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