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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?

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    $\begingroup$ It would help if you say what are the elements for which you are defining addition and multiplication, and even better if you quote the definitions. $\endgroup$ – Ittay Weiss Sep 10 '13 at 1:14
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    $\begingroup$ Do you have abstract definitions of addition and multiplication? As far as I know, no such definitions exist. $\endgroup$ – Michael Albanese Sep 10 '13 at 1:14
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    $\begingroup$ As far as operations on natural numbers go, you can make axioms for multiplication that don't depend on addition. See Skolem arithmetic. $\endgroup$ – Keshav Srinivasan Sep 10 '13 at 1:25
  • $\begingroup$ What do you mean by multiplication? Do you mean multiplication in a field? $\endgroup$ – anon Sep 10 '13 at 2:21
  • $\begingroup$ Multiplication in a field is where my question originated, but the concept of addition seems to always necessarily be derived from addition, despite the fact that I've read many arguments that multiplication is unique from addition. $\endgroup$ – user82004 Sep 10 '13 at 2:23
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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: $n+m = n+(m-1)+1$. It's just a natural hierarchy of operations.

All arithmetic for bigger number systems flows from this.

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  • $\begingroup$ The question wasn't clear to me, not knowing the "previous question" , answer withdrawn $\endgroup$ – imranfat Sep 10 '13 at 2:06
  • $\begingroup$ Who's to say that this multiplication isn't to be called addition though? What is the difference between multiplication and addition? $\endgroup$ – user82004 Sep 10 '13 at 2:11
  • $\begingroup$ In this example, can you explain how the cartesian product $m\times n$ corresponds to the arithmetic product of the numbers that sets $m$ and $n$ represent? I.E. how do you show that the sets $Succ(1)\times Succ(2)$ and $Succ(5)$ contain one another, and are thus equal? $\endgroup$ – Nicholas Cousar Dec 24 '20 at 9:43
  • $\begingroup$ [I'll denote successor of n as n' here.] So 1' × 2' = 2 × 3 = 6 = 5'. And the Cartesian product of sets 1' × 2' has the same number of elements as 5'. But it is not true that the sets 1' × 2' and 5' are equal (as sets). One can show there is an algorithm to turn such a Cartesian product into successors of 0. This amounts to some of peano's axioms: billcookmath.com/courses/math2510-spring2010/Proofs_in_PA.pdf $\endgroup$ – Bill Cook Dec 24 '20 at 13:50
  • $\begingroup$ I worked out your request (more or less) as an example here: billcookmath.com/courses/math2110-fall2020/… $\endgroup$ – Bill Cook Dec 24 '20 at 13:51
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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 $|PD| = c$.

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