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I have recently begun reading Terry Tao's Real Analysis 1. The Peano Axiom proofs in the book are very new to me. Because of this, I have little intuition as to whether my proofs are correct. In addition to this, solutions to the exercises are not available in the text and I don't know anyone who is well-versed in this. This is the motivation behind my question. Any feedback is welcome! Thank you very much for taking the time, I truly appreciate it.

Here are the axioms as stated in his book:

Axiom 2.1: $0$ is a natural number.

Axiom 2.2: If $n$ is a natural number, then $n++$ is also a natural number.

Axiom 2.3: $0$ is not the successor of any natural number; i.e., we have $n++\not= 0$ for every natural number $n.$

Axiom 2.4: Different natural numbers must have different successors; i.e., if $n,m$ are natural numbers and $n\not=m$, then $n++\not=m++$. Equivalently, if $n++ =m++$, then we must have $n=m.$

Axiom 2.5: Let $P(n)$ be any property pertaining to a natural number $n$. Suppose that $P(0)$ is true,and suppose that whenever $P(n)$ is true, $P(n++)$ is also true. Then $P(n)$ is true for every natural number $n.$

We can also assume that recursive definitions are well defined.

Definition of addition: Let $m$ be a natural number. To add zero to $m$, we define $0 +m:=m$. Now suppose inductively that we have defined how to add $n$ to $m$. Then we can add $n++$ to $m$ by defining $(n++ ) +m:= (n+m)++.$

Additionally, the following have been proven:

Lemma 2.2.2: $n+0=n$ for all natural numbers $n$.

Lemma 2.2.3: For any natural numbers $n$ and $m, n+(m++ ) = (n+m)++.$

Proposition 2.2.4: $n+m=m+n$ for all natural numbers $n,m.$

Given this, is the following proof right?

Theorem 1: For any natural numbers $a,b,c$ we have $(a+b)+c=a+(b+c).$

My proof: We induct on $c.$ First, we verify the base case: $$(a+b)+0$$ The above equals $a+b$ by lemma 2.2.2. Additionally, this can be written as $a+(b+0)$ by lemma 2.2.2. This concludes the base case. Next, assume that there exists a natural number $c$ such that $$(a+b)+c=a+(b+c)$$ for all natural numbers $a,b.$ Then, for $c++$ we have $$(a+b)+(c++)=((a+b)+c)++$$ by lemma 2.2.3. By the inductive hypothesis, this becomes $$(a+(b+c))++.$$ Using lemma 2.2.3, this becomes $$a+(b+c)++.$$ Applying lemma 2.2.3 once more yields $$a+(b+(c++))$$ as desired.

$\square$

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Good job, your proof is correct! Some comments:

  1. Your proof does not use, nor need, Proposition 2.2.4.
  2. Instead of "assume that there exists a natural number $c$ such that...", you should write "take any natural number $c$, and assume that for all $a,b$, $(a + b) + c = a + (b + c)$". It is under these assumptions that you are going to show that the statement "for all $a,b$, $(a + b) + c\!+\!\!+ = a + (b + c\!+\!\!+)$" holds for the same $c$.

edit: One more comment about style. Instead of the phrase "this becomes", you can always write chains of equalities. For example, you can write "by Lemma 2.2.3, $(a + (b + c))\!+\!\!+ = a + (b + c)\!+\!\!+$". This way the reader doesn't have to guess which side/subexpression of the original expression you mean, which may improve readability in more complicated arguments.

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    $\begingroup$ Thank you! I'll keep your second point in mind in the future. I stated 2.2.4 because it already proven. You're right though, I should have omitted it. $\endgroup$ Feb 20 '20 at 4:40

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