# Integral Multiple Distributes over Ring Addition

## Theorem

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

Let $a, b \in R$ and $m, n \in \Z$.

Then:

$(1): \quad \paren {m + n} \cdot a = \paren {m \cdot a} + \paren {n \cdot a}$
$(2): \quad m \cdot \paren {a + b} = \paren {m \cdot a} + \paren {m \cdot b}$

where $m \cdot a$ is as defined in integral multiple.

## Proof

We have that the additive group $\struct {R, +}$ is an abelian group.

$(1): \quad \paren {m + n} \cdot a = \paren {m \cdot a} + \paren {n \cdot a}$:

This is an instance of Powers of Group Elements: Sum of Indices when expressed in additive notation:

$\forall n, m \in \Z: \forall a \in R: m a + n a = \paren {m + n} a$

$\Box$

$(2): \quad m \cdot \paren {a + b} = \paren {m \cdot a} + \paren {m \cdot b}$:

This is an instance of Power of Product in Abelian Group when expressed in additive notation:

$\forall n \in \Z: \forall a, b \in R: n \paren {a + b} = n a + n b$

$\blacksquare$