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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?

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

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    $\begingroup$ You are correct! Does it make sense to you? It can be stated, equivalently, $x\geq -3$ and $x \leq 3$ $\endgroup$
    – amWhy
    Sep 19, 2014 at 23:57
  • $\begingroup$ 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. $\endgroup$ Sep 20, 2014 at 0:03
  • $\begingroup$ 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 $q$? $\endgroup$ Sep 20, 2014 at 0:09
  • $\begingroup$ $\neg P: x \ge -3$ $\neg Q: x \le 3$ $\endgroup$ Sep 20, 2014 at 0:20

1 Answer 1

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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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