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# Write each of the sets given below in the set builder form. (i). $A=\left\{ 1,\dfrac{1}{4},\dfrac{1}{9},\dfrac{1}{16},\dfrac{1}{25},\dfrac{1}{36},\dfrac{1}{49} \right\}$ (ii). $B=\left\{ \dfrac{1}{2},\dfrac{2}{5},\dfrac{3}{10},\dfrac{4}{17},\dfrac{5}{26},\dfrac{6}{37},\dfrac{7}{50} \right\}$  Hint: We can solve this question if we know that when we have to convert any set from a roster form to a set builder form, then we have to develop and identify a common relation among all the elements of the sets. Like, when we look at the terms in the first set, that is set A, we can see that they contain the squares of natural numbers in a specific pattern.

In this question, we are asked to convert the sets from roster form to set builder form. For that we need to find the common relation among the elements of the sets, so we will observe each element one by one. And we also require the range to write them in the set builder form.
(i). $A=\left\{ 1,\dfrac{1}{4},\dfrac{1}{9},\dfrac{1}{16},\dfrac{1}{25},\dfrac{1}{36},\dfrac{1}{49} \right\}$
Here we have been given set A. Now, to develop a relation, we will consider each of the terms in this set separately. So, we can write them as,
\begin{align} & 1=\dfrac{1}{{{\left( 1 \right)}^{2}}} \\ & \dfrac{1}{4}=\dfrac{1}{{{\left( 2 \right)}^{2}}} \\ & \dfrac{1}{9}=\dfrac{1}{{{\left( 3 \right)}^{2}}} \\ & \dfrac{1}{16}=\dfrac{1}{{{\left( 4 \right)}^{2}}} \\ & \dfrac{1}{25}=\dfrac{1}{{{\left( 5 \right)}^{2}}} \\ & \dfrac{1}{36}=\dfrac{1}{{{\left( 6 \right)}^{2}}} \\ & \dfrac{1}{49}=\dfrac{1}{{{\left( 7 \right)}^{2}}} \\ \end{align}
Here, we can see that there is a common nature among all the elements of the set A, that is they are reciprocal of the squares of natural numbers. Also, we can write them as, $x=\dfrac{1}{{{n}^{2}}}$. Now, if we look at the range, then we see that the first element has n = 1 and the last element has n = 7. So, we can say, $1\le n\le 7$.
Hence, we can write the set A as, $\left\{ x:x=\dfrac{1}{{{n}^{2}}},1\le n\le 7 \right\}$.
(ii). $B=\left\{ \dfrac{1}{2},\dfrac{2}{5},\dfrac{3}{10},\dfrac{4}{17},\dfrac{5}{26},\dfrac{6}{37},\dfrac{7}{50} \right\}$
Now, we will consider each element of the given set one by one to develop a relation among them. So, we can write them as,
\begin{align} & \dfrac{1}{2}=\dfrac{\left( 1 \right)}{{{\left( 1 \right)}^{2}}+1} \\ & \dfrac{2}{5}=\dfrac{\left( 2 \right)}{{{\left( 2 \right)}^{2}}+1} \\ & \dfrac{3}{10}=\dfrac{\left( 3 \right)}{{{\left( 3 \right)}^{2}}+1} \\ & \dfrac{4}{17}=\dfrac{\left( 4 \right)}{{{\left( 4 \right)}^{2}}+1} \\ & \dfrac{5}{26}=\dfrac{\left( 5 \right)}{{{\left( 5 \right)}^{2}}+1} \\ & \dfrac{6}{37}=\dfrac{\left( 6 \right)}{{{\left( 6 \right)}^{2}}+1} \\ & \dfrac{7}{50}=\dfrac{\left( 7 \right)}{{{\left( 7 \right)}^{2}}+1} \\ \end{align}
We can see a common nature among the elements of set B, that is they can be expressed as $x=\dfrac{n}{{{\left( n \right)}^{2}}+1}$. If we consider the range, then the first element has n = 1 and the last element has n = 7. So, we can say the range is $1\le n\le 7$.
Hence, we write set B as, $\left\{ x:x=\dfrac{n}{{{\left( n \right)}^{2}}+1},1\le n\le 7 \right\}$.

Note: We need to remember to develop a common relation between all the elements of the sets, while converting from roster to set builder form. The other thing that one should remember is that set builder form is always represented as, {f(x) : some relation, range}. We have an alternate way to represent set A given in the question, which is, {x : x is the reciprocal of the squares of natural numbers ranging from 1 to 7}.

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