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THERMO Spoken Here! ~ J. Pohl © ~ 2017 | www.ThermoSpokenHere.com (A0680-002) |

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.