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.

  • 3
    $\begingroup$ "not ($p$ and $q$)" is equivalent to "not $p$ or not $q$" $\endgroup$ Apr 26 '20 at 17:14
  • 3
    $\begingroup$ Maybe this (i.pinimg.com/736x/8f/b1/79/…) graphic can help you to visualize the statements $\endgroup$
    – Timmathy
    Apr 26 '20 at 17:16
  • $\begingroup$ 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". $\endgroup$ Apr 26 '20 at 17:21
  • $\begingroup$ 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. $\endgroup$ Apr 26 '20 at 17:25
  • $\begingroup$ In mathematics "not-A or not-B" means that A and B cannot both be true. $\endgroup$ 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).

  • $\begingroup$ You're welcome. $\endgroup$ Apr 26 '20 at 20:55

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