Q:

Ordering Integers

I am trying to make a definitive statement regarding positive integers $I_1<I_2<I_3$.  I can show that, by subtracting $I_1$ that
$$0<I_2-I_1<I_3-I_1$$
But I want to create a chain that includes the positive integer $I_3-I_2$ in comparison to the other difference integers.  If I were to subtract $I_2$ I would get 
$$I_1-I_2<0<I_3-I_2$$
I feel this issue cannot be resolved.  I could look at for example $2<3<8$ and I would have $1<5<6$ which would show that $I_2-I_1<I_3-I_2<I_3-I_1$, however, if I look at $2<7<8$ then it is the case that $1<5<6$ which shows that  $I_3-I_2<I_2-I_1<I_3-I_1$  
What is the correct way to handle this?  Do I just have to always look at both cases?  If looking at intervals involving these differences, do I just need to say that if
$$|I_3-I_2|<|I_2-I_1|\implies I_3-I_2<I_2-I_1<I_3-I_1$$
$$|I_2-I_1|<|I_3-I_2|\implies I_2-I_1<I_3-I_2<I_3-I_1$$
This seems like a very simple question and I think it is.  This question comes from the fact that I have finite summations involving the above bounds and I need to make precise statements about the order of the terms involved in the summations.  

A:

Given any positive integers $a,b,l$, you can define $(l_1,l_2,l_3)=(l,l+a,l+a+b)$. The only condition is that $0<a<a+b$ and $0<b<a+b$. You can't assert any relationship between $a$ and $b$.

