Can the following definition of the absolute value be considered as correct??

 

([math]x\geq 0\Longrightarrow |x|=x[/math]) and (x<0[math]\Longrightarrow |x|=-x[/math])

That works. So does saying that |x|=x for x>0, |x|=-x for x<=0. Or a three-way case,

 

[math]|x|=\begin{cases} \phantom{-}x & \text{if~}x>0 \\ -x &\text{if~} x<0\\ \phantom{-}0 & \text{if~}x=0\end{cases}[/math]

 

I like the latter as it explicitly identifies 0 is a special case and because of the symmetry. That's just personal preference.

 

There are many other ways to write it. For example

 

[math]|x|=\sqrt{x^2}[/math]

But ,however, logic dictates us the following:

 

If we put :

 

([math]x\geq 0[/math]): = p

 

 

(|x|=x) := r

 

(x<0): = q

 

 

(|x|= -x): = s...............then we have:

 

([math]p\Longrightarrow r[/math]) and ([math]q\Longrightarrow s[/math]) which logicaly implies:

 

[math]p\wedge q\Longrightarrow r\wedge s[/math]

 

But p&q means that we have : [math]x\geq 0[/math] and x<0 .

 

IS that possible??

Yes, in fact it's necessarily true. Since p^q is never true, p^q => r^s is a vacuous truth.

Remember that if a is false then [a=>b] is true regardless of b.

Yes but introducing false statements into a proof can lead us to disastrous consequences