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# f three sets are given as $A=\left\{ x:x\in W,3\le x\le 6 \right\},B=\{3,5,7\}$ and $C=\{2,4\}$ ; then find: $B-C$ .  Verified
For example: Consider the set $B=\{2,4,6,8,10\}$. Here $B$ is a set of elements that are positive and even numbers less than $12$. So, the property of the elements of the set is that they are positive even numbers less than $12$. So, the set $B$is represented in the set-builder form as $B=\{x:x\,is\,positive\,even\,number\,<12\}$ .
Now, the set $A$ is given as $A=\left\{ x:x\in W,3\le x\le 6 \right\}$. We can see that the property of the elements of the set $A$ is that they are whole numbers that are greater than or equal to $3$ and less than or equal to $6$. So, the set $A$ can be written as $A=\{3,4,5,6\}$.
Now, we will consider two sets $A$ and $B$ such that the set $A$ is given as $A=\{a,b,c,d,e\}$ , and the set $B$ is given as $B=\{d,e,f,g\}$. The difference of the two sets $A$ and $B$ is the set of all such elements, which are present in the set$A$ but not in the set $B$ . It is written as $A-B$ and is given as $A-B=\{a,b,c\}$.
Now, we are asked to find the value of $B-C$ , where the set $B$ is given as $B=\{3,5,7\}$ and, the set $C$ is given as $C=\{2,4\}$. We can see that all the elements in the set $B$ are not present in the set $C$.
So, the value of $B-C$ is given as $B-C=\{3,5,7\}$.
Note: The set builder form is read as “$A$ is the set of all $x$ such that $x$ is a natural number less than $7$.” Students generally get confused between set builder notation of sets and roster notation of sets. Both are different and hence, should not be confused. This confusion can lead to wrong answers. Also, while evaluating the difference of two sets, make sure to mention the elements which are present in the first set and not in the second set. Most of the students mention the elements that are present in the second set and not in the first set, which is wrong. Such mistakes should be avoided.