Difference of Two Squares/Algebraic Proof 1
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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
\(\ds \paren {x + y} \circ \paren {x + \paren {-y} }\) | \(=\) | \(\ds x \circ x + y \circ x + x \circ \paren {-y} + y \circ \paren {-y}\) | Distributivity of $\circ$ over $+$ in a ring | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ x + x \circ y + x \circ \paren {-y} + y \circ \paren {-y}\) | $R$ is a commutative ring | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ x + x \circ \paren {y + \paren {-y} } + \paren {-\paren {y \circ y} }\) | various ring properties | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ x + x \circ 0_R + \paren {-\paren {y \circ y} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x \circ x + \paren {-\paren {y \circ y} }\) |
$\blacksquare$