
Write the following sets in set builder form:
A. \[P = \left\{ {\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},.........\dfrac{9}{{10}},} \right\}\]
B. $Q = \left\{ {0,1,8,27,64,125,216} \right\}$
C. $R = \left\{ { - 36, - 30, - 24,..........42} \right\}$
D. $A = \left\{ {2,4,6,8} \right\}$
E. $B = \left\{ {3,9,27,81} \right\}$
F. $C = \left\{ {1,4,9,16,25} \right\}$
Answer
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Hint: First of all this is a very simple and a very easy problem. To solve this problem we need to have knowledge about how to write mathematically the set builder form. Set builder notation is a shorthand used to write sets, often for sets with an infinite number of elements. It is used with common types of numbers, such as integers, real numbers and natural numbers. This notation can also be used to express sets with an interval or an equation.
Complete answer:
Given in the question, many no. of sets, and we have to write the set builder form for each set.
Consider the first set as given below:
$ \Rightarrow P = \left\{ {\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},.........\dfrac{9}{{10}},} \right\}$
Here observe that each element in the set is following a particular pattern, that is the first term is $\dfrac{1}{2}$, and the next term is $\dfrac{2}{3}$, here the next term is obtained by adding 1 to both numerator and denominator.
Hence this pattern is followed throughout the set.
Hence the set builder form is given by:
$ \Rightarrow P = \left\{ {x:x = \dfrac{n}{{n + 1}};1 \leqslant n \leqslant 9} \right\}$
Now consider the second set as given below:
$Q = \left\{ {0,1,8,27,64,125,216} \right\}$
Here observe that each element in the set is following a particular pattern, that is the first term is ${0^3}$, the second term is ${1^3}$, and the next term is ${2^3}$, here the exponent is the same but the base of the exponent is changing. Every next base is obtained by adding 1 to it. Whereas the exponent remains the same.
Hence the set builder form is given by:
$ \Rightarrow Q = \left\{ {x:x = {n^3};0 \leqslant n \leqslant 6} \right\}$
Now consider the third set as given below:
$R = \left\{ { - 36, - 30, - 24,..........42} \right\}$
Here observe that each element in the set is following a particular pattern, that is the first term is $ - 36,$ which can be written as $6 \times - 6$ the second term is $ - 30,$ which can be written as $6 \times - 5$and the next term is $ - 24,$which can be written as $6 \times - 4$. The last term is $42 = 6 \times 7$. Hence every term is obtained by the multiplication of 6 with integers starting from -6 to 7.
Hence the set builder form is given by:
$ \Rightarrow R = \left\{ {x:x = 6n; - 6 \leqslant n \leqslant 7} \right\}$
Now consider the fourth set as given below:
$A = \left\{ {2,4,6,8} \right\}$
Here observe that each element in the set is following a particular pattern, that the first term is $2 = 2 \times 1$, the second term is $4 = 2 \times 2$, and the next term is $6 = 2 \times 3$.The last term is $8 = 2 \times 4$. Hence every term is obtained by the multiplication of 2 with integers starting from 1 to 4.
Hence the set builder form is given by:
$ \Rightarrow A = \left\{ {x:x = 2n;1 \leqslant n \leqslant 4} \right\}$
Now consider the fifth set as given below:
$B = \left\{ {3,9,27,81} \right\}$
Here observe that each element in the set is following a particular pattern, that the first term is $3 = {3^1}$, the second term is $9 = {3^2}$, and the next term is $27 = {3^3}$.The last term is $81 = {3^4}$. Hence every term is obtained by the adding 1 to the exponent of 3. Here the base remains constant which is 3, but the exponent changes from 1 to 4.
Hence the set builder form is given by:
$ \Rightarrow B = \left\{ {x:x = {3^n};1 \leqslant n \leqslant 4} \right\}$
Now consider the sixth set as given below:
$C = \left\{ {1,4,9,16,25} \right\}$
Here observe that each element in the set is following a particular pattern, that the first term is $1 = {1^2}$, the second term is $4 = {2^2}$, and the next term is $9 = {3^2}$.The last term is $25 = {5^2}$. Hence every term is obtained by the adding 1 to the base of the exponent, where the exponent is 2. Here the exponent remains constant which is 2, but the base changes from 1 to 5.
