[Math] Applying De Morgan’s Law

discrete mathematicslogic

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

  • $ \lnot P = \lnot ( x < -3) $ so $ \lnot P = x \geq -3 $ and
  • $ \lnot Q = \lnot( x > 3 ) $ so $ \lnot Q = x \leq 3 $

so your formula becomes: $ \lnot ( x \geq -3 \land x \leq 3 )$

(don't forget the $ \lnot$ )

and that can be simplified

GOOD LUCK

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