# Integral Multiple of Integral Multiple

## Theorem

Let $\struct {F, +, \times}$ be a field.

Let $a \in F$ and $m, n \in \Z$.

Then:

$\paren {m n} \cdot a = m \cdot \paren {n \cdot a}$

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

## Proof

We have that $\struct {F^*, \times}$ is the multiplicative group of $\struct {F, +, \times}$.

Let $a \in F^*$, that is, $a \in F: a \ne 0_F$, where $0_F$ is the zero of $F$.

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

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

$\Box$

Now suppose $a = 0_F$.

Then by definition of the zero element of $F$, we have that:

$\paren {m n} \cdot a = 0_F = m \cdot \paren {n \cdot a}$

$\blacksquare$