Help with writing a simple proof on inequalities

[thelannonmonk]

1. Let m, n, p, q \in Z
If 0 < m < n and 0< p \leq q, then mp < nq



2. Propositions/axioms I can use that relate to inequalities
2.4 Let m,n,p \in Z. If m < n and n < p, then m < p
2.5 For each n \in N there exists an m \in N such that m > n
2.6 Let m,n \in Z. If m \leq n \leq m then m=n
2.7 i)If m < n, then m+p < n+p
2.7 ii)If m < n and p < q then m+p < n+q

The Attempt at a Solution

So far, my first idea is to say if m < n, then sm < sn (s being an arbitrary integer). However, I don't have an axiom or proposition to reference for this step, so I don't even know if I can use it. If I could use it, then i would go on to say that if sm < sn, then if p<q, then pm<qn, but this is the thing they want me to prove! So I can't reference the proposition I am trying to prove, I am just stuck!

The other issue is that I can only use axioms an propositions from earlier in the book, so my options are limited, but the most obviously useful ones seem to be 2.7 (i) and (ii)

 

[dynamicsolo]

You may have to establish the proposition for 0 &lt; m &lt; n and s &gt; 0 , m &lt; n \Rightarrow sm &lt; sn as a lemma, if you were not given it. But you're in luck, because the "universe of discourse" is the integers and we are sticking to the positive ones. Use what you know about addition of positive integers and Prop. 2.7 for this.

Then you want to build a chain of inequalities starting from mp and changing one factor at a time until you have nq . (I have said enough...)

 

[lanedance]

how about
m<n
m+m<n+n (2.7)
2m<2n

so you should be able to build up to
s>0 then
sm<sn