Question

Given 4 sets A, B, C and D, let A = {x : x ϵ N}, B = {x : x = 2n, n ϵ N}, C = {x : x = 2n-1, n ϵ N } and D = {x : x is a prime natural number}. Find \[A\cap D\] , that is the intersection of the sets A and D.

Answer 1

Hint: We will solve this question by using different methods of set representation and some basic formulas of set theory like subset, intersection and union. If you have a prior knowledge of these concepts you will understand the solution easily.

Complete step-by-step answer:
Let us start with the basics of the representation of sets. There are two methods of representing the sets: Rooster method and the Set-builder method.
The representation in the question is the set-builder method in which the elements are not listed.
Now let us convert that into Rooster form in which the elements are listed because it will make our solution simpler.
We also need one more basic concept of subsets and union and intersection.
Always remember that if a set is a subset of another set, the intersection of these two sets will result in the smaller set that is the subset. The union of these sets will result in the bigger set that is the parent set.
Let us take an example.
Let X = {1, 2, 3, 4, 5, 6} and Y = {2, 3, 4}
Now, we can see that the set Y is the subset of X with common elements {2, 3, 4}.
Now let us take the intersection by taking the common elements. Thus, \[X\cap Y\text{ }=\text{ }\left\{ 2,\text{ }3,\text{ }4 \right\}\] .
As you can see the result is equal to Y.
Thus, \[X\cap Y\text{ }=\text{ }Y\] .
Using the same concept, we solve the given question.
Given,
 A = {x : x ϵ N}
In the Rooster method this is A = {1, 2, 3, 4, 5, 6…} , which is all the natural numbers.
Similarly, B = {x : x = 2n, n ϵ N}
In the Rooster form this is B = {2*1, 2*2, 2*3, 2*4, 2*5, 2*6…}.
$\Rightarrow $ B = {2, 4, 6, 8, 10, 12…}, which is a set of all even natural numbers.
In the same way, C = {x : x = 2n-1, n ϵ N }, which is in the Rooster form, C = {(2*1)-1, (2*2)-1, (2*3)-1, (2*4)-1, (2*5)-1, (2*6)-1}.
$\Rightarrow $ C = {1, 3, 5, 7, 9, 11…}, which is ultimately the set of all odd numbers.
Proceeding in the same way, we can write, D = {x : x is a prime natural number} as D = {2, 3, 5, 7, 11, 13…}, which is the set of all prime natural numbers.
Now let us find the common elements in the set A and set D.
The common elements are {2, 3, 5, 7, 11, 13…}
As you can see the elements of the set D are already present in the set A, which means the set D is the subset of set A.
Using the concept, we learnt before that if one set is the subset of the other, then their intersection is the smaller of those sets which is of course the subset.
Thus, the intersection of set A and set D is set D itself.
Thus,\[A\cap D\text{ }=\text{ }D\] .

Note: Always remember that the sets of all even natural numbers, all odd natural numbers and prime numbers are subsets of the set of natural numbers. Knowing this in advance would help you to solve this question without the Rooster method and in less time.