Understanding $(n \cdot 1)(m \cdot 1) = (nm) \cdot 1$

Question

I read that $(n \cdot 1)(m \cdot 1) = (nm) \cdot 1$ in a ring with unity $1$ because $\left(\underbrace{1+\cdots+1}_{n \textrm{ times}}\right)\left(\underbrace{1+\cdots+1}_{m \textrm{ times}}\right) = \underbrace{1+\cdots+1}_{nm \textrm{ times}}$ by the distributive law (of addition) of rings.

However, the left distributive law is $a \cdot (b+c) = (a \cdot b) + (a \cdot c)$ and the right distributive law is $(a+b) \cdot c = (a \cdot c) + (b \cdot c)$.

I don't understand how the equation above follows from either the left or the right distributive law. For instance, how is addition getting distributed (of course, $1$ above is not the same as the integer 1, but represents unity)? Thanks!

Answer 1

It’s easier to see if you temporarily abbreviate $n\cdot 1$ to $a$, say: then you have

$$a(\underbrace{1+\ldots+1}_m)=\underbrace{a+\ldots+a}_m$$

by repeated applications of the left distributive law. But

$$\begin{align*} \underbrace{a+\ldots+a}_m&=\underbrace{(\underbrace{1+\ldots+1}_n)+\ldots+(\underbrace{1+\ldots+1}_n)}_m\\ &=\underbrace{1+\ldots+1}_{nm}\;. \end{align*}$$

A really rigorous proof would proceed by a double induction on $m$ and $n$, but what’s going on is straightforward enough once you see it that the rigorous proof is a bit of overkill unless the point is to practise such arguments.

Answer 2

Say for example $n=2$ and $m=3$.

To understand how addition gets distributed, see this:

$(1+1)\cdot(1+1+1)=1\cdot(1+1+1)+1\cdot(1+1+1)$

$=1\cdot1+1\cdot1+1\cdot1+1\cdot1+1\cdot1+1\cdot1$.

Do you see how the first equality results from right distribution and the second from left?

Answer 3

Use induction (and a few parentheses). For example: \begin{align*} (1+1+1)\cdot(1+1+1)&=(1+1+1)\cdot(1+(1+1))\\ &=(1+1+1)\cdot1+(1+1+1)\cdot(1+1)\\ &=(1+1+1)+(1+1+1)\cdot1+(1+1+1)\cdot 1\\ &=(1+1+1)+(1+1+1)+(1+1+1). \end{align*} By grouping into pairs, you can use the distributive law to multiply across the pairs. When you do this, you reduce the number of $1$'s in a factor (and then induct).

Sketch: Proof by induction on $n$ and $m$. The statement is trivial when $n=1$ or $m=1$. Assume that the statement is true for $n$ and $m$ and consider $(n\cdot 1)((m+1)\cdot 1)$. Rewrite this as \begin{align*} (n\cdot 1)+(1+(m\cdot 1))&=(n\cdot 1)\cdot 1+(n\cdot 1)\cdot (m\cdot 1)\stackrel{IH}{=}(n\cdot 1)+(nm\cdot 1)\\&=((nm+n)\cdot 1)=(n(m+1)\cdot 1) \end{align*} The case for $((n+1)\cdot 1)(m\cdot 1)$ is similar.

Answer 4

You can quickly show with induction from the two-term distributive law (and associativity) that $a \cdot (b_1 + \cdots + b_n) = (a\cdot b_1) + \cdots + (a\cdot b_n)$ for any positive integer $n$, and similarly for the right-distributivity. Using J.W. Tanner's answer you can reach your conclusion.