# Basics of Set Theory : complement of an intersection.

My maths textbook says,

1) If x ∉ (A∩B)

=> x ∉ A or x ∉ B

2) If, A = {x:x is divisible by 3 and 5}

=> A' = {x:x is not divisible by 3 or x is not divisible by 5}

The italicised parts are troubling me. I am unable to visualise them or even comprehend them, especially the contextual meaning of the word "or" used in them. Please help me to understand these statements.

• "not ($p$ and $q$)" is equivalent to "not $p$ or not $q$" Apr 26 '20 at 17:14
• Maybe this (i.pinimg.com/736x/8f/b1/79/…) graphic can help you to visualize the statements Apr 26 '20 at 17:16
• In mathematics, unless otherwise stated, "or" always means the inclusive or. That is, "$p$ or $q$" by default means "at least one of $p,q$ is true". In this light, there isn't any contextual problems for "or". Apr 26 '20 at 17:21
• If $x \in A$ and $x \in B$ are both true, then $x$ will belong to both. Thus, if not, at least one of them must be false. Apr 26 '20 at 17:25
• In mathematics "not-A or not-B" means that A and B cannot both be true. Apr 26 '20 at 17:52

• The intersection of a set $$A$$ and of a set $$B$$ , namely the set $$A\cap B$$ , is defined using the logical operator " and " ( symbol : $$\land$$)

$$A\cap B$$ is the set of all objects that belong both to A and to B :

$$A\cap B= \{x| x \in A \land x\in B\}$$

• Consequently, in order to define the complement of the set $$A\cap B$$ we need to know what is the negation of an and-statement. And DeMorgan's law tells us that

"it is not the case that both sentence $$P$$ and sentence $$Q$$ are true"

is equivalent to

" either $$P$$ is false OR $$Q$$ is false".

Saying that " Peter is not both a pianist and a guitarist" means that " either Peter is not a pianist OR Peter is not a guitarist".

Note : you can check using a truth table that DeMorgan's law is actually a logical equivalence.

• If we apply this inside the set builder notation, with

$$A$$ = the set of all $$x$$ such that $$3$$ divides x

$$B$$= the set of all $$x$$ such that $$5$$ divides $$x$$ ,

we get :

$$(A\cap B)'$$

= the set of all $$x$$ that do NOT belong both to$$A$$ and to $$B$$.

$$= \{x| \neg (3|x \land 5|x)\}$$

$$= \{x| \neg (3|x) \lor \neg (5|x)\}$$

We also know that $$(P \implies Q) \iff (\lnot Q \implies \lnot P)$$. Say that it is not true that either $$x \not \in A$$ or $$x \not \in B$$. Then $$x \in A$$ and $$x \in B$$. Then $$x \in A \cap B$$. Therefore $$x \not \in (A \cap B) \implies (x \not \in A \lor x \not \in B)$$.

Now in this particular case, say that it is not true that $$x$$ is divisible by both $$3$$ and $$5$$. Then $$x$$ is either not divisible by $$3$$ or not divisible by $$5$$ (or both).

• You're welcome. Apr 26 '20 at 20:55