On trichotomy of order for natural numbers I am currently going through Chapter 2 of Tao's Analysis book, and I don't quite understand the proof of the Trichotomy of order for natural numbers (Proposition 2.2.13).
Let $a,b$ be natural numbers. Then exactly one of the following statements is true: $a<b,a=b,b<a$.
The proof goes as follows (reproduced from the book itself). First we show that we cannot have more than one of the statements holding at the same time. If $a<b$ then $a\neq b$ by definition, and if $a>b$ then $a\neq b$ by definition. If $a<b$ and $b<a$, then we would have that $a=b$, which is a contradiction. Thus, no more than one of the statements is true.
My question is in regard to the proof strategy that is being used for this part. I do understand what is going on, it makes sense to me. I see that the first two cases are to say that if one of the statements happen, then the other two cannot happen and the third case says that if the three happen at the same time, that's impossible.
I want to ask what's the proof strategy that is being followed (I'm relatively new to proof writing, so I need to have it clear in my head), because this is the first proof of this kind I have seen.
I appreciate your help.
 A: In the third edition he cites Prop. 2.2.12 as containing the statement which is used to conclude $a = b$ from $a < b \land b < a$. But to be honest I don't see any statement that immediately implies this. Prop. 2.2.12 does contain the anti-symmetry law, but anti-symmetry refers to $\leq$ and can therefore not be used on $a < b \land b < a$.
What is needed here is the transitivity of $<$ ; since $a < b \land b < a$ then implies $a < a $ which already gives the desired contradiction.
The overall idea of that first part is to show
$$
 \neg (a < b \land a = b)  ~\land~ \neg (b < a \land a = b) ~\land~ \neg (a < b \land b < a) 
$$
porving that at most one of the 3 parts of the disjuction $a < b \lor a = b \lor b < a$ is true. The second part of the proof (as found in the book) then shows by induction that the disjunction is true and therefore at least one of the parts must be true.
A: The basic proof strategy is this:
First: to establish that exactly one of the three claims hold, you typically show two things:

*

*That at least one of the two claims hold


*That at most one of the two claims hold
Please note that in your post, you only describe a proof for the second half, but I am sure the first half is provided in the book as well
Second, to establish that at most one of several claims hold, all you have to show is that you cannot have any of the two claims holding at once: if you show that, then clearly you also cannot have three or more claims holding.  This is what the proof does: it rules out all combinations of two claims: the first and the second can't hold together, the second and the third can't hold together, and the first and the third can't hold together.
Your analysis is actually a little off on that last one: it's not that the proof shows that you can't have all three, but simply that you can't have $a < b$ and $b < a$ at the same time, because if you do then, you have $a=b$, and now you have both $a < b$ and $a = b$, which some earlier theorem must already have ruled out.
