Field Product with Zero
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Theorem
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$.
Then:
- $a \times 0_F = 0_F$
That is, $0_F$ acts as a zero element of $F$, which justifies its name.
Proof
\(\ds a\) | \(=\) | \(\ds a \times 1_F\) | Field Axiom $\text M3$: Identity for Product: $1_F$ is the unity of $F$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a \times \paren {0_F + 1_F}\) | Field Axiom $\text A3$: Identity for Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds a \times 0_F + a \times 1_F\) | Field Axiom $\text D$: Distributivity of Product over Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds a \times 0_F + a\) | Field Axiom $\text M3$: Identity for Product | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a \times 0_F + a} + \paren {-a}\) | \(=\) | \(\ds a + \paren {-a}\) | adding $-a$ to both sides | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \times 0_F + \paren {a + \paren {-a} }\) | \(=\) | \(\ds a + \paren {-a}\) | Field Axiom $\text A1$: Associativity of Addition | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \times 0_F + 0_F\) | \(=\) | \(\ds 0_F\) | Field Axiom $\text A4$: Inverses for Addition | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \times 0_F\) | \(=\) | \(\ds 0_F\) | Field Axiom $\text A3$: Identity for Addition |
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 87 \alpha$
- 1973: C.R.J. Clapham: Introduction to Mathematical Analysis ... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $2 \ \text {(iii)}$