Question

List all the elements of each of the following sets given below:
(a). $A=\left\{ x:x=2n,n\in N\text{ }and\text{ }n\le 5 \right\}$
(b). $B=\left\{ x:x=2n+1,n\in W\text{ }and\text{ }n<5 \right\}$
(c). $C=\left\{ x:x=\dfrac{1}{n},n\in N\text{ }and\text{ }n<6 \right\}$
(d). $D=\left\{ x:x={{n}^{2}},n\in N\text{ }and\text{ 2}\le n\le 5 \right\}$
(e). $E=\left\{ x:x\in Z\text{ }and\text{ x=}{{\text{x}}^{2}} \right\}$
(f). $F=\left\{ x:x\in Z\text{ }and\text{ -}\dfrac{1}{2} < \text{x} < \dfrac{13}{2} \right\}$
(g). $G=\left\{ x:x=\dfrac{1}{2n-1},n\in N\text{ }and\text{ 1}\le n\le 5 \right\}$
(h). $H=\left\{ x:x\in Z,\left| x \right|\le 2 \right\}$

Answer 1

Hint: Listing the element of sets means showing all the elements of set into normal set form like if given a set $m=\left\{ x:x\in N,x<5 \right\}$ we can list the elements of set m as $\left\{ 1,2,3,4 \right\}$.

Complete step-by-step answer:

It is given in the question to list all the elements of given sets. We have set A as $A=\left\{ x:x=2n,n\in N\text{ }and\text{ }n\le 5 \right\}$. It means $x=2n$ where n belongs to a natural number less than 5.
For $n=1$ we get $x=2\times 1=2$,
For $n=2$ we get $x=2\times 2=4$
For $n=3$ we get $x=2\times 3=6$
For $n=4$ we get $x=2\times 4=8$
For $n=5$ we get $x=2\times 5=10$
Therefore set A can be written as $A=\left\{ 2,4,6,8,10 \right\}$.
Now, we have set B as $B=\left\{ x:x=2n+1,n\in W\text{ }and\text{ }n<5 \right\}$, which means $x=2n+1$ where n belongs to whole number less than 5. Since n belongs to the whole number, then we know that n begins from 0. Also, the value of n is less than 5, so we have the value of n = 0, 1, 2, 3 and 4. Now let us apply the condition, i.e $x=2n+1$.
For $n=0$, we have $x=2\times 0+1=1$
For $n=1$ , we have $x=2\times 1+1=3$
For $n=2$, we have $x=2\times 2+1=5$
For $n=3$, we have $x=2\times 3+1=7$
For $n=4$, we have $x=2\times 4+1=9$.
Therefore the B set can be written as $B=\left\{ 1,3,5,7,9 \right\}$.
We have set C as $C=\left\{ x:x=\dfrac{1}{n},n\in N\text{ }and\text{ }n<6 \right\}$, which means $x=\dfrac{1}{n}$ where n is any natural number less than 6. Since natural numbers start from 1, we have the values of n as 1, 2, 3, 4 and 5. Now, we can find the values of x by applying the condition.
For $n=1$ we have $x=\dfrac{1}{1}$
For $n=2$ we have $x=\dfrac{1}{2}$
For $n=3$ we have $x=\dfrac{1}{3}$
For $n=4$ we have $x=\dfrac{1}{4}$
For $n=5$ we have $x=\dfrac{1}{5}$ .
Therefore we get set C as $C=\left\{ 1,\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5} \right\}$.
Now, we have set D as $D=\left\{ x:x={{n}^{2}},n\in N\text{ }and\text{ 2}\le n\le 5 \right\}$, we have $x={{n}^{2}}$ where n belongs to natural numbers between 2 and 5 including 2 and 5. So, now let us apply the condition as
For $n=2$, we have $x={{2}^{2}}=4$
For $n=3$ we have $x={{3}^{2}}=9$
For $n=4$ we have $x={{4}^{2}}=16$
For $n=5$ we have $x={{5}^{2}}=25$.
Therefore we get set D as $D=\left\{ 4,9,16,25 \right\}$.
Now, we have set E as $E=\left\{ x:x\in Z\text{ }and\text{ x=}{{\text{x}}^{2}} \right\}$ which means that x is an integer and also $x={{x}^{2}}$. So, we have only two integers 0,1 that follow the given condition, that is
For $x=0$ we get $x={{0}^{2}}=0$ and for $x=1$ we get $x={{1}^{2}}=1$.
Therefore we get set E as $E=\left\{ 0,1 \right\}$.
Now, we have Set F as $F=\left\{ x:x\in Z\text{ }and\text{ -}\dfrac{1}{2} < \text{x} < \dfrac{13}{2} \right\}$ which means that x in an integer such that it lies between $-\dfrac{1}{2}< x <\dfrac{13}{2}$. Let us consider them in decimal form, then we know that $-\dfrac{1}{2}$ is -0.5 and $\dfrac{13}{2}$ is 6.5. So, the values of x must lie between -0.5 and 6.5.
Therefore we get set F as $F=\left\{ 0,1,2,3,4,5,6 \right\}$.
Now, we have set G as $G=\left\{ x:x=\dfrac{1}{2n-1},n\in N\text{ }and\text{ 1}\le n\le 5 \right\}$ we have $x=\dfrac{1}{2n-1}$ also n belongs to natural number between $1\le n\le 5$. Natural numbers start from 1 and as per the condition given, the range is between 1 and 5, including them. So, we have the values of n as 1, 2, 3, 4 and 5. Now, we can apply the condition to find values of x as
For $n=1$, we have $x=\dfrac{1}{2n-1}=\dfrac{1}{2\times 1-1}=\dfrac{1}{1}$
For $n=2$, we have $x=\dfrac{1}{2n-1}=\dfrac{1}{2\times 2-1}=\dfrac{1}{3}$
For $n=3$ we have $x=\dfrac{1}{2n-1}=\dfrac{1}{2\times 3-1}=\dfrac{1}{5}$
For $n=4$ we have $x=\dfrac{1}{2n-1}=\dfrac{1}{2\times 4-1}=\dfrac{1}{7}$
For $n=5$ we have $x=\dfrac{1}{2n-1}=\dfrac{1}{2\times 5-1}=\dfrac{1}{9}$.
Therefore we have set G as $G=\left\{ 1,\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9} \right\}$.
Now, we have set H as $H=\left\{ x:x\in Z,\left| x \right|\le 2 \right\}$, which means that x belongs to integers such that $\left| x \right|\le 2$. We know that integers contain both positive and negative numbers. Modulus of any number gives only positive value. So, as per the condition, when we remove the modulus, we must get a value of \[x\le 2\]. So, let us consider the possible positive values, then we will get \[x\text{ }=\text{ }0,\text{ }1,\text{ }2\] . Now, similarly, the modulus of the negative of these numbers will also result in value \[\le 2\], so we get x = -1, -2.
Therefore, we get set H as \[H=\left\{ 0,1,-1,2,-2 \right\}\].

Note: Many times students are stuck while converting set builder form into roster form. They must know how to apply the given conditions and also make sure that the values are in the range given in the set builder form. It is advised to do each step in an orderly manner so as to avoid any silly mistakes. Whenever n is an element of N, then the value of n starts from 1 and not 0, this is a common mistake committed by most students in a hurry to solve this type of question.