## SectionB.8Multiplication as a Binary Operation

We define a binary operation $$\times\text{,}$$ called multiplication, on the set of natural numbers. When $$m$$ and $$n$$ are natural numbers, $$m\times n$$ is also called the product of $$m$$ and $$n\text{,}$$ and it sometimes denoted $$m*n$$ and even more compactly as $$mn\text{.}$$ We will use this last convention in the material to follow. Let $$n\in \nonnegints\text{.}$$ We define

1. $$n0=0\text{,}$$ and

2. $$n(k+1)=nk +n\text{.}$$

Note that $$10=0$$ and $$01=00+0=0\text{.}$$ Also, note that $$11=10+1=0+1=1\text{.}$$ More generally, from (ii) and Lemma B.19, we conclude that if $$m,n\neq0\text{,}$$ then $$mn\neq0\text{.}$$

Let $$m,n\in \nonnegints\text{.}$$ Then

\begin{equation*} m(n+0)=mn = mn +0 = mn+ m0. \end{equation*}

Now assume $$m(n+k) = mn + mk\text{.}$$ Then

\begin{align*} m[n+(k+1)] \amp = m[(n+k)+1]=m(n+k)+m\\ \amp =(mn+mk)+m=mn+(mk+m)= mn+m(k+1). \end{align*}

Let $$m,n\in \nonnegints\text{.}$$ Then

\begin{equation*} (m+n)0 =0 = 0+0 = m0 + n0. \end{equation*}

Now assume $$(m+n)k = mk + nk\text{.}$$ Then

\begin{align*} (m+n)(k+1)\amp =(m+n)k+(m+n)= (mk+nk) +(m+n)\\ \amp =(mk+m)+(nk+n)=m(k+1)+n(k+1). \end{align*}

Let $$m,n\in \nonnegints\text{.}$$ Then

\begin{equation*} m(n0)= m0 = 0 = (mn)0. \end{equation*}

Now assume that $$m(nk)=(mn)k\text{.}$$ Then

\begin{equation*} m[n(k+1)]= m(nk + n)= m(nk) + mn =(mn)k + mn = (mn)(k+1). \end{equation*}

The commutative law requires some preliminary work.

The lemma holds trivially when $$n=0\text{.}$$ Assume $$k0= 0k=0\text{.}$$ Then

\begin{equation*} (k+1)0 =0 = 0+0= 0k+0=0(k+1). \end{equation*}

$$01=00+0=0 =10\text{.}$$ Assume $$k1=1k=k\text{.}$$ Then

\begin{equation*} (k+1)1=k1+11=1k+1=1(k+1). \end{equation*}

Let $$m\in \nonnegints\text{.}$$ Then $$m0=0m\text{.}$$ Assume $$mk=km\text{.}$$ Then

\begin{equation*} m (k+1) = mk +m = km+m= km +1m=(k+1)m. \end{equation*}