No ,no i mean what axioms for the order.For example one such system is the following:
1)
2)
3)
4)
5)
Now in a such axiomatic system, if in axiom (5) you put:
x= a , y= 0 and z=b we have the following result:
,which is exactly the theorem you are trying to prove ,since 0.a = 0
However in another axiomatic system for order we can have a different proof
Here is a translation of the axioms written in a formal language.
x<y implies for all x,y
x<y and y< z implies x< z for all x,y,z ( this is the transitive law)
implies x< y or y< x ,for all x,y ( this is another form of the trichotomy law)
y< z implies y+x < z+x ,for all x,y,z
0< x and y< z implies x.y < x.z ,for all x,y ,z ,
.
And if in this axiom we put x = a , y =0, z = b we have:
0< a and 0< b 0< a.b .
Which the theorem you want to prove.
However in another axiomatic system your theorem could be considered as an axiom