Subtraction has no Identity Element

Theorem

The operation of subtraction on numbers of any kind has no identity.

Proof

Aiming for a contradiction, suppose there exists an identity $e$ in one of the standard number systems $\GF$.

\(\ds \forall x \in \GF: \, \)

\(\ds x\)

\(=\)

\(\ds x - e\)

\(\ds \)

\(=\)

\(\ds e - x\)

\(\ds \leadsto \ \ \)

\(\ds x + \paren {-e}\)

\(=\)

\(\ds e + \paren {-x}\)

\(\ds \leadsto \ \ \)

\(\ds x + x\)

\(=\)

\(\ds e + e\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds e\)

That is:

$\forall x \in \GF: x = e$

But from Identity is Unique, if $e$ is an identity then there can be only one such.

From Proof by Contradiction it follows that $\GF$ has no such $e$.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.3$. Units and zeros: Example $72$