Answer
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Hint: Initially we check the given conditions given in the options. And then we need to see if there are any elements in the set which satisfy the given conditions. If there are no elements in a set satisfying those given conditions then the set is empty or a null set.
Complete step by step solution: First, we have a set of odd natural numbers divisible by 2.
As no odd number is divisible by 2.
We have, A set of odd natural numbers divisible by 2 is a null set.
The second option is a set of even prime numbers
As, we all know that 2 is a prime number, and 2 is also an even number.
Hence, A set of even prime numbers is not a null set because 2 is an even prime number.
Third option is { x : x is a natural numbers, \[x < 5\;\] and \[x > 7\]}
As, the natural numbers less than 5 are 1, 2, 3, 4 and the natural number greater than 7 are 8, 9, 10, 11,…
So, as we can clearly see that there is no common nature which is less than 5 and greater than 7.
So, {x: x is a natural number, \[x < 5\;\] and \[x > 7\]} is a null set.
The fourth option is {y: y is a point common to any two parallel lines}
As parallel lines do not intersect. Hence, they have no common point.
Hence, {y: y is a point common to any two parallel lines} is a null set.
Hence, option (i), (iii), (iv) are correct.
Note: Existence of null set can also be seen by axiomatic set theories. Axioms are the theories in mathematics which are self evidently true without having any proof. One such axiom is an axiom of choice. Using this, the null set existence can be proved.
Complete step by step solution: First, we have a set of odd natural numbers divisible by 2.
As no odd number is divisible by 2.
We have, A set of odd natural numbers divisible by 2 is a null set.
The second option is a set of even prime numbers
As, we all know that 2 is a prime number, and 2 is also an even number.
Hence, A set of even prime numbers is not a null set because 2 is an even prime number.
Third option is { x : x is a natural numbers, \[x < 5\;\] and \[x > 7\]}
As, the natural numbers less than 5 are 1, 2, 3, 4 and the natural number greater than 7 are 8, 9, 10, 11,…
So, as we can clearly see that there is no common nature which is less than 5 and greater than 7.
So, {x: x is a natural number, \[x < 5\;\] and \[x > 7\]} is a null set.
The fourth option is {y: y is a point common to any two parallel lines}
As parallel lines do not intersect. Hence, they have no common point.
Hence, {y: y is a point common to any two parallel lines} is a null set.
Hence, option (i), (iii), (iv) are correct.
Note: Existence of null set can also be seen by axiomatic set theories. Axioms are the theories in mathematics which are self evidently true without having any proof. One such axiom is an axiom of choice. Using this, the null set existence can be proved.
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Which of the following are examples of the null set?
(i) set of odd natural numbers divisible by 2.
(ii) set of even prime numbers
(iii) { x: x is a natural numbers, \[x < 5\;\] and \[x > 7\]}
(iv) { y: y is a point common to any two parallel lines}
(i) set of odd natural numbers divisible by 2.
(ii) set of even prime numbers
(iii) { x: x is a natural numbers, \[x < 5\;\] and \[x > 7\]}
(iv) { y: y is a point common to any two parallel lines}
![icon](https://i.ytimg.com/vi_webp/3CV7rFhN30M/hqdefault.webp)
Class 11 MATHS NCERT EXERCISE 1.2 (Question - 1) | Sets Class 11 Chapter 1 | NCERT | Ratan Kalra Sir
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