triclino Posted April 16, 2010 Share Posted April 16, 2010 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]) Link to comment Share on other sites More sharing options...
D H Posted April 16, 2010 Share Posted April 16, 2010 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] 1 Link to comment Share on other sites More sharing options...
triclino Posted April 16, 2010 Author Share Posted April 16, 2010 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?? Link to comment Share on other sites More sharing options...
the tree Posted April 16, 2010 Share Posted April 16, 2010 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. Link to comment Share on other sites More sharing options...
triclino Posted April 16, 2010 Author Share Posted April 16, 2010 Yes but introducing false statements into a proof can lead us to disastrous consequences Link to comment Share on other sites More sharing options...
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