1730.41 – Need More Money?

Tags:Problem Set 12GeometryAlgebra

Feeling tired and run down? Broke all the time? Need more money? Try a little algebra and enjoy as much money as you need.

Let mm be the money you have now and nn be the amount that will satisfy your needs. Now the average of the two is, as usual, A=m+n2A = \frac{m+n}{2}. We proceed as follows, employing powerful, timeless, well-tested tools of algebra. Watch carefully:

m+n=2A(m+n)(mn)=2A(mn)m2n2=2Am2Anm22Am=n22Anm22Am+A2=n22An+A2(mA)2=(nA)2mA=nAm=n.\begin{aligned} m+n &=& 2A \\ (m+n)(m-n) &=& 2A(m-n) \\ m^2 - n^2 &=& 2Am - 2An \\ m^2 - 2Am &=& n^2 - 2An \\ m^2 - 2Am + A^2 &=& n^2 - 2An + A^2 \\ (m-A)^2 &=& (n-A)^2 \\ m - A &=& n - A \\ m &=& n. \end{aligned}

Voila! It turns out that m=nm = n, and you have exactly as much money as you need.

Find the flaw in the reasoning.


The flaw is an old one: just because x2=(x)2x^2 = (-x)^2, it doesn't mean x=x.x = -x. Often xx and x-x are different. So you can't go from (mA)2=(nA)2(m-A)^2 = (n-A)^2 to mA=nAm - A = n - A .

(This doesn't mean that you can't trust algebra: you just have to be knowledgeable in how you use it.)

By the way, there is a similar argument for those who think they have more money than than they need.