# Proof related to rings $m \times (a + b) = m \times a + m \times b$

I have a trouble proving this sentence, as I don't know what assumption should start the proving implication. I know what are the characteristics of a ring, but I m, n are not in the ring.

    a, b are in ring (F, +, x).
m, n are whole numbers.
Proof that: m x (a + b) = m x a + m x b.


Let $G,+$ be an abelian group (with neutral element $\mathbf{0}$, just to make things clearer). The definition of $ma$, for $m$ integer and $a\in G$ is the same as for powers in multiplicative groups: \begin{cases} 0a=\mathbf{0} \\[6px] (m+1)a=ma+a & (m\ge0) \\[6px] ma=-(-m)a & (m<0) \end{cases}
This definition doesn't require the group to be abelian. The next property does: $$m(a+b)=ma+mb$$ Let's prove it by induction for $m\ge0$. The base case is obvious by definition. So assume we know $m(a+b)=ma+mb$; then \begin{align} (m+1)(a+b) &=m(a+b)+(a+b) &&\text{definition}\\ &=ma+mb+(a+b) &&\text{induction hypothesis}\\ &=ma+a+mb+b &&\text{commutativity}\\ &=(m+1)a+(m+1)b &&\text{definition} \end{align} For $m<0$, we have, setting $n=-m>0$ for simplicity \begin{align} m(a+b) &=-\bigl(n(a+b)\bigr) &&\text{definition}\\ &=-(na+nb) &&\text{above argument}\\ &=-(nb)-(na) &&\text{negative of a sum}\\ &=-(na)-(nb) &&\text{commutativity}\\ &=ma+mb &&\text{definition} \end{align}
For $m\in\mathbb{N}$ and $a\in F$, the product $ma$ is defined by induction as follows. First, we define $0\cdot a=0$. Next, assuming we have already defined $ma$, we define $(m+1)a=ma+a$. Intuitively, this means that $ma$ is just the sum $a+a+a+\dots+a$, where there are $m$ terms in the sum.
Since this definition is by induction, to prove that $m(a+b)=ma+mb$, you will have to use induction on $m$.