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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:

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    $\begingroup$ see in Peano axioms : Arithmetic for the standard recursive definition of $+$ and $\times$. You can find also them in any textbook of math log. $\endgroup$ – Mauro ALLEGRANZA Feb 7 '15 at 20:03
  • $\begingroup$ It looks like you did it already. Maybe replace $(m+1)$ with $sm$. $\endgroup$ – GEdgar Feb 7 '15 at 20:48
  • $\begingroup$ I have to give an answer using successor. Could you please elaborate what you were trying to say, Please! Thank you $\endgroup$ – user2884707 Feb 7 '15 at 20:52
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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 :

$$f_m(0)=0,$$

$$f_m(s(n))=f_m(n)+m.$$

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 $a \in A$

$f(s(n)) = g(n,f(n))$, forn any $n$

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 :

$$m \cdot 0 = 0$$

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


You can see also this post.

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