
Let A = {x: x is a digit in the number $3591$}, B = {x: $x \in N,x < 10$}. Which of the following is false?
A. $A \cap B = \{ 1,3,5,9\} $
B. $A - B = \phi $
C. $B - A = \{ 2,4,6,7,8\} $
D. $A \cup B = \{ 1,2,3,5,9\} $
Answer
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Hint: In this question, we have two different sets. Start the question by writing the two sets in roster form. Then solve each of the operations given in the option and then match it with the one set already given.
Complete step-by-step answer:
We are given two sets in set-builder form. First, we will convert them into roaster form for clarity.
A = $\{ 1,3,5,9\} $
B = $\{ 1,2,3,4,5,6,7,8,9\} $
Now, we will start by examining each option.
Option A- $A \cap B = \{ 1,3,5,9\} $
The sign ‘$ \cap $’ stands for intersection. It basically means all the common elements of the given sets. If we observe-$1,3,5,9$ are the common elements between $A$ and $B$.
Hence, option (A) is correct.
Option B- $A - B = \phi $
The sign ‘-‘stands for difference. It means- all the elements in A but not in B. Here, the only four elements that are there in A are also there in B. So, when the two sets are subtracted, nothing remains in the set A. Therefore, $A - B = \phi $.
This option is also correct.
Option C- $B - A = \{ 2,4,6,7,8\} $
This option is also in ‘difference’. But here it means- all the elements that are there in B but not in A. The set A includes four elements which will be subtracted from the set B. The only remaining elements in B will be- $2,4,6,7,8$.
Hence, this option is also correct.
Option D- $A \cup B = \{ 1,2,3,5,9\} $
The sign $' \cup '$ means union. It means all the elements in both the sets, written together. The common elements in both the sets are not repeated and are written only once. Following the rules, the resultant set would be- $\{ 1,2,3,4,5,6,7,8,9\} $ which is the set B, itself.
However, the option (D) is false.
Among the given options, option A,B and C are the correct answers.
Note: The order of the elements in the set does not matter. In this question, while expanding set A, we could have also written- $\{ 3,5,9,1\} $ or $\{ 9,5,1,3\} $. It is the same as $\{ 1,3,5,9\} $. These are just different ways of representing the same set. The elements of the set can be written in any order. The order of the elements is immaterial.
Complete step-by-step answer:
We are given two sets in set-builder form. First, we will convert them into roaster form for clarity.
A = $\{ 1,3,5,9\} $
B = $\{ 1,2,3,4,5,6,7,8,9\} $
Now, we will start by examining each option.
Option A- $A \cap B = \{ 1,3,5,9\} $
The sign ‘$ \cap $’ stands for intersection. It basically means all the common elements of the given sets. If we observe-$1,3,5,9$ are the common elements between $A$ and $B$.
Hence, option (A) is correct.
Option B- $A - B = \phi $
The sign ‘-‘stands for difference. It means- all the elements in A but not in B. Here, the only four elements that are there in A are also there in B. So, when the two sets are subtracted, nothing remains in the set A. Therefore, $A - B = \phi $.
This option is also correct.
Option C- $B - A = \{ 2,4,6,7,8\} $
This option is also in ‘difference’. But here it means- all the elements that are there in B but not in A. The set A includes four elements which will be subtracted from the set B. The only remaining elements in B will be- $2,4,6,7,8$.
Hence, this option is also correct.
Option D- $A \cup B = \{ 1,2,3,5,9\} $
The sign $' \cup '$ means union. It means all the elements in both the sets, written together. The common elements in both the sets are not repeated and are written only once. Following the rules, the resultant set would be- $\{ 1,2,3,4,5,6,7,8,9\} $ which is the set B, itself.
However, the option (D) is false.
Among the given options, option A,B and C are the correct answers.
Note: The order of the elements in the set does not matter. In this question, while expanding set A, we could have also written- $\{ 3,5,9,1\} $ or $\{ 9,5,1,3\} $. It is the same as $\{ 1,3,5,9\} $. These are just different ways of representing the same set. The elements of the set can be written in any order. The order of the elements is immaterial.
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