Hence the set builder form is given by:
$ \Rightarrow C = \left\{ {x:x = {n^2};1 \leqslant n \leqslant 5} \right\}$
Note:
While solving this problem we should be careful while writing the set builder form of the sets. As already discussed the set builder form is the short notation for denoting the set. Please note that while finding the set builder form of the set $R = \left\{ { - 36, - 30, - 24,..........42} \right\}$, here there may be a slight chance of confusion while writing the general term for the above set. Consider the first term -36 if it is written as $ - 6 \times 6$, then this would lead to a big confusion, and hence we wouldn’t be able to write down the correct set builder form.
Complete answer:
Given in the question, many no. of sets, and we have to write the set builder form for each set.
Consider the first set as given below:
$ \Rightarrow P = \left\{ {\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},.........\dfrac{9}{{10}},} \right\}$
Here observe that each element in the set is following a particular pattern, that is the first term is $\dfrac{1}{2}$, and the next term is $\dfrac{2}{3}$, here the next term is obtained by adding 1 to both numerator and denominator.
Hence this pattern is followed throughout the set.
Hence the set builder form is given by:
$ \Rightarrow P = \left\{ {x:x = \dfrac{n}{{n + 1}};1 \leqslant n \leqslant 9} \right\}$
Now consider the second set as given below:
$Q = \left\{ {0,1,8,27,64,125,216} \right\}$
Here observe that each element in the set is following a particular pattern, that is the first term is ${0^3}$, the second term is ${1^3}$, and the next term is ${2^3}$, here the exponent is the same but the base of the exponent is changing. Every next base is obtained by adding 1 to it. Whereas the exponent remains the same.
Hence the set builder form is given by:
$ \Rightarrow Q = \left\{ {x:x = {n^3};0 \leqslant n \leqslant 6} \right\}$
Now consider the third set as given below:
$R = \left\{ { - 36, - 30, - 24,..........42} \right\}$
Here observe that each element in the set is following a particular pattern, that is the first term is $ - 36,$ which can be written as $6 \times - 6$ the second term is $ - 30,$ which can be written as $6 \times - 5$and the next term is $ - 24,$which can be written as $6 \times - 4$. The last term is $42 = 6 \times 7$. Hence every term is obtained by the multiplication of 6 with integers starting from -6 to 7.
Hence the set builder form is given by:
$ \Rightarrow R = \left\{ {x:x = 6n; - 6 \leqslant n \leqslant 7} \right\}$
Now consider the fourth set as given below:
$A = \left\{ {2,4,6,8} \right\}$
Here observe that each element in the set is following a particular pattern, that the first term is $2 = 2 \times 1$, the second term is $4 = 2 \times 2$, and the next term is $6 = 2 \times 3$.The last term is $8 = 2 \times 4$. Hence every term is obtained by the multiplication of 2 with integers starting from 1 to 4.
Hence the set builder form is given by:
$ \Rightarrow A = \left\{ {x:x = 2n;1 \leqslant n \leqslant 4} \right\}$
Now consider the fifth set as given below:
$B = \left\{ {3,9,27,81} \right\}$
Here observe that each element in the set is following a particular pattern, that the first term is $3 = {3^1}$, the second term is $9 = {3^2}$, and the next term is $27 = {3^3}$.The last term is $81 = {3^4}$. Hence every term is obtained by the adding 1 to the exponent of 3. Here the base remains constant which is 3, but the exponent changes from 1 to 4.
Hence the set builder form is given by:
$ \Rightarrow B = \left\{ {x:x = {3^n};1 \leqslant n \leqslant 4} \right\}$
Now consider the sixth set as given below:
$C = \left\{ {1,4,9,16,25} \right\}$
Here observe that each element in the set is following a particular pattern, that the first term is $1 = {1^2}$, the second term is $4 = {2^2}$, and the next term is $9 = {3^2}$.The last term is $25 = {5^2}$. Hence every term is obtained by the adding 1 to the base of the exponent, where the exponent is 2. Here the exponent remains constant which is 2, but the base changes from 1 to 5.
Hence the set builder form is given by:
$ \Rightarrow C = \left\{ {x:x = {n^2};1 \leqslant n \leqslant 5} \right\}$
Note:
While solving this problem we should be careful while writing the set builder form of the sets. As already discussed the set builder form is the short notation for denoting the set. Please note that while finding the set builder form of the set $R = \left\{ { - 36, - 30, - 24,..........42} \right\}$, here there may be a slight chance of confusion while writing the general term for the above set. Consider the first term -36 if it is written as $ - 6 \times 6$, then this would lead to a big confusion, and hence we wouldn’t be able to write down the correct set builder form.
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