Prove that $mn  0$ and $n  0$. Prove that $mn &lt; 0$ if and only if $m &gt; 0$ and $n &lt; 0$ or $m &lt; 0$ and $n &gt; 0$. 
$m,n$ element of integers

Just starting out teaching myself discrete math still really bad at proofs, any help/advice on how to think/go about this would be greatly appreciated.
We need to make a few assumptions about how you have integers defined, but a very safe assumption is that you have defined $(\Bbb Z, +, \times, \leq)$ as an ordered commutative ring with identity constructed from the natural numbers in the usual way.

In particular, as part of the very definition of $(\Bbb Z,+,\times,\leq)$, we have the following (among many more properties, these are just the important ones necessary for my proof):

(In all of the following, $a,b,c$ and $x$ are arbitrary integers)



Properties of $+$


$0+a=a$ for any $a$
If $a$ and $b$ are integers, then $a+b$ is an integer




Properties of $\leq$ (and $&lt;$)


If $a\leq b$ then $a+c\leq b+c$
If $0\leq a$ and $0\leq b$ then $0\leq a\times b$
$0\leq 1$




Properties of $-$


$(-x)$ is the unique number such that $(-x)+x=0$
$-(-x) = x$




Properties of $\times$


$a\times b = b\times a$
$(-x)=(-1)\times x$
$a\times b = 0$ if and only if $a=0$ or $b=0$
If $a$ and $b$ is an integer, then $a\times b$ is an integer






Starting the desired proof, we approach via contrapositive for the forward implication.  That is, we wish to prove that if ($0\leq m$ and $0\leq n$) or ($m\leq 0$ and $n\leq 0$) then $0\leq m\times n$.  In doing so, we will have proven that $m\times n&lt;0$ implies $m&lt;0$ and $0&lt;n$ or $0&lt;m$ and $n&lt;0$.

Suppose $0\leq m$ and $0\leq n$.  Then by definition $0\leq m\times n$, so we are done.

Suppose $m\leq 0$ and $n\leq 0$.  Then $m+(-m)\leq 0+(-m)$ and therefore $0\leq -m$.  Similarly $0\leq -n$.  Then $0\leq (-m)\times (-n) = ((-1)\times m)\times ((-1)\times n) = (-1)\times (-1)\times m\times n = m\times n$.

Thus if $m\leq 0$ and $n\leq 0$ or if $0\leq m$ and $0\leq n$, we have $0\leq m\times n$

For the remaining direction for the proof, we wish to show that if $0&lt;m$ and $n&lt;0$ or if $m&lt;0$ and $0&lt;n$ then $m\times n &lt; 0$

Suppose $0&lt;m$ and $n&lt;0$.  Then $n+(-n)&lt;0+(-n)=-n$.  Thus $0&lt;m\times (-n)$.  Thus $0+(m\times n) &lt; m\times (-n)+(m\times n) = m\times (-1)\times n + m\times n = -(m\times n)+(m\times n) = 0$.  Therefore $m\times n&lt;0$

The proof for $m&lt;0$ and $0&lt;n$ is identical by relabeling which was $m$ and which was $n$.
1) If m &lt; 0 and n >0 or m>0 and n &lt; 0,then mn &lt; 0.

2) If mn &lt; 0, suppose m > 0 and n > 0,this gives mn > 0 which is contradiction.Hence ,one of them should definitely be less than 0.If both are,then contradiction arises again as product would be greater than 0