THERMO Spoken Here! ~ J. Pohl © (A0680~1/15) | (A0920 ~ Theorem of Pythagoras) |

The Greeks understood the totality of anything to always 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.

The Greeks understood the totality of anything to equal the sum of its parts.

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.

Premise presently unwritten!