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Q3(vii):
Taking the set of natural numbers as the universal set, write down the complements of the following sets: (vii) $\{ x : x$ is a perfect cube\}

Solution :

Step 1: Formal Definition of the Universal Set and Given Set

Given the Universal Set $U$ is the set of all natural numbers:

$U = \mathbb{N} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...\}$ [Per the standard definition of Natural Numbers in set theory].

Let $A$ be the set defined in the problem:

$A = \{x : x \text{ is a perfect cube}\}$

Listing the elements of $A$ based on $n^3$ where $n \in \mathbb{N}$:

$A = \{1^3, 2^3, 3^3, 4^3, ...\} = \{1, 8, 27, 64, ...\}$

Step 2: Defining the Complement of a Set

The complement of a set $A$, denoted by $A^c$ or $A'$, is defined as the set of all elements in the universal set $U$ that are not in $A$.

Mathematically: $A^c = \{x : x \in U \text{ and } x \notin A\}$ [Per the definition of the Complement of a Set].

Step 3: Deriving the Complement Set

To find $A^c$, we subtract the set of perfect cubes from the set of natural numbers:

$A^c = \mathbb{N} \setminus \{x : x = n^3, n \in \mathbb{N}\}$

This implies that $A^c$ contains all natural numbers except those that can be expressed as $n^3$ for some integer $n \ge 1$.

Visual Representation

Universal Set U (Natural Numbers) Set A (Cubes) A' (Non-cubes)

Step 4: Expressing the Result in Set-Builder Notation

Using the logical condition $x \neq n^3$, we represent the complement set as follows:

$A^c = \{x : x \in \mathbb{N} \text{ and } x \neq n^3 \text{ for any } n \in \mathbb{N}\}$

Category Description
Universal Set ($U$) $\{1, 2, 3, 4, 5, 6, 7, 8, ...\}$
Given Set ($A$) $\{1, 8, 27, 64, ...\}$
Complement ($A^c$) $\{2, 3, 4, 5, 6, 7, 9, 10, ...\}$

Final Solution: The complement of the set $\{x : x \text{ is a perfect cube}\}$ is $\{x : x \in \mathbb{N}, x \text{ is not a perfect cube}\}$.


More Questions from Class 11 Mathematics Sets EXERCISE 1.5


CBSE Solutions for Class 11 Mathematics Sets


Chapters in CBSE - Class 11 Mathematics


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