Give a recursive definition of the operation of multiplication of natural numbers using the operations s and addition.

Question

Give a recursive definition

Basis:

m,n subset of N(natural numbers) A contains N (0,m) subset of A

If n = 0, 0*m = 0

(m + 1) * x = mx + x

Recursive Steps:

Answer 1

We assume that you are searching for the (elementary) set-theoretic definition of multiplication, having available the sum ($+$).

If so, we need the successor function, defined as :

$s(x) = x \cup \{ x \}$.

Thus, the recursive definition of multiplication needs a function $f_m : \omega \to \omega$, for each $m \in \omega$, such that :

$$f_m(0)=0,$$

<p>$$f_m(s(n))=f_m(n)+m.$$</p>

Then we can apply the Recursion theorem :

Given a set $A$ and a function $g : \omega \times A \to A$, there exists a unique function $f : \omega \to A$ that satisfies the following recursive definition:

$f(0) = a$, for $a \in A$

<p>$f(s(n)) = g(n,f(n))$, forn any $n$</p>

with $\omega$ as $A$, $0$ as $a$ and $+$ as $g$.

We have that $\{ \langle \langle m,n \rangle, f_m(n) \rangle \mid \langle m,n \rangle \in \omega^2 \}$ is single-valued in the second projection, for $\langle m,n \rangle = \langle m',n' \rangle$ implies $m = m'$, and hence also $f_m = f_{m'}$; but $\langle m,n \rangle = \langle m',n' \rangle$ implies also $n = n'$ and thus $f_m(n) = f_{m'}(n')$, showing that the $f_m$ are "well-defined" functions.

Thus, we can call the set $\{ \langle \langle m,n \rangle, f_m(n) \rangle \mid m,n \in \omega^2 \}$ with “$\cdot$” and we have - as expected - that $\cdot : \omega \times \omega \to \omega$ satisfies the conditions :

$$m \cdot 0 = 0$$

$$m \cdot s(n) = (m \cdot n) + m.$$

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