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

# Prove: (A - B)2 = A2 - 2AB + B2

```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 = A2 + 2 AB + B2

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:

A2 = (A - B)2 + 2(A - B)B + B2

A2 = (A - B)2 + 2AB - 2B2 + B2

Now, rearrange and collect the above equation to obtain:

(A - B)2 = A2 - 2AB + B2   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

```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!