what rational number with the denominator of 7 is between √8 and √9

5 answers

To find a rational number with a denominator of 7 that is between \( \sqrt{8} \) and \( \sqrt{9} \), we first calculate the values of these square roots:

\[ \sqrt{8} \approx 2.828 \] \[ \sqrt{9} = 3 \]

Now we are looking for a rational number in the form \( \frac{n}{7} \) such that:

\[ 2.828 < \frac{n}{7} < 3 \]

To find the corresponding values for \( n \), we can multiply the inequalities by 7:

\[ 7 \times 2.828 < n < 7 \times 3 \]

Calculating these products:

\[ 19.796 < n < 21 \]

Since \( n \) must be an integer, the possible integer value for \( n \) is 20.

Now, we can check this value:

\[ \frac{20}{7} \approx 2.857 \]

Since \( 2.828 < 2.857 < 3 \), we find that \( \frac{20}{7} \) is indeed between \( \sqrt{8} \) and \( \sqrt{9} \).

Thus, the rational number with the denominator of 7 that is between \( \sqrt{8} \) and \( \sqrt{9} \) is:

\[ \frac{20}{7} \]

ok can you make it a bit more less of work

Sure! To find a rational number with a denominator of 7 that falls between \( \sqrt{8} \) and \( \sqrt{9} \):

  1. We know \( \sqrt{8} \approx 2.828 \) and \( \sqrt{9} = 3 \).
  2. We need \( \frac{n}{7} \) such that \( 2.828 < \frac{n}{7} < 3 \).
  3. Thus, we multiply by 7 to find \( n \):
    • Multiply \( 2.828 \) by 7: approximately \( 19.796 \).
    • Multiply \( 3 \) by 7: \( 21 \).
  4. The integer \( n \) must be \( 20 \) (since it's the only integer between \( 19.796 \) and \( 21 \)).

Therefore, the rational number is:

\[ \frac{20}{7} \]

This is your answer in a more concise format!

to make it even easier, consider that √9 = 3. so just subtract 1/7 from that, giving 2 6/7 = 20/7

You're absolutely right!

Starting with \( \sqrt{9} = 3 \), you can simply subtract \( \frac{1}{7} \) to find a rational number just below it:

\[ 3 - \frac{1}{7} = \frac{21}{7} - \frac{1}{7} = \frac{20}{7} \]

So, \( \frac{20}{7} \) is indeed a rational number between \( \sqrt{8} \) and \( \sqrt{9} \).

Thank you for pointing out the simpler method!