# Applying De Morgan's Law

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?

$$-3 \le x \le 3$$

• You are correct! Does it make sense to you? It can be stated, equivalently, $x\geq -3$ and $x \leq 3$ Sep 19, 2014 at 23:57
• I'm understand how to get the solution, I just don't understand the process of arriving at it. I just derived this answer by looking at the solution in the other thread and working it out. Sep 20, 2014 at 0:03
• Let $p: x<-3$, and $q:x>3$. Then the negation of $p \lor q$ would be $\neg p \land \neg q$. Now what are the negations of $p$ and $q$? Sep 20, 2014 at 0:09
• $\neg P: x \ge -3$ $\neg Q: x \le 3$ Sep 20, 2014 at 0:20

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 $\lnot (\lnot P \land \lnot Q )$

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