Prove ##a + b = b + a## using Peano postulates

[issacnewton]

Homework Statement

Prove $a + b = b + a$ using Peano postulates

Relevant Equations

Peano postulates

Following is a set of Peano postulates I am using as defined in the book "Th real numbers and real analysis" by Ethan Bloch.

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 $G = \mathbb{N}$I have to prove the Commutative Law for addition $a + b = b + a$ using Peano postulates, given that $a, b\in \mathbb{N}$. Now define the set

$$ G = \{ x \in \mathbb{N} |\forall\; y \in \mathbb{N} \quad (x + y) = (y + x) \} $$

I have proven previously, that for $a \in \mathbb{N}$, we have,

$$ 1 + a = s(a) = a + 1 \cdots\cdots (1) $$

So, for some $y \in \mathbb{N}$, we get $1 + y = y + 1$. That proves that $1 \in G$. Now, suppose that $r \in G$. This means that

$$ \forall\; y \in \mathbb{N} \quad (r + y) = (y + r) \cdots\cdots (2) $$

We need to prove that

$$ \forall\; y \in \mathbb{N} \quad s(r) + y = y + s(r) $$

Let $y \in \mathbb{N}$ be arbitrary. From $(2)$, we get $(r + y) = (y + r)$. It follows that $s(r + y) = s(y + r)$. Now, addition function is defined as follows in this book

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

$$ n + 1 = s(n) $$
$$ n + s(m) = s(n + m) $$

Using this, we get, $s(r + y) = y + s(r)$. Now, using $(1)$, we have, $1 + (r + y) = y + s(r)$. I have also previously proven Associative Law for Addition. So, using that, we get, $(1 + r) + y = y + s(r)$. Again using $(1)$, this becomes $s(r) + y = y + s(r)$. Since $y \in \mathbb{N}$ is arbitrary, $s(r) \in G$. So, for some $r \in G$ it implies that $s(r) \in G$. From Peano postulates, this means that $G = \mathbb{N}$.

Now let $a, b \in \mathbb{N}$ be arbitrary. So, $a \in G$. It follows that $(a + b) = (b + a)$

Is this a sound proof ?
Thanks

 

[nuuskur]

So, for some $y \in \mathbb{N}$, we get $1 + y = y + 1$. That proves that $1 \in G$.

Be more precise. The equality $1+y=y+1$ must hold for all $y\in\mathbb N$, then it follows that $1\in G$. You can show the identity holds by induction using $s(1+n) = 1+s(n)$.

Using this, we get, $s(r + y) = y + s(r)$. Now, using $(1)$, we have, $1 + (r + y) = y + s(r)$. I have also previously proven Associative Law for Addition.

Show explicitly how $1+(r+y) = y+s(r)$ follows from associativity. Otherwise, you make a remark about associativity and it's up to the reader to realise that's how you concluded that equality. Leave that ambiguous style to math authors, you have the luxury of writing things out in detail.

Since $y \in \mathbb{N}$ is arbitrary, $s(r) \in G$. So, for some $r \in G$ it implies that $s(r) \in G$.

You have already fixed $r\in G$, there is no need to requantify it. It's confusing the reader.One might condense the induction argument as follows, label the equalities and explain them separately. Keep things concise. It holds that $1\in G$. Let $r\in G$ and $y\in\mathbb N$. We have that
<br /> s(r)+y \overset{1.}= (1+r)+y \overset{2.}= 1+(r+y) \overset{3.}= (r+y)+1 \overset{4.}= (y+r)+1 \overset{5.}= y + s(r).<br />

1. Definition of $s(n)$
2. Associativity of $+$
3. $1\in G$
4. Induction assumption
5. Associativity of $+$ and definition of $s(n)$

 

[PeroK]

Is this a sound proof ?
Thanks

I agree with @nuuskur. This isn't as good as your previous proof of the associativity of $+$. You should try to be more positive and direct.

 

[issacnewton]

Thanks nuuskur. I will modify the proof.