What is the least integer whose square root is an irrational number between 5 and 7?

A. 26
B. 13
C. 25
D. 6

1 answer

To find the least integer whose square root is an irrational number between 5 and 7, we first identify the range of numbers we want to consider:

  1. The square root of a number \( n \) is irrational between the values \( 5 \) and \( 7 \).
  2. Squaring the bounds gives us:
    • Lower limit: \( 5^2 = 25 \)
    • Upper limit: \( 7^2 = 49 \)

Thus, we are looking for integers \( n \) in the range \( 25 < n < 49 \).

Next, we need \( n \) to be an integer that is not a perfect square because the square root must be irrational. The perfect squares between 25 and 49 are:

  • \( 25 \) (since \( 5^2 = 25 \))
  • \( 36 \) (since \( 6^2 = 36 \))
  • \( 49 \) (since \( 7^2 = 49 \))

Now, we identify the integers in the interval \( (25, 49) \) that are not perfect squares. The integers between 25 and 49 are: 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, and 48.

Next, we can see which of these integers is the least integer that satisfies the requirement:

  • 26 is not a perfect square.

To confirm, we take the square root of 26: \[ \sqrt{26} \approx 5.099 \] This value is indeed an irrational number between 5 and 7.

Thus, the answer is: \[ \boxed{26} \]

Similar Questions
  1. Name ALL sets of numbers to which each number belongsReal Rational Irrational Integer Whole Natural 0 Whole Integer Square root
    1. answers icon 1 answer
  2. Products of Irrational Square Roots Quick Check1 of 51 of 5 Items Question Rewrite the irrational square root √48 as the
    1. answers icon 1 answer
  3. Products of Irrational Square Roots Quick Check3 of 53 of 5 Items Question Rewrite the irrational cube root ^3√48 as a product
    1. answers icon 1 answer
    1. answers icon 1 answer
more similar questions