Proving that if $a < b$ and $c < 0$, then $bcLet $a$, $b$, and $c$ be real numbers with $a < b$ and $c < 0$. 

Prove that $bc < ac$.

Proof: Let $a$, $b$, and $c$ be real numbers with $a < b$ and $c < 0$. Since $a < b$, then $a-b < 0$. 
Then I am stuck. My professor told me that I need to prove why a negative times a negative is a positive. 
 A: $ac<bc \Rightarrow ac-bc<0 \Rightarrow c(a-b)<0$.
Now we examine the factors. $c<0$ (given) and $a<b \Rightarrow a-b<0$.
$\underbrace{c}_{<0}(\underbrace{a-b}_{<0})<0$
Both factors are negative, and since negative times negative is positive, we have
$c(a-b)>0$ or $bc<ac$.
A: Note that $bc-ac=(b-a)c$. 
The RHS has two factors: $b-a$ wich is positive (since $a<b$) and $c$ wich is negative. 
This results in a negative product, so $bc-ac<0$ or equivalently $bc<ac$.

Underlying:
If $x>0$ and $y<0$ then $x(-y)>0$ as product of positive factors. 
This with $xy+x(-y)=x(y+(-y))=x.0=0$. 
Then $xy=-x(-y)<0$. 
This shows that the product of a positive and negative factor yields a negative product.
A: Since $-c\gt 0, b-a\gt 0$, we have
$$ac-bc=(-c)(b-a)\gt 0\Rightarrow ac\gt bc.$$
Here, the fact that "positive $\times$ positive is positive" is used.
A: Let $d>0$, then $a+d=b \Rightarrow ca+cd=cb \Rightarrow cb-ca=cd$, 
since $c<0,\ d>0 \Rightarrow cd<0 \Rightarrow cb-ca<0 \Rightarrow cb<ca$
A: $c<0$ and $a-b<0$ implies that $c(a-b)>0$, why $ac>bc$.
