Difference of Two Squares/Geometric Proof 2

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Theorem

Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.

Let $x, y \in R$.


Then:

$x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$


When $R$ is one of the standard sets of numbers, that is $\Z, \Q, \R$, and so on, then this translates into:

$x^2 - y^2 = \paren {x + y} \paren {x - y}$


Proof

Difference-of-Two-Squares.png

Let $\Box ABCD$ be a square of side length $x$.

Let $\Box DEFG$ be a square of side length $y$ where $y < x$

Let $EF$ be produced to $H$.


The area of $\Box ABCD$ is seen to be equal to the sum of:

the area of the rectangle $AEHB$
the area of the rectangle $FGCH$
the area of the square $DEFG$


From Area of Square:

the area of $\Box ABCD$ is equal to $x^2$
the area of $\Box DEFG$ is equal to $y^2$


From Area of Rectangle:

the area of $AEHB$ is equal to $x \paren {x - y}$
the area of $FGCH$ is equal to $y \paren {x - y}$


Hence:

\(\ds x^2 - y^2\) \(=\) \(\ds x \paren {x - y} + y \paren {x - y}\)
\(\ds \) \(=\) \(\ds \paren {x + y} \paren {x - y}\) Real Multiplication Distributes over Addition

$\blacksquare$