# My favourite proof

My younger daughter Lottie is 16 today, and I gave her a square card with \(2^4\) on one side and \(4^2\) on the other.

\[2^4 = 4^2\]

This is the only solution to \(a^b = b^a\) for whole numbers \(a\) and \(b\). The proof of this statement is especially elegant. I can't remember where I read it first, but it was when I was doing my A-levels.

Starting with \[a^b = b^a\] taking (natural) logs \[b \log a = a \log b\] and rearranging, we get \[\frac{\log a}{a} = \frac{\log b}{b}\]

Now consider the graph \[y = \frac{\log x}{x}\]

plotted here in red (thanks to Desmos):

The notable thing about the graph is that it intercepts the \(x\) axis at \(x=1\), rises to a maximum between \(x=2\) and \(x=3\), then steadily decreases as \(x\) increases, but never reaches the \(x\) axis.

The blue line shows the solution above graphically, since \[\frac{\log 2}{2} = \frac{\log 4}{4} \approx 0.35\]

Now, this graph shows that \(a=2\), \(b=4\) is the *only* solution.

To see why, imagine that there is another solution, \(a>4\). Then, draw a horizontal line through the point on the graph where \(x=a\), shown here in green:

The green line intercepts the red curve strictly between \(x=1\) and \(x=2\), which shows that \(x\) *is not a whole number*. Therefore, there are no whole number solutions for \(a>4\).

Finally, we can see that \(a=3\) does not have any solutions, for the same reason (the other intercept is strictly between \(x=2\) and \(x=3\) and again is not a whole number).

The thing I like about this proof is that it proves something about *whole numbers*, by operating in the domain of *real numbers*, and by using simple arguments about the properties of the graph of a function.

[One last pleasing thing. It's easy to show (by differentiation and setting to zero), that the curve is at its maximum at \(x=e\).]