What is the least Integra whose square root is an irrational number between 5 and 7
The least integer whose square root is an irrational number between 5 and 7 can be found by trial and error. We can start by finding the square roots of integers between 5 and 6, and checking if they are irrational by determining if they are non-repeating and non-terminating decimals:
√25 = 5 (rational)
√26 ≈ 5.1 (irrational)
√27 ≈ 5.2 (irrational)
√28 ≈ 5.3 (irrational)
√29 ≈ 5.4 (irrational)
√30 ≈ 5.5 (irrational)
√31 ≈ 5.6 (irrational)
√32 ≈ 5.7 (irrational)
√33 ≈ 5.7 (irrational)
Therefore, the least integer whose square root is an irrational number between 5 and 7 is 27.
Show an example of a rational number that is between 10 and 11
An example of a rational number between 10 and 11 can be found by taking the average of the two numbers.
Average of 10 and 11: (10 + 11) / 2 = 21 / 2 = 10.5
Therefore, 10.5 is a rational number between 10 and 11.
To find the least integer whose square root is an irrational number between 5 and 7, we need to look for the smallest perfect square that is larger than 25 (5^2) but smaller than 49 (7^2).
Starting from 5^2 = 25, we can see that the next perfect square is 6^2 = 36, which is larger than 25. However, the square root of 36 is not an irrational number between 5 and 7.
Moving on, the next perfect square is 7^2 = 49. The square root of 49 is 7, which is equal to the upper limit of our search. Therefore, 49 does not meet the criteria.
The next perfect square would be 8^2 = 64, which is larger than 49 and outside our desired range.
Continuing in this manner, we can see that the smallest perfect square that satisfies the given conditions is 6^2 = 36. The square root of 36 is 6, which is an integer, so it does not satisfy the requirement of an irrational number.
Thus, there is no integer whose square root is an irrational number between 5 and 7.