Integral Multiple of Integral Multiple

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({F, +, \times}\right)$ be a field.

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


Then:

$\left({m n}\right) \cdot a = m \cdot \left({n \cdot a}\right)$

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


Proof

We have that $\left({F^*, \times}\right)$ is the multiplicative group of $\left({F, +, \times}\right)$.

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:

$\forall m, n \in \Z: \left({m n}\right) a = m \left({n a}\right)$

$\Box$

Now suppose $a = 0_F$.

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

$\left({m n}\right) \cdot a = 0_F = m \cdot \left({n \cdot a}\right)$

$\blacksquare$


Sources