# Ring is Commutative iff Opposite Ring is Itself

## Theorem

Let $\struct {R, +, \times}$ be a ring.

Let $\struct {R, +, *}$ be the opposite ring of $\struct {R, +, \times}$.

Then $\struct {R, +, \times}$ is a commutative ring if and only if:

$\struct {R, +, \times} = \struct {R, +, *}$

## Proof

By definition of the opposite ring:

$\forall x, y \in R: x * y = y \times x$

### Necessary Condition

Let $\struct {R, +, \times}$ be a commutative ring.

Then:

 $\ds \forall x, y \in R: \,$ $\ds x \times y$ $=$ $\ds y \times x$ $\ds \leadsto \ \$ $\ds x \times y$ $=$ $\ds x * y$ Definition of opposite ring $\ds \leadsto \ \$ $\ds \struct {R, +, \times}$ $=$ $\ds \struct {R, +, *}$ Equality of Algebraic Structures

$\blacksquare$

### Sufficient Condition

Let $\struct {R, +, \times} = \struct {R, +, *}$.

Then:

 $\ds \forall x, y \in R: \,$ $\ds x \times y$ $=$ $\ds x * y$ Equality of Algebraic Structures $\ds \leadsto \ \$ $\ds x \times y$ $=$ $\ds y \times x$ Definition of opposite ring

Thus by definition $\struct {R, +, \times}$ is a commutative ring.

$\blacksquare$