# Faustian bargain and the geometric and arithmetic means

Reading a random article this week I came across a quote I’ve seen reprinted many times before. It is a play on the Faustian bargain where Faust in Goethes story makes a pact with the devil (Méphistophélès) for worthy pleasures and gains in exchange for his most precious possession but

Algebra is the offer made by the devil to the mathematician. The devil says: `I will give you this powerful machine, it will answer any question you like. All you need to do is give me your soul: give up geometry and you will have this marvellous machine.’

I believe it is mostly to Michael Atiyah but it’s rarely quoted explicitly and tends to just be used as a fun entry to some critique of the decline in the education of classical geometry and the trends of constructive and purely symbolic algebra without any visual or geometric connection. Usually accompanied by imlication that you have lost some understanding of the problem while gaining an answer… I suppose that was the angle I too had in mind here.

The case which came to mind in this case was the super simple algebraic proof of the so called inequality of geometric and arithmetic means. Recall that symbolically the arithmetic mean of two numbers $a$ and $b$ is $(a + b)/2$ and the geometric mean is $\sqrt{ab}$ and are essentially two different ways of finding ‘ a number which ‘lies between’ the two numbers and they’re algebraically connected to eachother in many cute ways but the one I would like to discuss here is the inequality

$\sqrt{ab} \leq \frac{a + b}{2}$.

That is the geometric mean is always less than the arithmetic mean. Appealing to the devil we can prove this fact using algebra in a single step and some rearranging. From some god we are told to look at the expression

$(\sqrt{a} - \sqrt{b})^2 \geq 0$

which we know is true because squares are positive and expanding using algebra

$0 \leq (\sqrt{a} - \sqrt{b})^2 = (\sqrt a)^2 + 2\sqrt{a}\sqrt{b} + (\sqrt{b})^2 = a^2 + b^2 - 2\sqrt{ab}$

and adding $2 \sqrt{ab}$ to both sides we get the original inequality. Easy as pie.

It’s of course convincing and easy to trace out all the steps but why I would look at this in the first place is magic the first time and many of the steps such as $\sqrt{a}\sqrt{b} = \sqrt{ab}$ linger in the minds of many students without any clear reason as to why it should be true.

To this we may contrast the algebraic proof of the inequality with a geometric one. It relies on mapping the problem onto one of lengths on a plane and relies on a particular way of constructing the geometric mean of two lengths and is not one which I will explain but which you can check out at http://www.cut-the-knot.org/pythagoras/GeometricMean.shtml

The charm I feel is that the inequality is reduced to visual fact. It is quite clear that the geometric mean less than the arithmetic simply by looking at it and to my  own sensibilities this sort of elaboration fosters a greater understanding than simply looking at the algebra of it. Though no one can deny the efficiency of the machine.

## 2 thoughts on “Faustian bargain and the geometric and arithmetic means”

1. But the geometric mean as BE took four lines of algebra to establish. maybe I don’t see it right !
There is an in-between way as follows:
Consider the two relationships (a + b)/2 = 1 and root(ab) = 1 (so ab = 1 as well)
Plot them, b against a, to get a straight line passing through (1,1) and a rectangular hyperbola (they don’t need to know the term) also passing through (1,1).
The curve is always above the straight line, so for any value of a the geometric mean of a and some number b to have the same value (1 in this case) as the arithmetic mean of a and some other number b’ it is required that b is larger than b’. (except when a = b = b’).
What I like about this is that the formal math is simple, but the logical argument is more complex. Variety is all.
Did you see my post on this ?
https://howardat58.wordpress.com/2015/01/28/the-mean-which-mean-with-interesting-ratios/

2. Well the geometric proof took 4 lines and a circle to complete so in terms of number of operations they’re effectively equal ; ) .
I see what you mean, and it’s a great visualization, but I feel the step from illustration to proof is a bit greater than straight algebra or straight construction but as you say; variation is the key to understanding.
With regards to the mean-value post. No. Not before now, to which I’d like to add that also in the extended cases the geometric constructions provide a neat way to see the ordering (hierarchy) of the different means: Harm < Geo < Arith < RMS (or equal) through: http://jwilson.coe.uga.edu/EMAT6680Fa08/Wisdom/EMAT6690/Investigate%20Means/meancircle.gif