Let's analyze each of the given statements step by step to determine which one is true:
1. Statement 1: [tex]\(\sqrt{3}\)[/tex] is a rational number.
- A number is rational if it 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 square root of 3, [tex]\(\sqrt{3}\)[/tex], is an irrational number because it cannot be expressed as a fraction of two integers. Its decimal representation is non-terminating and non-repeating.
- [tex]\(\sqrt{3}\)[/tex] is not a rational number.
2. Statement 2: [tex]\(-6.1\overline{33}\)[/tex] is an irrational number.
- The notation [tex]\( -6.1\overline{33} \)[/tex] indicates that the decimal 33 repeats indefinitely.
- A number with a repeating decimal expansion is a rational number, as it can be expressed as a fraction.
- Therefore, [tex]\(-6.1\overline{33}\)[/tex] is a rational number, not an irrational number.
3. Statement 3: [tex]\(\sqrt{4}\)[/tex] is a rational number and an integer.
- The square root of 4 is [tex]\(\sqrt{4} = 2\)[/tex].
- 2 is an integer and it can be expressed as a fraction [tex]\(\frac{2}{1}\)[/tex].
- Hence, [tex]\(\sqrt{4}\)[/tex] is both a rational number and an integer.
4. Statement 4: 0 is neither a rational number nor an irrational number.
- 0 can be expressed as the fraction [tex]\(\frac{0}{1}\)[/tex], where 0 and 1 are integers and the denominator is not zero.
- Therefore, 0 is a rational number.
- Consequently, the statement that 0 is neither rational nor irrational is false.
After analyzing all the statements, we can see:
- Statement 1 is false because [tex]\(\sqrt{3}\)[/tex] is irrational.
- Statement 2 is false because [tex]\(-6.1\overline{33}\)[/tex] is rational.
- Statement 3 is true because [tex]\(\sqrt{4}\)[/tex] is 2, which is a rational number and an integer.
- Statement 4 is false because 0 is a rational number.
Thus, the true statement is:
Statement 3: [tex]\(\sqrt{4}\)[/tex] is a rational number and an integer.
So the answer is 3.
