Question

What is a rational number with a denominator of 7 that is between 7–√ and 8–√ .

Answer 1

To find a rational number with a denominator of 7 that is between \(7 - \sqrt{7}\) and \(8 - \sqrt{8}\), we first need to approximate the values of \(7 - \sqrt{7}\) and \(8 - \sqrt{8}\).

1. Calculate \( \sqrt{7} \):
\[
\sqrt{7} \approx 2.64575
\]
Therefore,
\[
7 - \sqrt{7} \approx 7 - 2.64575 \approx 4.35425.
\]

2. Calculate \( \sqrt{8} \):
\[
\sqrt{8} = 2\sqrt{2} \approx 2.82843
\]
Thus,
\[
8 - \sqrt{8} \approx 8 - 2.82843 \approx 5.17157.
\]

Now we need a rational number with a denominator of 7 that lies between approximately \(4.35425\) and \(5.17157\). A rational number can be expressed in the form \(\frac{n}{7}\), where \(n\) is an integer.

To find \(n\):
1. Multiply the lower bound by 7:
\[
4.35425 \times 7 \approx 30.48175
\]
So, \(n\) (for the lower bound) should be greater than approximately 30.48175, which means \(n \geq 31\).

2. Multiply the upper bound by 7:
\[
5.17157 \times 7 \approx 36.201
\]
Thus, \(n\) (for the upper bound) should be less than approximately 36.201, meaning \(n \leq 36\).

Therefore, \(n\) can take on values of \(31, 32, 33, 34, 35, 36\):

Checking these values:
- For \(n = 31\):
\[
\frac{31}{7} \approx 4.42857 \quad (\text{in between})
\]
- For \(n = 32\):
\[
\frac{32}{7} \approx 4.57143 \quad (\text{in between})
\]
- For \(n = 33\):
\[
\frac{33}{7} \approx 4.71429 \quad (\text{in between})
\]
- For \(n = 34\):
\[
\frac{34}{7} \approx 4.85714 \quad (\text{in between})
\]
- For \(n = 35\):
\[
\frac{35}{7} \approx 5 \quad (\text{in between})
\]
- For \(n = 36\):
\[
\frac{36}{7} \approx 5.14286 \quad (\text{in between})
\]

All the rational numbers \( \frac{31}{7}, \frac{32}{7}, \frac{33}{7}, \frac{34}{7}, \frac{35}{7}, \text{ and } \frac{36}{7} \) fall within the range.

Thus, any of these could be your answer, but a common choice is:

\[
\frac{34}{7}
\]

This is a rational number with a denominator of 7 that lies between \(7 - \sqrt{7}\) and \(8 - \sqrt{8}\).