
Describe the following sets in Roster form:
a. {x: x is a letter before e in the English alphabet}
b. {x$ \in $N: x2 < 25}
c. {x$ \in $ N: x is a prime number, 10 < x < 20}
d. {x$ \in $N: x = 2 n, n $ \in $N}
e. {x$ \in $R: x > x}
f. {x: x is a prime number which is a divisor of 60}
g. {x: x is a two – digit number such that the sum of its digit is 8}
h. The set of all letters in the word ‘trigonometry’.
i. The set of all letters in the word ‘Better’.
Answer
510.9k+ views
Hint: We will use the definition of the roster form of the sets which says that the way of describing the content of a set by listing the elements in a set of curly brackets separated by commas is called the roster form of the set.
Complete step by step answer:
a. {x: x is a letter before e in the English alphabet}
Now, if we see the alphabets, there are only 4 alphabets before e namely a, b, c, and d.
$\therefore ${a, b, c, d} is the roster form of this set.
b. {x$ \in $N: x2 < 25}
if we take square root both sides in this inequality x2 < 25, we get
$ \Rightarrow $$x = \sqrt {25} = 5$
Hence, the natural numbers less than 5 are 1, 2, 3 and 4.
$\therefore ${1 2, 3, 4} is the roster form of this set.
c. {x$ \in $ N: x is a prime number, 10 < x < 20}
All the prime numbers lying between 10 and 20 are 11, 13, 17 and 19.
$\therefore ${11, 13, 17, 19} is the roster form of this set.
d. {x$ \in $N: x = 2 n, n $ \in $N}
for x$ \in $N, x = 1, 2, 3, 4, 5, …
and n$ \in $N, n = 1, 2, 3, 4, 5, …
but here, x = 2n $ \Rightarrow $n = $\dfrac{x}{2}$. Hence, x should be a multiple of 2 in order that n$ \in $N as $\dfrac{1}{2},\dfrac{3}{2},\dfrac{5}{2},...$$ \notin $N
therefore, for x = 2, 4, 6, 8, …, n = 1, 2, 3, 4, …
$\therefore ${2, 4, 6, 8, …} is the roster form of this set.
e. {x$ \in $R: x > x}
There is no such real number which is greater than itself. Hence, this set will be a null set and denoted as: {} or $\phi $.
f. {x: x is a prime number which is a divisor of 60}
The numbers which are divisor of 60 can be listed as 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Amongst them, only 2, 3, and 5 are prime numbers.
$\therefore ${2, 3, 5} is the roster form of this set.
g. {x: x is a two-digit number such that the sum of its digit is 8}
The total 2 – digit numbers whose sum of its digit is 8 can be listed as 17, 26, 35, 44, 53, 62, 71, and 80.
$\therefore ${17, 26, 35, 44, 53, 62, 71, 80} is the roster form of this set.
h. The set of all letters in the word ‘trigonometry’.
All the letters in the word ‘TRIGONOMETRY’ can be listed as T, R, I, G, O, N, M, E, and Y.
$\therefore ${T, R, I, G, O, N, M, E, Y} is the roster form of this set.
i. The set of all letters in the word ‘Better’.
All the letters in the word better can be listed as B, E, T, and R.
$\therefore ${B, E, T, R} is the roster form of this set.
Note: In such questions, you must know which form is the roster and which is set – builder form of the set. You may go wrong while obtaining the elements of each set as you are provided with the set-builder form only. You may go wrong in the last two parts because, in them, there is no need to repeat the same word because no matter how many times it occurs, it will be treated as a single element of the set.
Complete step by step answer:
a. {x: x is a letter before e in the English alphabet}
Now, if we see the alphabets, there are only 4 alphabets before e namely a, b, c, and d.
$\therefore ${a, b, c, d} is the roster form of this set.
b. {x$ \in $N: x2 < 25}
if we take square root both sides in this inequality x2 < 25, we get
$ \Rightarrow $$x = \sqrt {25} = 5$
Hence, the natural numbers less than 5 are 1, 2, 3 and 4.
$\therefore ${1 2, 3, 4} is the roster form of this set.
c. {x$ \in $ N: x is a prime number, 10 < x < 20}
All the prime numbers lying between 10 and 20 are 11, 13, 17 and 19.
$\therefore ${11, 13, 17, 19} is the roster form of this set.
d. {x$ \in $N: x = 2 n, n $ \in $N}
for x$ \in $N, x = 1, 2, 3, 4, 5, …
and n$ \in $N, n = 1, 2, 3, 4, 5, …
but here, x = 2n $ \Rightarrow $n = $\dfrac{x}{2}$. Hence, x should be a multiple of 2 in order that n$ \in $N as $\dfrac{1}{2},\dfrac{3}{2},\dfrac{5}{2},...$$ \notin $N
therefore, for x = 2, 4, 6, 8, …, n = 1, 2, 3, 4, …
$\therefore ${2, 4, 6, 8, …} is the roster form of this set.
e. {x$ \in $R: x > x}
There is no such real number which is greater than itself. Hence, this set will be a null set and denoted as: {} or $\phi $.
f. {x: x is a prime number which is a divisor of 60}
The numbers which are divisor of 60 can be listed as 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Amongst them, only 2, 3, and 5 are prime numbers.
$\therefore ${2, 3, 5} is the roster form of this set.
g. {x: x is a two-digit number such that the sum of its digit is 8}
The total 2 – digit numbers whose sum of its digit is 8 can be listed as 17, 26, 35, 44, 53, 62, 71, and 80.
$\therefore ${17, 26, 35, 44, 53, 62, 71, 80} is the roster form of this set.
h. The set of all letters in the word ‘trigonometry’.
All the letters in the word ‘TRIGONOMETRY’ can be listed as T, R, I, G, O, N, M, E, and Y.
$\therefore ${T, R, I, G, O, N, M, E, Y} is the roster form of this set.
i. The set of all letters in the word ‘Better’.
All the letters in the word better can be listed as B, E, T, and R.
$\therefore ${B, E, T, R} is the roster form of this set.
Note: In such questions, you must know which form is the roster and which is set – builder form of the set. You may go wrong while obtaining the elements of each set as you are provided with the set-builder form only. You may go wrong in the last two parts because, in them, there is no need to repeat the same word because no matter how many times it occurs, it will be treated as a single element of the set.
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