Here are two squares that are the same size. In the left square, we will label the blue square

*a*² and the red square

*b*². The two white rectangles both have area

*ab*. Together, they show the famous "middle term" representation of

(

*a*+

*b*)² =

*a*² + 2

*ab*+

*b*²

The square on the right has four white right triangles that are the equivalent of the two white rectangles sliced diagonally, where the diagonal is the hypotenuse, which we usually label

*c*. The yellow square with the gap in it has area

*c*². Since the squares are the same size we get

*c*² + 4(½

*ab*) =

*a*² + 2

*ab*+

*b*² Next step: get rid of the parentheses

*c*² + 2

*ab*=

*a*² + 2

*ab*+

*b*² Next step: subtract 2

*ab*from both sides

*c*² =

*a*² +

*b*² and we are done.

Second proof just using the yellow square and the gap inside.

The gap in the middle of the yellow square is (

*a*-

*b*)² =

*a*² - 2

*ab*+

*b*². The four yellow triangles are 2

*ab*, exactly the same as the four white triangles. That means this picture tells us

*c*² =

*a*² - 2

*ab*+

*b*² + 2

*ab*

So we combine like terms to get

*c*² =

*a*² +

*b*² and once again, Q.E.D., the Latin abbreviation for "that which has been demonstrated".

Tomorrow: number theory and Pythagorean Theorem.

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