Which groups do the following numbers belong to?
\dfrac{\sqrt{3}}{2}
It is clear that \dfrac{\sqrt{3}}{2} is not an integer, so we need to verify if it is rational or irrational.
Notice that the square root of a non-perfect integer is always irrational. In this question, since 3 is not a perfect square, we have:
\sqrt{3} \in \mathbb{R} \setminus \mathbb{Q}
On the other hand, a non-zero rational number times an irrational number will always be irrational. Therefore:
\dfrac{\sqrt{3}}{2} = \dfrac{1}{2} \times \sqrt{3}\in \mathbb{R} \setminus \mathbb{Q}
\dfrac{\sqrt{3}}{2} \in \mathbb{R} \setminus \mathbb{Q}
-2
The set of all integers is denoted by \mathbb{Z}. Since -2 is an integer, we have -2 \in \mathbb{Z}.
-2 \in \mathbb{Z}
\dfrac{{2}}{3}
A rational number is of the form \dfrac{a}{b}, where a and b are integers. The set of all rational numbers is denoted by \mathbb{Q}. Therefore, \dfrac{2}{3} \in \mathbb{Q}.
\dfrac{2}{3} \in \mathbb{Q}
5
The set of all positive integers is denoted by \mathbb{N}. Since 5 is a positive integer, we have 5 \in \mathbb{N}.
5 \in \mathbb{N}
0
The set of all integers is denoted by \mathbb{Z}. Since 0 is an integer, we have 0 \in \mathbb{Z}. Because the set of all nonnegative integers is denoted by \mathbb{N}, we also have 0\in \mathbb{N}
0 \in \mathbb{Z} and 0 \in \mathbb{N}
2.\overline{3}
We know that a number is rational if and only if its decimal representation is repeating or terminating. Therefore, 2.\overline{3} \in \mathbb{Q}.
2.\overline{3} \in \mathbb{Q} .
\pi
We know that a number is irrational if its decimal representation is neither repeating nor terminating. Since the decimal representation is neither repeating nor terminating, we can conclude that it is an irrational number.
\pi \in \mathbb{R} \setminus\mathbb{Q} .