Product with Field Negative
Jump to navigation
Jump to search
Theorem
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $a, b \in F$.
Then:
- $-\paren {a \times b} = a \times \paren {-b} = \paren {-a} \times b$
Corollary
- $\paren {-1_F} \times a = \paren {-a}$
Proof
\(\ds a \times b + a \times \paren {-b}\) | \(=\) | \(\ds a \times \paren {b + \paren {-b} }\) | Field Axiom $\text A1$: Associativity of Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds a \times 0_F\) | Field Axiom $\text A4$: Inverses for Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_F\) | Field Product with Zero | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -\paren {a \times b}\) | \(=\) | \(\ds a \times \paren {-b}\) | Definition of Ring Negative |
Similarly:
\(\ds \paren {-a} \times b + a \times b\) | \(=\) | \(\ds \paren {\paren {-a} + a} \times b\) | Field Axiom $\text A1$: Associativity of Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_F \times a\) | Field Axiom $\text A4$: Inverses for Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_F\) | Field Product with Zero | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -\paren {a \times b}\) | \(=\) | \(\ds \paren {-a} \times b\) | Definition of Ring Negative |
$\blacksquare$
Sources
- 1973: C.R.J. Clapham: Introduction to Mathematical Analysis ... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $2 \ \text {(iv)}$