $p+q = 1$, show that $p q < 1/2$. Can a stronger statement be made?

Solution

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 $1/2$. In other words, we want to solve the inequality,

\begin{aligned} p - p^2 &<& \frac{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:

$p = \frac{2 ± \sqrt{4 - 8}}{4} = \frac{2 ± \sqrt{-4}}{4}.$

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

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

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.