
Which of the following is correct?
A) $\left\{ {x:{x^2} = - 1,x \in Z} \right\} = \phi $
B) $\phi = 0$
C) $\phi = \left\{ 0 \right\}$
D) $\phi = \left\{ \phi \right\}$
Answer
496.8k+ views
Hint: In this question we have to check which option is correct. We can see that we have to check which of them is an empty set. We know that a set that has no element is called an empty set. It is denoted by $\phi $ . So we will check each of the given options and see which of them is an empty set.
Complete step by step answer:
As we know, the set of real numbers is made by combining the set of rational numbers and irrational numbers. The real numbers include natural numbers or counting numbers, whole numbers, integers, rational and irrational numbers.
In the first option we have,
$\left\{ {x:{x^2} = - 1,x \in Z} \right\} = \phi $
Here we have
${x^2} = - 1$
By taking the square root to the both sides of the equation,
$x = \sqrt { - 1} $ .
We know this is the value of iota, i.e.
$i = \sqrt { - 1} $ .
But iota $(i)$ is not a real number, it is a complex number.
Hence we can say that option (a) is correct because this set is an empty set.
Now in the second option we have
$\phi = 0$
We know that $0$ is a whole number, so it cannot be an empty set.
Hence the second option is wrong.
In the third option we have
$\phi = \left\{ 0 \right\}$
In this expression we can see that this set contains $0$ as an element and from the above we can see that zero is not an empty set. It is a whole number.
So again this is not an empty set, hence this option is wrong.
Now we will see the fourth option:
$\phi = \left\{ \phi \right\}$
We should know that this is also not an empty set, because this is a set that contains a null set. It has an element i.e. $\phi $ .
Therefore this is also not an empty set.
Hence the correct answer is option (A) $\left\{ {x:{x^2} = - 1,x \in Z} \right\} = \phi $.
Note:
We know that every set is a subset of itself. Hence the empty set will also have one subset i.e. $\phi $ .
Therefore it can also be written as
$P(\phi ) = \phi $ .
We should not get mistaken that $\left\{ \phi \right\}$ is an empty set. It represents a set that contains an empty set as an element and hence it has a cardinality one.
Complete step by step answer:
As we know, the set of real numbers is made by combining the set of rational numbers and irrational numbers. The real numbers include natural numbers or counting numbers, whole numbers, integers, rational and irrational numbers.
In the first option we have,
$\left\{ {x:{x^2} = - 1,x \in Z} \right\} = \phi $
Here we have
${x^2} = - 1$
By taking the square root to the both sides of the equation,
$x = \sqrt { - 1} $ .
We know this is the value of iota, i.e.
$i = \sqrt { - 1} $ .
But iota $(i)$ is not a real number, it is a complex number.
Hence we can say that option (a) is correct because this set is an empty set.
Now in the second option we have
$\phi = 0$
We know that $0$ is a whole number, so it cannot be an empty set.
Hence the second option is wrong.
In the third option we have
$\phi = \left\{ 0 \right\}$
In this expression we can see that this set contains $0$ as an element and from the above we can see that zero is not an empty set. It is a whole number.
So again this is not an empty set, hence this option is wrong.
Now we will see the fourth option:
$\phi = \left\{ \phi \right\}$
We should know that this is also not an empty set, because this is a set that contains a null set. It has an element i.e. $\phi $ .
Therefore this is also not an empty set.
Hence the correct answer is option (A) $\left\{ {x:{x^2} = - 1,x \in Z} \right\} = \phi $.
Note:
We know that every set is a subset of itself. Hence the empty set will also have one subset i.e. $\phi $ .
Therefore it can also be written as
$P(\phi ) = \phi $ .
We should not get mistaken that $\left\{ \phi \right\}$ is an empty set. It represents a set that contains an empty set as an element and hence it has a cardinality one.
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