To determine which set of numbers the real number [tex]\(\sqrt{23}\)[/tex] belongs to, let's explore the properties of [tex]\(\sqrt{23}\)[/tex]:
1. Integers: Integers are whole numbers including negative numbers, zero, and positive numbers. Examples include [tex]\(-3, 0, 4\)[/tex]. Since [tex]\(23\)[/tex] is not a perfect square (there is no integer [tex]\(n\)[/tex] such that [tex]\(n^2 = 23\)[/tex]), [tex]\(\sqrt{23}\)[/tex] is not an integer.
2. Whole Numbers: Whole numbers start from 0 and include all positive integers (i.e., [tex]\(0, 1, 2, 3, \ldots\)[/tex]). Since [tex]\(\sqrt{23}\)[/tex] is not an integer, it cannot be a whole number either.
3. Rational Numbers: Rational numbers are numbers that can be expressed as a fraction [tex]\(\frac{a}{b}\)[/tex] where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers, and [tex]\(b \neq 0\)[/tex]. The decimal expansion of rational numbers either terminates or repeats periodically. However, [tex]\(\sqrt{23}\)[/tex] cannot be expressed as a ratio of two integers because its decimal representation is non-terminating and non-repeating. This means [tex]\(\sqrt{23}\)[/tex] is not a rational number.
4. Irrational Numbers: Irrational numbers are numbers that cannot be written as a simple fraction, and their decimal expansions are non-terminating and non-repeating. Since [tex]\(\sqrt{23}\)[/tex] cannot be expressed as a ratio of two integers and its decimal expansion is non-terminating and non-repeating, it is classified as an irrational number.
Therefore, [tex]\(\sqrt{23}\)[/tex] belongs to the set of:
Irrational numbers
The correct choice is:
irrational numbers
