2782.11 – Minding Your p's and q's

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

Solution

Since q=1−pq = 1 - p, we can work exclusively with pp. We want to know if the quantity pq=p(1−p)=p−p2p q = p (1 - p) = p - p^2 is less than 1/21/2. In other words, we want to solve the inequality,

p−p2<122p−2p2<1f(p)=2p2−2p+1>0. \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)f(p) is always positive. Well, f(p)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)=2p2−2p+1=0.f(p) = 2 p^2 - 2 p + 1 = 0.

According to the quadratic formula the solution is:

p=2±4−84=2±−44. p = \frac{2 ± \sqrt{4 - 8}}{4} = \frac{2 ± \sqrt{-4}}{4}.

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

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