Square of Difference/Algebraic Proof 1
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Theorem
- $\forall x, y \in \R: \paren {x - y}^2 = x^2 - 2 x y + y^2$
Proof
| \(\ds \paren {x - y}^2 =\) | \(=\) | \(\ds \paren {x - y} \cdot \paren {x - y}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x \cdot \paren {x - y} - y \cdot \paren {x - y}\) | Real Multiplication Distributes over Addition | |||||||||||
| \(\ds \) | \(=\) | \(\ds x \cdot x - x \cdot y - y \cdot x + y \cdot y\) | Real Multiplication Distributes over Addition | |||||||||||
| \(\ds \) | \(=\) | \(\ds x^2 - 2xy + y^2\) |
$\blacksquare$