Question
What is a rational number with a denominator of 7 that is between 7–√ and 8–√. Write your answer as an improper fraction.
Answers
Answered by
oobleck
AAAaannndd the bot gets it wrong yet again!
√7 ≈ 2.645 and 4/7 < 2.645 < 5/7
√8 ≈ 2.828 and 5/7 < 2.828 < 6/7
so √7 < 19/7 < √8
√7 ≈ 2.645 and 4/7 < 2.645 < 5/7
√8 ≈ 2.828 and 5/7 < 2.828 < 6/7
so √7 < 19/7 < √8
Answered by
wawa
Describe the relationship between rational numbers and whole numbers. Are all rational numbers also whole numbers? Are all whole numbers also rational numbers? Give examples to support your answer.(4 points)
Answered by
GPT 3.5
To find a rational number between \(7-\sqrt{2}\) and \(8-\sqrt{2}\) with a denominator of 7, we can first express \(7-\sqrt{2}\) and \(8-\sqrt{2}\) with a common denominator of 7.
\[
7-\sqrt{2} = \frac{7}{1} - \frac{\sqrt{2}}{1} = \frac{7}{1} - \frac{\sqrt{2}}{1} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{7}{1} - \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{1} = \frac{7}{1} - \frac{\sqrt{4}}{1} = \frac{7}{1} - \frac{2}{1} = \frac{5}{1}
\]
\[
8-\sqrt{2} = \frac{8}{1} - \frac{\sqrt{2}}{1} = \frac{8}{1} - \frac{\sqrt{2}}{1} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{8}{1} - \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{1} = \frac{8}{1} - \frac{\sqrt{4}}{1} = \frac{8}{1} - \frac{2}{1} = \frac{6}{1}
\]
So, we need to find a rational number between \(\frac{5}{1}\) and \(\frac{6}{1}\) with a denominator of 7. To find such a number, we can take the average of \(\frac{5}{1}\) and \(\frac{6}{1}\) which gives us \(\frac{5+6}{2} = \frac{11}{2}\).
To write this as an improper fraction, we divide the numerator 11 by the denominator 2, which gives us:
\[
\frac{11}{2} = \boxed{ \frac{11}{2}}
\]
\[
7-\sqrt{2} = \frac{7}{1} - \frac{\sqrt{2}}{1} = \frac{7}{1} - \frac{\sqrt{2}}{1} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{7}{1} - \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{1} = \frac{7}{1} - \frac{\sqrt{4}}{1} = \frac{7}{1} - \frac{2}{1} = \frac{5}{1}
\]
\[
8-\sqrt{2} = \frac{8}{1} - \frac{\sqrt{2}}{1} = \frac{8}{1} - \frac{\sqrt{2}}{1} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{8}{1} - \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{1} = \frac{8}{1} - \frac{\sqrt{4}}{1} = \frac{8}{1} - \frac{2}{1} = \frac{6}{1}
\]
So, we need to find a rational number between \(\frac{5}{1}\) and \(\frac{6}{1}\) with a denominator of 7. To find such a number, we can take the average of \(\frac{5}{1}\) and \(\frac{6}{1}\) which gives us \(\frac{5+6}{2} = \frac{11}{2}\).
To write this as an improper fraction, we divide the numerator 11 by the denominator 2, which gives us:
\[
\frac{11}{2} = \boxed{ \frac{11}{2}}
\]
Answered by
GPT 3.5
Apologies for the previous incorrect response. You are absolutely right.
To find a rational number between \(7 - \sqrt{2}\) and \(8 - \sqrt{2}\) with a denominator of 7, we can compare these numbers to the rational numbers with a denominator of 7:
\(\sqrt{7} \approx 2.645\) and \(\sqrt{8} \approx 2.828\).
We know that \(\frac{4}{7} < 2.645 < \frac{5}{7}\) and \(\frac{5}{7} < 2.828 < \frac{6}{7}\).
Therefore, a rational number with a denominator of 7 that lies between \(7 - \sqrt{2}\) and \(8 - \sqrt{2}\) is \(\frac{19}{7}\).
So, the answer is \(\frac{19}{7}\).
To find a rational number between \(7 - \sqrt{2}\) and \(8 - \sqrt{2}\) with a denominator of 7, we can compare these numbers to the rational numbers with a denominator of 7:
\(\sqrt{7} \approx 2.645\) and \(\sqrt{8} \approx 2.828\).
We know that \(\frac{4}{7} < 2.645 < \frac{5}{7}\) and \(\frac{5}{7} < 2.828 < \frac{6}{7}\).
Therefore, a rational number with a denominator of 7 that lies between \(7 - \sqrt{2}\) and \(8 - \sqrt{2}\) is \(\frac{19}{7}\).
So, the answer is \(\frac{19}{7}\).