I am reading Tao's analysis. in the book the proof of Trichotomy of natural numbers use proposition 2.2.12 and Definition 2.2.11.
The proof is divided in to two parts
- Not more than one statement is true
- Atleast one statment is true.
For second part Tao uses induction. Is the following a valid proof which is much simpler.
From Definition 2.2.11 (Ordering of natural numbers) we know $a \ge b$ which is equvivalent to $a = b+m$ for some natural number $m$. Now either $m=0$ or $m \neq 0$. If $m=0$ then $a=b$. If $m \neq 0$ then $a \gt b$, again by definition of ordering of natural numbers. Similary we can show for $b \ge a$. Hence Proved at least one statement is true.
[[ This is mostly a Comment which is too long to Put in Comment Box ]]
Initially that Proof is looking good but there are Difficulties ....
There 3 Issues with that Proof :
(1) Your Proof uses the Lemma : "$a \ge b$" or "$b \ge a$" : Was the Lemma known or Proved ?
(2) How do we know that the Lemma is "Mutually Exclusive" ? It is not "Mutually Exclusive" here.
(3) How can we state that Both true will give $a=b$ ? Otherwise , we might get Both true giving "$a>b$ AND $b>a$" : we have to show that is not Possible.
[[ Elaborating Point 3 : "$X \lor Z$" AND "$Y \lor Z$" Both true can occur when $Z$ is true : It can even occur when $X$ AND $Y$ are true !
In other words : "A greater than OR EQUAL to B" AND "B greater than OR EQUAL to A" can occur when "A EQUAL B" : It can occur even when "A greater than B" AND "B greater than A" ! ]]