Yes. For example, the square root of 1 is 1, and the square root of 4 is 2.
No. The square root of an integer is either an integer, or an irrational number.
The square root of an integer is a CYCLOTOMIC integer.
Actually, as a non-integer square root of a number, 82 is an irrational number. It has no end.
no, the square root of an integer will not always be another integer. take the integer 27, for example. the square root of 27 is about 5.1961, which is not an integer.No eg square root of 17 is 4.1231056...
Square root of 25 = 251/2 = 5, which is an integer. So the square root of 25 is integer.
No, the square root of 24 is NOT an integer. An integer is a whole number, and the square root of 24 is 4.89897948557.
Yes, a perfect square is a number that has an integer square root.
No it is not.
-- If the square root is an integer, then add ' 1 ' to it. -- If the square root is not an integer, then there isn't any.
The square of that integer.
No. The square root of an integer is either an integer, or an irrational number.
No. The square root of an integer is always either an integer or an irrational number.
No, the square root of an integer will not always be an integer. It may or may not be. The square root of nine is three, both of which are integers. The square root of two is not an integer. In fact, it is not even a rational number.
The square root of 48 is not an integer which means a whole number.
The square root of 27 is an irrational number
The square root of 61 is an irrational number
A number that has an integer square root is a perfect square, which is an integer that can be expressed as the square of another integer. For example, 16 is a perfect square because its square root is 4, which is an integer. Other examples include 1 (√1 = 1), 4 (√4 = 2), and 25 (√25 = 5). In general, any number of the form ( n^2 ), where ( n ) is an integer, will have an integer square root.