
If $A$ and $B$ are two disjoint sets, then which one of the following is correct?
A. $A-B=A-(A\cap B)$
B. $B-{{A}^{c}}=A\cap B$
C. $A\cap B=(A-B)\cap B$
D. All of these
Answer
163.5k+ views
Hint: In this question, we are to find the correct statements from the given statements. Since the given sets are disjoint, we can write $A\cap B=\varnothing $. By using this we can verify the above statements.
Formula Used:Set: Collection of objects in which it is possible to decide whether a given object is belongs to the collection or not is said to be a set. Those objects are nothing but the elements in the set.
Sets are represented in two ways: The roaster method and the Set builder form
Roaster method:
The set of vowels – \[\left\{ a,\text{ }e,\text{ }i,\text{ }o,\text{ }u \right\}\]
The set of natural numbers – \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots. \right\}\]
Set builder form:
The set of vowels – \[\left\{ x:x\text{ }is\text{ }a\text{ }vowel\text{ }in\text{ }English\text{ }alphabet \right\}\]
The set of natural numbers – \[\left\{ x:x\text{ }is\text{ }a\text{ }natural\text{ }number \right\}\]
Some of the important mathematical sets:
$N$ - the set of natural numbers - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
Complete step by step solution:Given that, $A$ and $B$ are two disjoint sets.
So, we can write $A\cap B=\varnothing $
Then,
\[A-B=\{x:x\in A,x\notin B\}\] and $A-B\subset A$
$A-(A\cap B)=A-\varnothing =A$
Thus,
$A-B=A-(A\cap B)$ is true.
\[B-{{A}^{c}}=B-(\bigcup -A)=\varnothing \] and we know that $A\cap B=\varnothing $
So, $B-{{A}^{c}}=A\cap B$ is also true.
$A-B=\{x:x\in A,x\notin B\}$
Then, w can write $A-B\subset A$
Thus,
$(A-B)\cap B=\varnothing $
So, $A\cap B=(A-B)\cap B$ is also true.
Therefore, all the statements are correct.
Option ‘D’ is correct
Note: Since the given sets are disjoint, there are no common elements between them. Hence, the given statements are true.
Formula Used:Set: Collection of objects in which it is possible to decide whether a given object is belongs to the collection or not is said to be a set. Those objects are nothing but the elements in the set.
Sets are represented in two ways: The roaster method and the Set builder form
Roaster method:
The set of vowels – \[\left\{ a,\text{ }e,\text{ }i,\text{ }o,\text{ }u \right\}\]
The set of natural numbers – \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots. \right\}\]
Set builder form:
The set of vowels – \[\left\{ x:x\text{ }is\text{ }a\text{ }vowel\text{ }in\text{ }English\text{ }alphabet \right\}\]
The set of natural numbers – \[\left\{ x:x\text{ }is\text{ }a\text{ }natural\text{ }number \right\}\]
Some of the important mathematical sets:
$N$ - the set of natural numbers - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
Complete step by step solution:Given that, $A$ and $B$ are two disjoint sets.
So, we can write $A\cap B=\varnothing $
Then,
\[A-B=\{x:x\in A,x\notin B\}\] and $A-B\subset A$
$A-(A\cap B)=A-\varnothing =A$
Thus,
$A-B=A-(A\cap B)$ is true.
\[B-{{A}^{c}}=B-(\bigcup -A)=\varnothing \] and we know that $A\cap B=\varnothing $
So, $B-{{A}^{c}}=A\cap B$ is also true.
$A-B=\{x:x\in A,x\notin B\}$
Then, w can write $A-B\subset A$
Thus,
$(A-B)\cap B=\varnothing $
So, $A\cap B=(A-B)\cap B$ is also true.
Therefore, all the statements are correct.
Option ‘D’ is correct
Note: Since the given sets are disjoint, there are no common elements between them. Hence, the given statements are true.
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