Inverse of Field Product
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Theorem
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $a, b \in F$ such that $a \ne 0$ and $b \ne 0$.
Then:
- $\paren {a \times b}^{-1} = b^{-1} \times a^{-1}$
Proof
We are given that $a \ne 0$ and $b \ne 0$.
From Field has no Proper Zero Divisors and Rule of Transposition, we have:
- $a \times b \ne 0$
By Field Axiom $\text M4$: Inverses for Product we have that $\paren {a \times b}^{-1}$ exists.
Then we have:
\(\ds \paren {b^{-1} \times a^{-1} } \times \paren {a \times b}\) | \(=\) | \(\ds b^{-1} \times \paren {a^{-1} \times a} \times b\) | Field Axiom $\text M1$: Associativity of Product | |||||||||||
\(\ds \) | \(=\) | \(\ds b^{-1} \times 1_F \times b\) | Field Axiom $\text M4$: Inverses for Product | |||||||||||
\(\ds \) | \(=\) | \(\ds b^{-1} \times b\) | Field Axiom $\text M3$: Identity for Product | |||||||||||
\(\ds \) | \(=\) | \(\ds 1_F\) | Field Axiom $\text M4$: Inverses for Product |
Hence the result by definition of multiplicative inverse.
$\blacksquare$
Sources
- 1973: C.R.J. Clapham: Introduction to Mathematical Analysis ... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $3 \ \text {(vii)}$