
Which of the following is not an integer?
$\left( a \right) 0$
$\left( b \right) -1$
$\left( c \right) -1.5$
$\left( d \right) 1$
Answer
445.8k+ views
Hint: An integer is a number that can be written without a fractional component. The set of integers include the negative numbers, zero and the positive numbers. The set of integers is a subset of the set of rational numbers.
Complete step-by-step solution:
We are asked to find the number that is not an integer among the given numbers.
Before finding out the non-integer number from these options, we should know what an integer is.
An integer is a number that cannot be written as a fraction. It can be zero. It can be any natural number as well as the additive inverses of the natural numbers.
So, the integers are $...,-2,-1,0,1,2,...$
Also, we can understand that the numbers with decimal points do not belong to the set of integers.
That is, the numbers $2.5,1.33,-0.5,...$ do not belong to the set of integers.
The fractions like $\dfrac{3}{2},\dfrac{1}{2},-\dfrac{5}{2},...$ are not integers.
But the numbers $\dfrac{4}{2}=2,\dfrac{6}{3}=2,\dfrac{4}{1}=4,...$ are integers.
Also, the numbers $\sqrt{2},\sqrt{3},\sqrt{5},...$ do not belong to the set of integers.
But we know that $\sqrt{4}=2,\sqrt{9}=3,...$
Therefore, these numbers are integers.
Let us consider the given options.
Option $\left( a \right)$ is the number $0.$
So far, we have seen that zero belongs to the set of integers.
Thus, $0$ is an integer.
Option $\left( b \right)$ is the number $-1.$
Since $1$ is a natural number, $-1$ is an integer.
Option $\left( c \right)$ is the number $-1.5.$ This is a decimal number.
We have said that the decimal numbers do not belong to the set of integers.
Therefore, $-1.5$ is a non-integer.
Now, the option $\left( d \right)$ is the natural number $1$ and so it is an integer.
Hence, among these numbers, option $\left( c \right) -1.5$ is the number that is not an integer.
Note: The set of integers is denoted by $\mathbb{Z}.$ The set of natural numbers $\mathbb{N}$ is a proper subset of the set of integers $\mathbb{Z}.$ The set of integers $\mathbb{Z}$ is a subset of the set of rational numbers $\mathbb{Q},$ the set of real numbers $\mathbb{R}$ and thus the set of complex numbers $\mathbb{C}.$
Complete step-by-step solution:
We are asked to find the number that is not an integer among the given numbers.
Before finding out the non-integer number from these options, we should know what an integer is.
An integer is a number that cannot be written as a fraction. It can be zero. It can be any natural number as well as the additive inverses of the natural numbers.
So, the integers are $...,-2,-1,0,1,2,...$
Also, we can understand that the numbers with decimal points do not belong to the set of integers.
That is, the numbers $2.5,1.33,-0.5,...$ do not belong to the set of integers.
The fractions like $\dfrac{3}{2},\dfrac{1}{2},-\dfrac{5}{2},...$ are not integers.
But the numbers $\dfrac{4}{2}=2,\dfrac{6}{3}=2,\dfrac{4}{1}=4,...$ are integers.
Also, the numbers $\sqrt{2},\sqrt{3},\sqrt{5},...$ do not belong to the set of integers.
But we know that $\sqrt{4}=2,\sqrt{9}=3,...$
Therefore, these numbers are integers.
Let us consider the given options.
Option $\left( a \right)$ is the number $0.$
So far, we have seen that zero belongs to the set of integers.
Thus, $0$ is an integer.
Option $\left( b \right)$ is the number $-1.$
Since $1$ is a natural number, $-1$ is an integer.
Option $\left( c \right)$ is the number $-1.5.$ This is a decimal number.
We have said that the decimal numbers do not belong to the set of integers.
Therefore, $-1.5$ is a non-integer.
Now, the option $\left( d \right)$ is the natural number $1$ and so it is an integer.
Hence, among these numbers, option $\left( c \right) -1.5$ is the number that is not an integer.
Note: The set of integers is denoted by $\mathbb{Z}.$ The set of natural numbers $\mathbb{N}$ is a proper subset of the set of integers $\mathbb{Z}.$ The set of integers $\mathbb{Z}$ is a subset of the set of rational numbers $\mathbb{Q},$ the set of real numbers $\mathbb{R}$ and thus the set of complex numbers $\mathbb{C}.$
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