I am stuck on problem 9 from Berberian- A First Course in Real Analysis which deals with inequalites in an ordered field.
Problem: Show that in an ordered field, $a< ar + (1-r)b< b$ whenever $a<b$ and $0<r<1$.
Proof Attempt: Suppose $a<b$ and $0<r<1$. By multiplying $0<r<1$ by $a$ we have $0<ra<a$. By adding $-1$ to $0<r<1$ and then multiplying by $b$, we have $-b<(r-1)b<0$ or equivalently $b>(1-r)b>0$.
I am unsure how to manipulate the inequalites to proceed furthur. Any hints would be appreciated.