Basic Thermodynamics ~ J. Pohl © | www.THERMOspokenhere.com (4-A112) |

Prove: (A - B)^{2} = A^{2} - 2AB + B^{2}
The Greeks understood the totality of anything

to equal the sum of its parts.

We begin with the slightly easier task; a proof regarding (A + B)^{2}. (This proof was made by Euclid around 300 B.C.)

By inspection of the figure to the right, it is apparent (add the areas).

(A + B)^{2} = A^{2} + 2 AB + B^{2}

But we need (A - B)². So Let's use the identical square areas but assign new lengths for the previous A and B such that (A - B) arises as a factor. The second sketch, below right, shows these choices.

Again the area of the outer square will equal the sum of its areas interior:

A^{2} = (A - B)^{2} + 2(A - B)B + B^{2}

A^{2} = (A - B)^{2} + 2AB - 2B^{2} + B^{2}

Now, rearrange and collect the above equation to obtain:

(A - B)^{2} = A^{2} - 2AB + B^{2} Q.E.D.

A technique used in geometry is also used in thermodynamics. Break things apart, solve, then put the pieces back together.

Prove: (A - B)^{ 2} = A^{ 2} - 2AB + B^{ 2}

An easy way to precede is to begin by with a slightly easier task regarding **( A + B )²**. The common result was first obtained by Euclid around 300 B.C.