Condition for Difference of Field Elements to be Zero
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Theorem
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a, b \in F$
Then:
- $a - b = 0_F$
- $a = b$
where $a - b$ denotes subtraction.
Proof
Necessary Condition
Let $a = b$.
Then:
\(\ds a - b\) | \(=\) | \(\ds a + \paren {-b}\) | Definition of Field Subtraction | |||||||||||
\(\ds \) | \(=\) | \(\ds b + \paren {-b}\) | as $a = b$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_F\) | Field Axiom $\text A4$: Inverses for Addition |
$\Box$
Sufficient Condition
Let $a - b = 0_F$.
Then:
\(\ds a + \paren {-b}\) | \(=\) | \(\ds 0_F\) | Definition of Field Subtraction | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a + \paren {-b} } + b\) | \(=\) | \(\ds 0_F + b\) | adding $b$ to both sides | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a + \paren {\paren {-b} + b}\) | \(=\) | \(\ds 0_F + b\) | Field Axiom $\text A1$: Associativity of Addition | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a + 0_F\) | \(=\) | \(\ds 0_F + b\) | Field Axiom $\text A4$: Inverses for Addition | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds b\) | Field Axiom $\text A3$: Identity for Addition |
$\blacksquare$
Sources
- 1973: C.R.J. Clapham: Introduction to Mathematical Analysis ... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $2 \ \text {(ii)}$