Can multiplication be defined without addition?

Question

I'm struggling to understand how to define multiplication and addition, now that I've been told that multiplication is not just repeated addition.

It seems that the axioms for the two are identical, save that multiplication is said to not have an inverse for the additive identity.

Doesn't this imply that multiplication cannot be defined without an addition?

Answer 1

Can you define multiplication with out addition? For natural numbers (i.e. $0,1,2,\dots$), sure. I'm not sure why you would want to do this, but it could be done as follows:

Let $0=\phi=\{\}$ (the empty set), $1 = \{0\}$, $2 = 1 \cup \{1\} = \{0,1\}$, $3 = 2 \cup \{2\} = \{0,1,2\}$, $\dots$

Define $m \times n = \{ (i,j) \;|\; i \in m \mbox{ and } j \in n\}$ (the Cartesian product of $m$ and $n$ -- all ordered pairs). Then set $m \times n$ can be put into 1-1 correspondence with some natural number (set). Call this number $m \cdot n$. Voila! Multiplication.

Ok. Now for a reality check. When defining the natural numbers themselves I've used succession (i.e. plus one): $n+1 = n \cup \{n\}$ (so there's addition hiding in the very definition of a natural number).

Next, if I were to actually prove that $m \times n$ is in 1-1 correspondence with some natural number, I would almost certainly end up developing addition to do so.

So can you avoid addition? Yes and no. But necessarily in an unnatural way.

Exponentiation is repeated multiplication: $m^n = m^{n-1}m$. Multiplication is repeated addition: $mn = m(n-1)+m$. Addition is repeated succession: $m+n = m+(n-1)+1$. It's just a natural hierarchy of operations.

All arithmetic for bigger number systems flows from this.

Answer 2

You can define multiplication without using addition by using similar triangles in Euclidean geometry.

Once a unit length is specified, you can get $a*b$ by solving $\dfrac{a}{c} =\dfrac{1}{b}$. To do this, have two lines from a point P. The first has a point $A$ with $|PA| = a$. (Note: $|UV|$ is the distance between points $U$ and $V$.) The second line has points $B$ and $C$ with $|PB|=1$ and $|PC| = b$. Draw a line through $C$ parallel to $AB$. This intersects the first line at a point $D$ such $|PD| = c$.