Question
What is the least integer whose square root is an irrational number between 5 and 7?(1 point)
Responses
13
13
6
6
25
25
26
26
Responses
13
13
6
6
25
25
26
26
Answers
GPT-4o mini
To find the least integer whose square root is an irrational number between 5 and 7, we need to identify the range of squares of 5 and 7:
- The square of 5 is \( 5^2 = 25 \).
- The square of 7 is \( 7^2 = 49 \).
This means we are looking for irrational numbers \( x \) such that:
\[
25 < x < 49
\]
To find the least integer in this range, we start with the smallest integer greater than 25, which is 26.
Now we can check whether 26 is irrational. The square root of 26 (\( \sqrt{26} \)) is indeed an irrational number since it cannot be expressed as a simple fraction.
Let's confirm that \( \sqrt{26} \) lies between 5 and 7:
- \( \sqrt{25} = 5 \) and \( \sqrt{49} = 7 \), so:
\[
5 < \sqrt{26} < 7
\]
Thus, the square root of 26 is an irrational number between 5 and 7.
Therefore, the least integer whose square root is an irrational number between 5 and 7 is:
**26**
- The square of 5 is \( 5^2 = 25 \).
- The square of 7 is \( 7^2 = 49 \).
This means we are looking for irrational numbers \( x \) such that:
\[
25 < x < 49
\]
To find the least integer in this range, we start with the smallest integer greater than 25, which is 26.
Now we can check whether 26 is irrational. The square root of 26 (\( \sqrt{26} \)) is indeed an irrational number since it cannot be expressed as a simple fraction.
Let's confirm that \( \sqrt{26} \) lies between 5 and 7:
- \( \sqrt{25} = 5 \) and \( \sqrt{49} = 7 \), so:
\[
5 < \sqrt{26} < 7
\]
Thus, the square root of 26 is an irrational number between 5 and 7.
Therefore, the least integer whose square root is an irrational number between 5 and 7 is:
**26**