To determine which number is irrational, an integer, and a real number, let's analyze each option in detail:
1. Irrational Numbers:
- An irrational number cannot be expressed as a fraction of two integers (it cannot be written in the form [tex]\(\frac{a}{b}\)[/tex] where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers and [tex]\(b \neq 0\)[/tex]).
- Irrational numbers have non-repeating, non-terminating decimal expansions.
2. Integer:
- An integer is any whole number, which can be positive, negative, or zero.
3. Real Number:
- A real number includes all rational and irrational numbers. All integers are real numbers as well.
Now let's evaluate each given option:
Option A: There is no such number.
- This statement suggests that there cannot be a number that is simultaneously irrational, an integer, and a real number.
Option B: [tex]\(\frac{4}{3}\)[/tex]
- [tex]\(\frac{4}{3}\)[/tex] is a rational number because it can be expressed as a fraction of two integers. Therefore, it is not irrational.
Option C: 0
- 0 is an integer and a real number, but it is not an irrational number. It can be exactly expressed as a fraction ([tex]\(\frac{0}{1}\)[/tex]).
Option D: [tex]\(\sqrt{4}\)[/tex]
- [tex]\(\sqrt{4} = 2\)[/tex], which is an integer and a real number, but not irrational, as it terminates and can be expressed as [tex]\(\frac{4}{2}\)[/tex].
Upon analyzing all the options, we can see that:
- A number cannot be both an integer and irrational simultaneously because integers are by definition rational.
Therefore, the correct answer is:
A. There is no such number.
