$\sqrt{ab} = \sqrt{a} \sqrt{b}$ I'm trying to show that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ where $a, b > 0$
For contradiction assume $$\sqrt{ab} \not= \sqrt{a} \sqrt{b}$$
$$
\sqrt{ab} - \sqrt{a} \sqrt{b} \not=0
$$
$$
\left(\sqrt{ab} + \sqrt{a} \sqrt{b}\right) \left(\sqrt{ab} - \sqrt{a} \sqrt{b}\right) \not=0
$$
$$
 ab - ab \not=0
$$
$$
 ab \not= ab 
$$
If this logic is incomplete/ wrong, could you post a link of the correct proof?
 A: The proof you have assumes that $\sqrt{ab}+\sqrt{a}\sqrt{b}\neq 0$ without explicitly stating/proving that. But if you want to work from that contradiction, squaring both sides of the (in)equation 
$$
\sqrt{ab} \neq \sqrt{a} \sqrt{b}
$$
and using commutativity of multiplication would do it, so the reason for that whole business isn't really clear.
A: You can prove it only if you know the following fact:
If $a\geq 0$ then there exists exactly one real number $\xi \geq 0$ (called the square root of $a$) such that $\xi^2=a$
If $\xi\geq 0, \eta\geq 0$ such that $\xi^2=a, \eta^2=b$ then
$$
(\xi \eta)^2= \xi^2\eta^2=ab
$$
and since you know that $\xi \eta \geq 0$ this is (!) the square root of $ab$
A: This is partly a comment on the answer given by Blah, but also an answer in its own right.  The OP's proof is basically correct; it lacks only the explicit statement that $\sqrt{ab}+\sqrt a\sqrt b\not=0$ if $a,b\gt0$.  Moreover the OP's proof essentially contains a proof of the fact Blah says you need to know.  
Let me re-express what's being asked for.  Suppose $p$ is a positive number such that $p^2=ab$.  Likewise suppose $r$ and $s$ are positive numbers such that $r^2=a$ and $s^2=b$.  We wish to prove $p=rs$.  The OP's proof starts by assuming $p\not=rs$, which means $p-rs\not=0$.  Multiplying this by the positive (hence non-zero) quantity $p+rs$, we get
$$(p+rs)(p-rs)=p^2-(rs)^2\not=0$$
But
$$p^2-(rs)^2=p^2-r^2s^2=ab-ab=0$$
which is a contradiction.  Hence we must have $p=rs$.  (In particular, by setting $b=1$ and taking $s=1$ to be "a" square root of $b$, we find that $p=r$ if $p^2$ and $r^2$ both equal $a$.)
One final remark (mostly to the OP).  Note that instead of a string of formulas, I wrote things out in more or less complete sentences, making explicit the logical connections.  I would strongly encourage the OP to do so as well, especially when first learning to write proofs (which I assume is the case here).  
