Prove ##a \cdot 1 = a = 1 \cdot a## for ##a \in \mathbb{N}##

[issacnewton]

Homework Statement

Prove $a \cdot 1 = a = 1 \cdot a$ for $a \in \mathbb{N}$ using Peano postulates

Relevant Equations

Peano postulates

I have to prove $a \cdot 1 = a = 1 \cdot a$ for $a \in \mathbb{N}$.
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

$$ G = \{ z \in \mathbb{N} | z \cdot 1 = z = 1 \cdot z \} $$

We want to prove that $G = \mathbb{N}$. For this purpose, we will use part 3) of Peano postulates given above. Obviously, $G \subseteq \mathbb{N}$. From the multiplication definition given above, we have $1 \cdot 1 = 1 = 1 \cdot 1$. And since $1 \in \mathbb{N}$, it follows that $1 \in G$. Suppose $r \in G$. This means that $r \in \mathbb{N}$ and

$$ r \cdot 1 = r = 1 \cdot r $$

Now, from part a) of multiplication operation given above, it follows that

$$ s(r) \cdot 1 = s(r) \cdots\cdots (1)$$

Now from part b) of multiplication operation given above, it follows that

$$ 1 \cdot s(r) = (1 \cdot r) + 1 $$

And since $r \in G$, we have $1 \cdot r = r$. Hence

$$ 1 \cdot s(r) = r + 1 $$

And using addition operation definition, $r + 1 = s(r)$, so

$$ 1 \cdot s(r) = s(r) \cdots\cdots (2) $$

Collecting equation 1 and 2, we have

$$ s(r) \cdot 1 = s(r) = 1 \cdot s(r) $$

Since $s(r) \in \mathbb{N}$, it follows 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 $a \in \mathbb{N}$, $a \in G$ and it follows that

$$ a \cdot 1 = a = 1 \cdot a $$

Is the proof correct ?

Thanks

 

[PeroK]

Looks good to me.