# Trichotomy proof

1. Jan 25, 2012

### Dansuer

1. The problem statement, all variables and given/known data
I was trying to prove trichotomy for natural numbers.

2. Relevant equations
$n > m \Leftrightarrow \exists k \neq 0 (n = m + k)$
$n < m \Leftrightarrow \exists k \neq 0 (m = n + k)$

Trichotomy means in a set
$\forall n \forall m ( n = m \vee n> m \vee n<m)$

3. The attempt at a solution
I need an hint, a push in the right direction.

thank you :)

2. Jan 25, 2012

### HallsofIvy

Using what basis? There are many different ways of defining the natural numbers and different ways of defining "<". Typically, "Trichotomy" is taken as part of the definitionof "<" for the natural numbers. How are you defining "<"?

A standard method is "<" is a binary relation on the natural numbers satisfying
1) If m< n and p is any natural number then m+p< n+p.
2) If m< n and 0< p, then mp< np.
3) If m and n are natural numbers then one and only one must be true:
a) m= n.
b) m< n.
c) n< m.

An equivalent definition is
"There exist a subset of the natural numbers, P, satifying
1) If m and n are both in P then mn is in P.
2) If m and n are both in P then m+ n is in P.
3) if m is a natural number then one and only one must be true:
a) m= 0.
b) m is in P.
c) -m is in P.

It is (3), of course, that is equivalent to "trichotomy". If you have that definition of "order", you define "<" by "a< b if and only if b- a is in P.

Last edited by a moderator: Jan 25, 2012
3. Jan 25, 2012

### Dansuer

I define < and > like i've written in the Relevant Equations section.
I define the natural numbers in a non rigorous way, just as an intuitive concept.
Addition is a binary relation with those property.

$a+b \in N$
$a+b = b+a$
$(a+b)+c = a+(b+c)$
There is one only element, and it's 0, for which $a+0 = a$

I'm trying to prove some facts starting from those definitions. Trichotomy would be nice to prove, since i can use it to prove other things.

Last edited: Jan 25, 2012
4. Jan 25, 2012

### Dansuer

I just realized that my definition of trichotomy is wrong, or incorrect. Usually $\vee$ is the inclusive or, but in this case should be the exclusive or since all 3 propositions can't be true, and not even two of them can.