Taking the set of natural numbers as the universal set,write down the complement of the following set:
$A = \{ x : x \text{ is a positive multiple of } 3 \}$
$A = \{ x : x \text{ is a positive multiple of } 3 \}$
(N/A) Let $U = N$ be the set of natural numbers.
The complement of a set $A$,denoted by $A'$,is defined as $A' = \{ x : x \in U \text{ and } x \notin A \}$.
Given $A = \{ x : x \text{ is a positive multiple of } 3 \}$.
Therefore,the complement $A' = \{ x : x \in N \text{ and } x \text{ is not a multiple of } 3 \}$.
The complement of a set $A$,denoted by $A'$,is defined as $A' = \{ x : x \in U \text{ and } x \notin A \}$.
Given $A = \{ x : x \text{ is a positive multiple of } 3 \}$.
Therefore,the complement $A' = \{ x : x \in N \text{ and } x \text{ is not a multiple of } 3 \}$.
Explore More
Similar Questions
If $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,$A = \{2, 4, 6, 8\}$,and $B = \{2, 3, 5, 7\}$,verify that $(A \cap B)^{\prime} = A^{\prime} \cup B^{\prime}$.
Medium
View Solution$A$ and $B$ are events such that $P(A)=0.42$,$P(B)=0.48$ and $P(A \cap B)=0.16$. Determine $P(\text{not } B)$.
Easy
View SolutionIf $P(A) = 0.65$ and $P(B) = 0.15$,then $P(\bar{A}) + P(\bar{B}) = $
Medium
View SolutionTaking the set of natural numbers as the universal set,write down the complement of the following set:
$A = \{ x : x \text{ is an odd natural number} \}$
$A = \{ x : x \text{ is an odd natural number} \}$
Easy
View SolutionThe number of subsets of $\{1, 2, 3, \ldots, 9\}$ containing at least one odd number is
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo