Question

What is a rational number with a denominator of 7 that is between 7√ and 8√. Write your answer as an improper fraction

Answer 1

To find a rational number with a denominator of 7 that is between \( 7\sqrt{} \) and \( 8\sqrt{} \), we first need to clarify the limits given. Let's assume you meant \( 7\sqrt{2} \) and \( 8\sqrt{2} \).

1. Calculate \( 7\sqrt{2} \) and \( 8\sqrt{2} \):
\[
7\sqrt{2} \approx 7 \times 1.414 \approx 9.899
\]
\[
8\sqrt{2} \approx 8 \times 1.414 \approx 11.313
\]

So we need a rational number with a denominator of 7 that lies between approximately 9.899 and 11.313.

2. The two bounds in fraction form (keeping the denominator of 7) would be:
\[
x = \frac{a}{7}
\]
where:
\[
9.899 < \frac{a}{7} < 11.313
\]

3. Multiply the inequalities by 7 to eliminate the denominator:
\[
9.899 \times 7 < a < 11.313 \times 7
\]
Calculate:
\[
9.899 \times 7 \approx 69.293
\]
\[
11.313 \times 7 \approx 79.191
\]

4. Thus we have:
\[
69.293 < a < 79.191
\]

5. The integer values of \( a \) within this range are 70, 71, 72, ..., 79.

Hence, we can choose any of these integers for \( a \). For a simple example, let’s choose \( a = 72 \).

The corresponding fraction would then be:
\[
\frac{72}{7}
\]

This is an improper fraction representing a rational number with a denominator of 7 that lies between \( 7\sqrt{2} \) and \( 8\sqrt{2} \).