4
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Here is the proposed theory:

Definition: Let $M$ be a nonempty set with a binary operation $+$ satisfying the following properties:

P-0: The operation $+: M \times M \to M$ is both associative and commutative.

P-1: $\text{For every } x,y,z \in M \text{, if } z + x = z + y \, \text{ then } \, x = y$.

P-2: $\text{For every } x,y,z \in M \text{, if } z = x + y \, \text{ then } \, z \ne x$.

P-3: $\text{For every } x,y \in M \text{, if } x \ne y \, \text{ then } \, [\exists u \; | \, x = y +u] \text{ or } [\exists u \; | \, y = x +u]$.

P-4: $\text{For all } X \subset M \text{ such that } X \ne \emptyset $
$\quad \exists \, x_0 \in X \text{ such that }$
$\quad \forall x \in X \; \; [\,x = x_0 \text{ or } (\exists u \in M \text{ such that } x_0 + u = x)\,]$

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Proposition 1: There exist a unique element, call it $1 \in M$, such that any other element $x \in M$ can be written uniquely in the form $x = 1 +u$.
Proof
Apply $\text{P-4}$ to the universal set $M$. $\quad \blacksquare$

Theorem 2: Let $N$ be a subset of $M$ closed under addition. If $1 \in N$ then $N = M$.
Proof
To get a contradiction, suppose $M$ has elements not in $N$. Then applying $\text{P-4}$ and proposition 1 to $M\setminus N$ the following must hold

$\tag 1 1 + \alpha = \beta \text{ where } \beta \in M\setminus N$ $\quad \text{ and every other element of } M\setminus N \text{ is 'larger than' } \beta$

If $\alpha \in N$, then by $\text{(1)}$ the element $\beta$ would also belong to $N$. But $\text{(1)}$ also implies that $\alpha$ is 'less than' $\beta$, a contradiction.$\quad \blacksquare$

Theorem 3: Let $D$ be any subset of $M$ satisfying the following two properties:

$\quad 1 \in D$
$\quad \text{If } d \in D \text{ then } d + 1 \in D$

Then $D = M$.

Proof
To arrive at a contradiction, apply $\text{P-4}$ to get the 'least' element $m$ of $M$ that is not $D$. By proposition 1 we can write $1 + u = m$. Now $u$ must be in $D$ (it is 'smaller' than $m$), but then so is $m$, a contradiction.$\quad \blacksquare$

We can also formulate the strong induction formulation of theorem 3, but we simply assume it in what follows.

Now let $K$ be the carrier set for an algebraic structure defined by a binary operation $\times$. No properties on $\times$ need be postulated to prove the following result.

Theorem 4: For any two morphisms

$\quad f: (M,1,+) \to (K,\times)$
$\quad g: (M,1,+) \to (K,\times)$

if $f(1) = g(1)$ then $f = g$.

Proof
Let $D$ be the set of elements in $M$ where the two functions agree. Using induction (theorem 3), we see that the morphisms are equal. $\quad \blacksquare$

Theorem 5: If $(M,1,+)$ and $(N,1,+)$ are two are additive systems satisfying $\text{P-0}$ thru $\text{P-4}$ then there exist one and only one isomorphism between them.
Proof (sketch)
We have the two 'increment by 1' functions, $\sigma_M$ and $\sigma_N$. The construction of a bijective mapping $\phi$ is defined by

$\quad \phi: 1 \mapsto 1$
$\quad (\phi \circ \sigma_M)(m) = (\sigma_N \circ \phi)(m)$

and this mapping is the only viable morphism. To show $\phi$ is indeed a morphism, use induction, so we can checkoff

$\quad \phi ((u+1) + v) = (\phi \circ \sigma_M) (u + v) = (\sigma_N \circ \phi)(u+v)=$
$\quad \quad \sigma_N(\phi(u) + \phi(v))= (\phi(u) + \phi(1)) + \phi(v) = \phi(u + 1) + \phi(v) $
etc.
So the same relation defining the bijection is used to show it is a morphism. $\quad \blacksquare$

Are these valid arguments and logical constructions?

If they are, we now get addition 'free of charge' and can use inductive arguments. Of course we still need to define multiplication, but that can be done by analyzing the morphisms $\mu: M \to M$.

By theorem 4 any self-morphism of $M$ is completely specified by knowing where the generator $1$ goes. So do all morphisms of the form

$\tag 2 \mu_n: 1 \mapsto n \text{ with } n \in M$

actually exist?

The identity mapping $\mu_1$ is a morphism and it commutes with itself.

Assume that for $n \in M$ we have a a morphism $\mu_n$ and that it commutes with all morphisms $\mu_k$ where $k + u = n$.

For $n+1$ define

$\tag 3 \mu_{n+1} = \mu_{n} + \mu_{1}$

It is immediate that $\mu_{n+1}$ is a morphism that commutes with itself and also commutes with $\mu_{n}$. Moreover, if $\mu_k$ commutes with $\mu_{n}$ then

$\tag 4 \mu_{n+1} \circ \mu_k = (\mu_{n} + \mu_{1}) \circ \mu_k = (\mu_{n} \circ \mu_k) + (\mu_{1} \circ \mu_k) =$ $\quad \quad \quad \quad \quad \quad \quad \quad (\mu_{k} \circ \mu_n) + \mu_k = \mu_{k} \circ (\mu_n + \mu_1) = \mu_k \circ \mu_{n+1}$

We have defined a commutative binary operation $*$, call it multiplication, on $M$.

Theorem 6: In a natural way, $M$ is the carrier set for two binary operations $(M,1,+,*)$ with multiplication distributing over addition.

The above simple arguments, once verified, provide an alternative to using the Dedekind–Peano axioms.

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  • $\begingroup$ @Joe Induction = Theorem 3 and P-4 makes it work. $\endgroup$ – CopyPasteIt Oct 30 '18 at 14:29
  • $\begingroup$ Sorry, I read more closely so I deleted my comment. $\endgroup$ – Joe Oct 30 '18 at 14:29
  • $\begingroup$ Theorem 4 is almost certainly not true without some unstated assumptions on $K$ and the morphism to it. What reason do we have to suppose that $f(n)=g(n)$ implies $f(n+1)=g(n+1)$? $\endgroup$ – Malice Vidrine Nov 1 '18 at 7:24
  • $\begingroup$ @MaliceVidrine IF $f(1) = g(1)$ Then $f(2) = f(1+1) = f(1) + f(1) = g(1) + g(1) = g(1 + 1) = g(2) \quad |||$ IF $[f(1) = g(1) \land f(2) = g(2)]$ Then $f(3) = f(2+1) = f(2) + f(1) = g(2) + g(1) = g(2 + 1) = g(3) \quad |||$ ...keep going with (strong) induction... $\endgroup$ – CopyPasteIt Nov 1 '18 at 10:47
  • $\begingroup$ What reason have you given to suppose that $f$ or $g$ preserve the binary operation? $\endgroup$ – Malice Vidrine Nov 1 '18 at 16:18

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