Negating statements / Finding $(A \cap B)',A \oplus B$ if $A=\{x \in\Bbb R \mid -3\le x\le0\}$ and $B=\{x \in \Bbb R\mid -1 < x < 2\}$

Question

I am a bit new on this field and I am trying to solve some questions. I don't really think they are hard but there are some key points that I don't get it or I am stuck.

Lets see.

1) Write the negation of the sentences:
1a: "If the teacher is missing, some students will not do their homework."

My approach: Let $p=$ The teacher is missing; $q=$ some students; $r=$ do their h/w. That would be $(p \to (q \land \neg r) )$. I searched for "De Morgan's law", but didn't find anything about what to do with the negation of "$\to$".

1b. Find the negation of: $(\forall x \in D)(x+4 \le 8)$.

I really have no idea, but the obvious would be $(\not\exists x \notin D)(x+4 \le 8)$.

Also the negation of $\exists$ would be $\not\exists$, right?

2) $A=\{x \in\Bbb R \mid -3\le x\le0\}$ and $B=\{x \in \Bbb R\mid -1 < x < 2\}$

2a) Find $(A \cap B)'$.

I think this is simple, so the answer would be all the numbers $\in \Bbb R$, except the numbers $-1 < x \le 0$ (That would be their common numbers). That makes sense to me, I mean a lot of sense.

2b) $A \oplus B$.

No I cant even visualize this. I know the truth table of xor at least, but this cant really help me to continue.

I would really appreciate some input, at least for my answers if possible.

Thanks in advance!

Answer 1

Since Peter already addressed question 1, let me comment on 2.

Your idea for 2a. is correct. So let us try to be a little more specific. We currently have:

$$(A\cap B)' = \{x \in \Bbb R\mid \text{not}(-1 < x \le 0)\}$$

The statement $\text{not}(-1<x\le 0)$ can be simplified: it must be that either $-1 \not <x$, i.e. $-1 \ge x$, or $x \not\le 0$, i.e. $x > 0$. In conclusion:

$$(A\cap B)' = \{x \in \Bbb R \mid x \le -1 \text{ or } x > 0\}$$

---

Now for 2b. By definition of $\oplus$, we have that: $$A \oplus B = \{x \mid x \in A \text{ or } x \in B, \text{ but not both}\}$$

which can be written as $$A \oplus B = (A\cup B)\setminus (A \cap B)$$ with $\setminus$ denoting set difference, $C \setminus D = \{x \in C: x \notin D\}$.

We had already determined $A \cap B$. Can you also do $A \cup B$ and after that, $A \oplus B$? I advise you to stay close to the definitions, write out everything meticulously, and simplify your answer as much as possible. That's the best way to get to grips with it.

Answer 2

Re Ia) "Some students" is not a proposition is it? So you can't use a propositional variable to represent that phrase!

If the teacher is missing, some students will not do their homework

can be rendered

$(P \to Q)$

where $Q$ represents the proposition "some students will not do their homework", and $\to$ represents the conditional construction. The issue of the negation of conditionals is a vexed one. If the conditional can be read as the material truth-functional conditional then

$\neg(P \to Q)$

is equivalent to

$(P \land \neg Q)$.

But if not, then not!

1b)

$(\forall x \in D) (x+4 \leq 8) $

is short for

$\forall x(x \in D \to x+4 \leq 8) $

where this time the material conditional is definitely intended. Which is equivalent to

$\neg\exists x\neg(x \in D \to x+4 \leq 8) $

So its negation is

$\exists x\neg(x \in D \to x+4 \leq 8) $

which is equivalent to

$\exists x (x \in D \land \neg(x+4 \leq 8)) $

or

$\exists x (x \in D \land x+4 > 8) $

which abbreviates to

$(\exists x \in D)(x+4 > 8) $

which is -- if you think about it -- intuitively the right answer.