Is this proof involving order theory correct?

Question

I need to prove that if $a<b$ and $c<0$ then $ac>bc$. This is how I tried to prove it: We know that $a<b$ so it follows that $-b<-a$ and therefore $c\boldsymbol{\cdot}-b>c\boldsymbol{\cdot}-a$ and finally This is equivalent to $c\boldsymbol{\cdot}b<c\boldsymbol{\cdot}a$. I think I'm missing something in that last step, but I'm not sure. I'm starting to prove things in math, so I would happily hear any tips on how to improve my proof-writing skills. Thanks a lot in advance!.

Answer 1

Given $a<b$, we know $a-b < 0$. Multiply $c$ to $a-b$, we can see that since $c<0$ and $a-b<0$, $(a-b)c$ is a product of two negative value, so $(a-b)c >0$, which imples $ac-bc >0$, hence $ac >bc$.

All operations in the proof should follow the ordered field axioms, which can be found here: http://homepages.math.uic.edu/~kauffman/axioms1.pdf