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 between $7\sqrt{7}$ and $8\sqrt{7}$, we can find a common denominator and create two fractions with that denominator. We can then find a rational number between these two fractions by finding a fraction $\frac{a}{b}$ such that $7\sqrt{7} < \frac{a}{b} < 8\sqrt{7}$. To do this, we can look for a fraction $\frac{a}{b}$ such that $49 \cdot 7 < a^2 < 64 \cdot 7$. This inequality is true if and only if $\sqrt{49 \cdot 7} < a < \sqrt{64 \cdot 7}$. Simplifying these square roots gives $7\sqrt{7} < a < 8\sqrt{7}$. We have found a fraction between $7\sqrt{7}$ and $8\sqrt{7}$: $7\sqrt{7} < \sqrt{49 \cdot 7}< \frac{8}{1}\sqrt{7}$. Since $\sqrt{343} = 7\sqrt{7}$, we can rewrite this as $343 < \frac{a^2}{b^2} < 448$. To find a fraction $\frac{a}{b}$ in lowest terms such that $343 < \frac{a^2}{b^2} < 448$, we can start with a fraction very close to $\sqrt{343 \cdot 7} = 7\sqrt{49} = 49$. Indeed, we can use $\frac{50}{1}$. Converting to lowest terms, this gives $\boxed{\frac{50}{1}}$.