Field Product with Non-Zero Element yields Unique Solution

Theorem

Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $a, b, x \in F$ such that $b \ne 0_F$.

Let:

$b \times x = a$

Then:

$x = a b^{-1}$

That is:

$x = \dfrac a b$

where $\dfrac a b$ denotes division.

Proof

\(\ds b \times x\)

\(=\)

\(\ds a\)

with $b \ne 0_F$

\(\ds \leadsto \ \ \)

\(\ds b^{-1} \times \paren {b \times x}\)

\(=\)

\(\ds b^{-1} \times a\)

multiplying both sides by $b^{-1}$, which exists because $b \ne 0$

\(\ds \leadsto \ \ \)

\(\ds \paren {b^{-1} \times b} \times x\)

\(=\)

\(\ds b^{-1} \times a\)

Field Axiom $\text M1$: Associativity of Product

\(\ds \leadsto \ \ \)

\(\ds 1_F \times x\)

\(=\)

\(\ds b^{-1} \times a\)

Field Axiom $\text M4$: Inverses for Product

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds b^{-1} \times a\)

Field Axiom $\text M3$: Identity for Product

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Fields: $\S 14$. Definition of a Field
• 1973: C.R.J. Clapham: Introduction to Mathematical Analysis ... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $3 \ \text {(iii)}$