Question

Let be two sets E and F such that:E = {x € R: -4 ≤ x ≤ 4}F = = xWhat is the complement of E?Make the product between the complementary of E and F

Answer 1

Recall that:

`E^c=\mleft\lbrace x\in\R\colon x\\otin E\mright\rbrace\text{.}`

Since

`E=\mleft\lbrace x\in\R\colon-4\le x\le4\mright\rbrace\text{.}`

Therefore:

`\begin{gathered} E^c=\mleft\lbrace x\in\R\colon x\\otin E\mright\rbrace=\lbrace x\in\R\colon x<-4\text{ or x>4}\} \\ =\mleft\lbrace x\in\R\colon x<-4\mright\rbrace\cup\lbrace x\in\R\colon x>4\rbrace\text{.} \end{gathered}`

Now, the cartesian product between the complementary of E and F is:

`E^c* F=(\lbrace x\in\R\colon x<-4\rbrace\cup\lbrace x\in\R\colon x>4\rbrace)*(\mleft\lbrace x\in\R\colon\mright|x|=x\})\text{.}`

Now, recall that:

`\begin{gathered} |x|=x\text{ if and only if x}\ge0, \\ (A\cup B)* C=A* C\cup B* C. \end{gathered}`

Therefore:

`E^c* F=(\lbrace x\in\R\colon x<-4\rbrace*\lbrace x\in\R\colon x\ge0\})\cup(\lbrace x\in\R\colon x>4\rbrace)*\lbrace x\in\R\colon x\ge0\})\text{.}`

Answer:

`\begin{gathered} E^c=\lbrace x\in\R\colon x<-4\rbrace\cup\lbrace x\in\R\colon x>4\rbrace\text{.} \\ E^c* F=(\lbrace x\in\R\colon x<-4\rbrace*\lbrace x\in\R\colon x\ge0\})\cup(\lbrace x\in\R\colon x>4\rbrace)*\lbrace x\in\R\colon x\ge0\})\text{.} \end{gathered}`