## 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*}