To find the least integer whose square root is an irrational number between 5 and 7, we need to find the smallest perfect square that is greater than 7 squared.
The square of 7 is 49, and the square of 8 is 64. Since 7 squared is 49 and 8 squared is 64, any number between 49 and 64 will have an irrational square root between 5 and 7.
Therefore, the least integer whose square root is an irrational number between 5 and 7 is 50.
What is the least integer whose square root is an irrational number between 5 and 7?
5 answers
What is the least integer whose square root is an irrational number between 5 and 7?
To find the least integer whose square root is an irrational number between 5 and 7, we need to consider the square of the smaller number (5) and the square of the larger number (7).
The square of 5 is 25, and the square of 6 is 36. Since 5 squared is 25 and 6 squared is 36, any number between 25 and 36 will have an irrational square root between 5 and 6.
Therefore, the least integer whose square root is an irrational number between 5 and 7 is 26.
The square of 5 is 25, and the square of 6 is 36. Since 5 squared is 25 and 6 squared is 36, any number between 25 and 36 will have an irrational square root between 5 and 6.
Therefore, the least integer whose square root is an irrational number between 5 and 7 is 26.
Show an example of a rational number that is between 10−−√ and 11−−√
To find a rational number between $\sqrt{10}$ and $\sqrt{11}$, we can take the average of these two numbers.
\begin{align*}
\frac{\sqrt{10} + \sqrt{11}}{2} &= \frac{\sqrt{10} + \sqrt{11}}{2} \cdot \frac{\sqrt{10} - \sqrt{11}}{\sqrt{10} - \sqrt{11}} \\
&= \frac{(\sqrt{10})^2 - (\sqrt{11})^2}{2(\sqrt{10} - \sqrt{11})} \\
&= \frac{10 - 11}{2(\sqrt{10} - \sqrt{11})} \\
&= \frac{-1}{2(\sqrt{10} - \sqrt{11})} \\
&= \frac{-1}{2\sqrt{10} - 2\sqrt{11}}
\end{align*}
So, an example of a rational number between $\sqrt{10}$ and $\sqrt{11}$ is $\frac{-1}{2\sqrt{10} - 2\sqrt{11}}$.
\begin{align*}
\frac{\sqrt{10} + \sqrt{11}}{2} &= \frac{\sqrt{10} + \sqrt{11}}{2} \cdot \frac{\sqrt{10} - \sqrt{11}}{\sqrt{10} - \sqrt{11}} \\
&= \frac{(\sqrt{10})^2 - (\sqrt{11})^2}{2(\sqrt{10} - \sqrt{11})} \\
&= \frac{10 - 11}{2(\sqrt{10} - \sqrt{11})} \\
&= \frac{-1}{2(\sqrt{10} - \sqrt{11})} \\
&= \frac{-1}{2\sqrt{10} - 2\sqrt{11}}
\end{align*}
So, an example of a rational number between $\sqrt{10}$ and $\sqrt{11}$ is $\frac{-1}{2\sqrt{10} - 2\sqrt{11}}$.
1 answer