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

[issacnewton]

Homework Statement

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

Relevant Equations

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

I am also going to assume following results for this proof.
Identity law for multiplication for $a \in \mathbb{N}$
$$ a \cdot 1 = a = 1 \cdot a $$

Distributive law

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

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

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

We want to prove that $G = \mathbb{N}$. For this purpose, we will use part 3) of Peano postulates given above. Obviously, $G \subseteq \mathbb{N}$. Let $x, y \in \mathbb{N}$ be arbitrary. Using Identity law for multiplication, we have $(x \cdot y) \cdot 1 = (x \cdot y)$ and $x \cdot (y \cdot 1) = x \cdot y = (x \cdot y)$. Since $1 \in \mathbb{N}$ from Peano postulates, it follows that $1 \in G$. Suppose $r \in G$. So $r \in \mathbb{N}$ and

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

Now, let $x, y \in \mathbb{N}$ be arbitrary. Now, using addition definition part a), $(x \cdot y) \cdot s(r) = (x \cdot y) \cdot (r+1)$. Now using Distributive law, we have $(x \cdot y) \cdot s(r) = (x \cdot y) \cdot r + (x \cdot y) \cdot 1$. Since $r \in G$, we have , from equation (1), that $(x \cdot y) \cdot r = x \cdot (y \cdot r)$. So, we have $(x \cdot y) \cdot s(r) = x \cdot (y \cdot r) + (x \cdot y) \cdot 1$. And using Identity law for multiplication, $(x \cdot y) \cdot 1 = (x \cdot y)$. It means that $(x \cdot y) \cdot s(r) = x \cdot (y \cdot r) +(x \cdot y)$. Now, using distributive law, $(x \cdot y) \cdot s(r) = x \cdot (y \cdot r + y)$. And, using definition of multiplication part b), $y \cdot r + y = y \cdot s(r)$. So, we have $(x \cdot y) \cdot s(r) = x \cdot (y \cdot s(r))$. And since $x, y \in \mathbb{N}$ are arbitrary, it proves that

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

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

Now, if $a,b,c \in \mathbb{N}$, we have $c \in G$ and it follows that $(a\cdot b)\cdot c =a\cdot (b \cdot c)$.

Is this a valid proof ?

Thanks

 

[PeroK]

Looks good.