We have now seen that for any

1.

2.

3.

In fact, exactly one of those three statements is true. (This is called the law of trichotomy.)

Proof:

Let

Clearly

1 ∈

Suppose

So

Now let

Let

We have already shown that 1 ∈

Suppose

Case 1. If

Case 2. If

Case 3. If

In all three cases,

Thus

So we have

[edited to add: But there's probably a simple one-line proof that a smarter person would have written.]

*m*,*n*∈ ℕ, at most one of the following three statements is true:1.

*m*=*n*,2.

*m*<*n*,3.

*m*>*n*.In fact, exactly one of those three statements is true. (This is called the law of trichotomy.)

Proof:

Let

*A*= {*p*∈ ℕ | 1 ≤*p*}.Clearly

*A*⊆ ℕ.1 ∈

*A*, trivially.Suppose

*n*∈*A*. If*n*= 1, then*n** = 1 + 1 > 1, so*n** ∈*A*. On the other hand, if*n*> 1, then 1 +*m*=*n*for some*m*∈ ℕ, so*n** = (1 +*m*)* = 1 +*m**, so again*n** ∈*A*.So

*A*= ℕ by induction.Now let

*m*∈ ℕ.Let

*B*= {*p*∈ ℕ |*p*<*m*or*p*=*m*or*p*>*m*}.We have already shown that 1 ∈

*B*.Suppose

*n*∈*B*.Case 1. If

*n*<*m*, then either*m*=*n*+ 1 =*n**, in which case*n** ∈*B*, or*m*=*n*+*q*for some*q*∈ ℕ where*q*> 1, in which case*q*= 1 +*r*=*r** for some*r*∈ ℕ, so*m*=*n*+*r** = (*n*+*r*)* = (*r*+*n*)* =*r*+*n**, so*n** <*m*, so again*n** ∈*B*.Case 2. If

*n*=*m*, then*n** =*m*+ 1, so*n** >*m*, so again*n** ∈*B*.Case 3. If

*n*>*m*, then*n*=*m*+*q*for some*q*∈ ℕ, so*n** = (*m*+*q*)* =*m*+*q**, so yet again*n** >*m*and*n** ∈*B*.In all three cases,

*n** ∈*B*.Thus

*B*= ℕ by induction.So we have

*n*<*m*or*n*=*m*or*n*>*m*for all*n*∈ ℕ, given our choice of*m*. But*m*was arbitrary, so we're done.[edited to add: But there's probably a simple one-line proof that a smarter person would have written.]