I'm working on my assignment for Discrete Math and I'm not fully understanding how to do this question for it so I was wondering if anyone here could help show me how to do it properly;
Use De Morgan’s Laws to state the negations of the following
i. Either x < -3 or x > 3
I understand what De Morgan's Laws are:
$$\neg(P \vee Q) \equiv (\neg P \wedge \neg Q)$$
$$\neg(P \wedge Q) \equiv (\neg P \vee \neg Q)$$
I'm just unsure of how to apply De Morgan's Laws to this question. I saw in another thread someone asking a similar question and tried to work it out myself by just guessing, so would this be correct?
$$-3 \le x \le 3$$
Best Answer
Lets do it step by step
But first of all there is a problem with "either"
Logic always works with "inclusive" or so $P \lor Q$ is also true if P and Q are both true.
In "Either $ x < -3 $ or $ x > 3 $ "the two propositions have some problem to be both true, but propositional just logic doesn't look that deep,
luckely we we can just treat it as $ x < -3 $ (inclusive or) $ x > 3 $
using P as meaning $ x < -3 $ and Q as meaning $ x > 3 $
we get
$ P \lor Q $
And this is equivalent to $ \lnot (\lnot P \land \lnot Q ) $
So that should be your answer.
but i guess you need to go a bit deeper
I guess you may assume
so your formula becomes: $ \lnot ( x \geq -3 \land x \leq 3 )$
(don't forget the $ \lnot$ )
and that can be simplified
GOOD LUCK