Verifying proof for the following question I wanted to check my proof if it is correct or not. Using these field axioms:
(i)Trichotomy Property:  Exactly one of $x<y$, $y<x$, or $x=y$ hold.
(ii)Transitivity: if $x<y$ and $y<z$ (which we could write in shorthand as $x<y<z$), then $x<z$.
(iii)If $x<y$ then $x+z < y+z$.
(iv) If $x<y$ and $z>0$ then $xz<yz$.

  
*
  
*(A1) Addition is commutative
  
*(A2) Addition is associative
  
*(A3) Addition has a neutral element $0$
  
*(A4) Any element has an additive inverse
  
*(A5) Multiplication is commutative
  
*(A6) Multiplication is associative
  
*(A7) Multiplication has a neutral element $1$
  
*(A8) Any non-zero element has a multiplicative inverse
  
*(A9) Multiplication distributes over addition
  



*

*Prove that for all $a,\, b,\, c\in\mathbb{R}$, if $a<b$ and $c<0$ then $bc < ac$.


Since $a<b$, by (A4), we have that $$
0 = a + (-a) < b + (-a) = b - a.$$ 
Since $b - a > 0$ and $c < 0$, then $c(b-a) < 0$. By (A9), this is equivalent to $cb-ca<0$. By (A3), $$
cb=cb-ca+ca<0+ca=ca.$$
Thus $cb<ca$. 
 A: "Since b−a>0 and c<0, then c(b−a)<0"
Why do you state this. Either this is what you wish to prove, or (and I suspect this) you are using  iv) but not saying so.
So you should say so.
Since $(b-a)> 0$ and $c < 0$ then by iv) $c(b-a) < 0\cdot (b-a)$ 
But notice. You don't have an axiom that $0\cdot x = 0$(!!!!)
You have to prove that.
Hint 1: $0d = (0+0)d$ 
Hint 2: $-0d$ must exist.
....
Once you prove that you have 
$c(b-a) <0$
So by A9 $c(b-a) = c(b+(-a)) = cb +c(-a)$
So $cb + c(-a) < 0$ and 
....
Now we need to prove $c(-a) = -ca$.
Hint: $c(-a) + ca = c(-a + a)$.
...
Now we have $cb - ca <0$ and ... you can finish your proof.
A: Your proof has a gap.  How do you know that $b-a \gt 0 \land c \lt 0 \Rightarrow c(b-a) \lt 0$?  I'm not saying you're wrong -- I'm just saying you haven't yet proved it.
Edit to add:
Lemma:  If $c \lt 0 \land x \lt y, \text{ then } xc \gt yc$.
Proof: $c+(-c) = 0 \land c \lt 0 \Rightarrow -c \gt 0$.  (Use property (iii), adding $-c$ to both sides of the inequality.)
Thus, $-xc \lt -yc$.  Add $xc + yc$ to both sides:  $yc  \lt xc$ and the Lemma is proved.
