To find an irrational number \( n \) that satisfies the inequality \( 4.5 \leqslant n \leqslant 5.5 \), let's explore some common irrational numbers and their properties.
Irrational numbers are those that cannot be expressed as a fraction of two integers. Examples include numbers such as \(\sqrt{2}\), \(\pi\), and \(\text{e}\). These numbers have non-terminating and non-repeating decimal expansions.
A suitable irrational number for this problem can be found by considering the square roots of certain whole numbers. Let's consider \(\sqrt{5}\):
1. The square root function is known to produce irrational numbers when its argument is not a perfect square.
2. \(\sqrt{5}\) is an irrational number because 5 is not a perfect square.
Now, compute the decimal value of \(\sqrt{5}\):
[tex]\[ \sqrt{5} \approx 2.23606797749979 \][/tex]
However, \(\sqrt{5}\) does not fall within the interval \( 4.5 \leqslant n \leqslant 5.5 \).
To adjust our approach, observe the numbers between 4.5 and 5.5. Let's consider \(\sqrt{20}\). This is appropriate because:
1. \(\sqrt{20}\) simplifies to \( 2\sqrt{5} \), and since \(\sqrt{5}\) is irrational, multiplying an irrational number by an integer (that is, 2) results in another irrational number.
2. The simplified form is equal to \( 2 \times 2.23606797749979 \approx 4.47213595499958 \).
We need to find a value that lies strictly within the desired range. So, considering \(\sqrt{24}\), which simplifies to \( 2\sqrt{6} \):
1. \(\sqrt{24}\) lies between the numbers 4.5 and 5.5.
2. As \( 6 \) is not a perfect square, \( \sqrt{6} \approx 2.449 \)
3. Simplified:
[tex]\[ \sqrt{24} \approx 4.89898 \][/tex]
Thus, the number \(\sqrt{24}\) satisfies the inequality:
[tex]\[ 4.5 \leqslant \sqrt{24} \approx 4.89898 \leqslant 5.5 \][/tex]
Therefore, an irrational number [tex]\( n \)[/tex] satisfying [tex]\( 4.5 \leqslant n \leqslant 5.5 \)[/tex] can indeed be [tex]\(\sqrt{24}\)[/tex].
