Prove ##(a+b)\cdot c=a\cdot c+b\cdot c## using Peano postulates

[issacnewton]

Homework Statement

Prove $(a+b)\cdot c=a\cdot c+b\cdot c$ using Peano postulates where $a,b,c \in \mathbb{N}$

Relevant Equations

Peano postulates

I want to prove that $(a+b)\cdot c=a\cdot c+b\cdot c$ using Peano postulates where $a,b,c \in \mathbb{N}$.
The book I am using ("The real numbers and real analysis" by Ethan Bloch ) defines Peano postulates little differently.
Following is a set of Peano postulates I am using. (Axiom 1.2.1 in Bloch's book)

There exists a set $\mathbb{N}$ with an element $1 \in \mathbb{N}$ and a function $s: \mathbb{N} \rightarrow \mathbb{N}$ that satisfy the following three properties.

1) There is no $n \in \mathbb{N}$ such that $s(n) = 1$
2) The function $s$ is injective.
3) Let $G \subseteq \mathbb{N}$ be a set. Suppose that $1 \in G$, and that if $g \in G$ then $s(g) \in G$. Then $G = \mathbb{N}$

And addition operation is given in Theorem 1.2.5 as follows

There is a unique binary operation $+: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ that satisfies the following two properties for all $n, m \in \mathbb{N}$

a. $n + 1 = s(n)$
b. $n + s(m) = s(n + m)$

Multiplication operation is given in Theorem 1.2.6 as follows

There is a unique binary operation $\cdot: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ that satisfies the following two properties for all $n, m \in \mathbb{N}$

a. $n \cdot 1 = n$
b. $n \cdot s(m) = (n \cdot m) + n$

with this background, we proceed to the proof. Let us define a set

$$ G = \{ z \in \mathbb{N} | \mbox{if } x, y \in \mathbb{N} \; (x+y)\cdot z = x \cdot z + y \cdot z \} $$

We want to prove that $G = \mathbb{N}$. For this purpose, we will use part 3) of Peano postulates given above. Obviously, $G \subseteq \mathbb{N}$. Now, I am going to use another result I have proven earlier. If $a \in \mathbb{N}$, we have $a \cdot 1 = a = 1 \cdot a$. So, if $x, y \in \mathbb{N}$ are arbitrary, it will follow that $(x + y) \cdot 1 = (x + y) = x \cdot 1 + y \cdot 1$. Since $1 \in \mathbb{N}$ from Peano postulates, it follows that $1 \in G$. Now suppose $r \in G$. It means that $r \in \mathbb{N}$ and

$$ \forall x, y \in \mathbb{N} \bigl[ (x + y) \cdot r = x \cdot r + y \cdot r \bigr] \cdots\cdots (1)$$

Now, let $x, y \in \mathbb{N}$ be arbitrary. Now using multiplication operation part b), it will follow that

$$ (x + y) \cdot s(r) = ((x + y) \cdot r) + (x + y) $$

Using equation 1), it will follow that

$$ (x + y) \cdot s(r) = (x \cdot r + y \cdot r) + (x + y) $$

I have already proven associative law for addition. So, using that,

$$ (x + y) \cdot s(r) = ((x \cdot r + y \cdot r) + x) + y $$

$$ (x + y) \cdot s(r) = (x \cdot r + (y \cdot r + x)) + y $$

I have also proven commutative law for addition earlier. So, we use that now

$$ (x + y) \cdot s(r) = (x \cdot r + (x + y \cdot r)) + y $$

Using associative law for addition again,

$$ (x + y) \cdot s(r) = ((x \cdot r + x) + y \cdot r) + y $$

Now using part b) of the addition operation, $x \cdot s(r) = x \cdot r + x$. So, we have

$$ (x + y) \cdot s(r) = (x \cdot s(r) + y \cdot r) + y $$

Using associative law for addition again,

$$ (x + y) \cdot s(r) = x \cdot s(r) + (y \cdot r + y) $$

And, again using part b) of the addition operation, $y \cdot s(r) = y \cdot r + y$. So, we have

$$ (x + y) \cdot s(r) = x \cdot s(r) + y \cdot s(r) $$

Since $x,y$ are arbitrary, it follows that

$$ \forall x, y \in \mathbb{N} \bigl[ (x + y) \cdot s(r) = x \cdot s(r) + y \cdot s(r) \bigr] $$

And $s(r) \in \mathbb{N}$. So, $s(r) \in G$. So, $r \in G$ implies that $s(r) \in G$.
Using part 3) of the Peano postulates, it follows that $G = \mathbb{N}$.

Now $a,b,c, \in \mathbb{N}$. So, $c \in G$. And it will follow that

$$ (a+b)\cdot c=a\cdot c+b\cdot c $$

Is this a valid proof ?

Thanks

 

[fresh_42]

It is a valid proof.

Suggestion (for future proofs like this): Maybe you could list all equations and axioms you have used at the end of such a proof, then put this list at the beginning with a note of what has been already proven, and what is an axiom, and then refer to them in the actual proof by their number in the list.

The way you write the proof is a bit of a miracle for us readers: whenever you need a property it falls from the sky with a note (postulate, earlier proven, definition). That is ok if you have the book at hand, but mysterious for us who do not. Listing them at the beginning, as you did with the definition of multiplication and addition btw., makes the structure of such a proof more transparent.

However, that is only an idea and my personal opinion. Your proof was fine.

 

[PeroK]

... as your proofs get more challenging, you may have to be more concise. And, as suggested above, you ought to put the Peano axioms,, definitions of $+$ and $\cdot$ and previous results in the "relevant equations. The proof itself should then be slicker and easier to follow. For example:

We will use Peano axiom 3) for the set
$$ G = \{ z \in \mathbb{N} | \mbox{if } x, y \in \mathbb{N} \; (x+y)\cdot z = x \cdot z + y \cdot z \} $$Let $x, y \in \mathbb N$. From the multiplication property a), applied to $(x +y), x$ and $y$ we have, :$$(x + y) \cdot 1 = x + y = (x \cdot 1) + (y \cdot 1)$$Hence $1 \in G$.

Assume that $r \in G$. Using multiplication property a) and associativity and commutivity of addition and the inductive hypothesis on $r$ we have:
$$(x + y) \cdot s(r) = (x + y) \cdot r + (x + y) = x \cdot r + y \cdot r + x + y = (x \cdot r + x) + (y \cdot r + y)$$And, using multiplicative property b) for $x, r$ and $y, r$ we have:
$$(x \cdot r + x) + (y \cdot r + y) = x \cdot s(r) + y \cdot s(r)$$Hence $s(r) \in G$ and we have shown that $G =\mathbb N$ QED

Now, you might want to add a bit more formality. But, perhaps you should start letting the inescapable logic speak for itself!

 

[issacnewton]

Thanks for the input fresh_42. Even I noticed that. I was wondering how could I improve the readability of my proof. I will try to follow your advice from now on.