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
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:
From Area of Square:
From Area of Rectangle:
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 |
$\blacksquare$