Product of Field Negatives

Theorem

Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.

Let $a, b \in F$.

Then:

$-\paren {a \times b} = \paren {-a} \times \paren {-b} = a \times b$

Proof

\(\ds \paren {-a} \times \paren {-b}\)

\(=\)

\(\ds -\paren {a \times \paren {-b} }\)

Product with Field Negative

\(\ds \)

\(=\)

\(\ds -\paren {-\paren {a \times b} }\)

Product with Field Negative

\(\ds \)

\(=\)

\(\ds a \times b\)

Negative of Field Negative

$\blacksquare$

Sources

• 1973: C.R.J. Clapham: Introduction to Mathematical Analysis ... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $2 \ \text {(v)}$