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Yes. The square root of a natural number (positive integer) is either an integer, or an irrational number.
Irrational.
This is an easy question : root 2, root 3, root 5. Any square root of an integer which is not integral itself is irrational.
Yes, the square root of a whole number is either an integer or an irrational number. If the whole number is a perfect square (like 0, 1, 4, 9, etc.), its square root is an integer. However, if the whole number is not a perfect square (like 2, 3, 5, etc.), its square root is an irrational number.
26 is the least integer whose square root is an irrational number between 5 and 7. This is apparent as the square root of the previous integer (25) is a rational number and since the division method for calculating the square root produces a decimal that continues infinitely without repetition.
The square root of 25 is 5, which is a whole number, an integer, a natural number and a rational number.
no.No. The square root of 5 is an irrational number. The two closest numbers with integer square roots are 4 (with a square root of 2) and 9 (with a square root of 3). Since there are no integers between 2 and 3 and 5 lies between 4 and 9, it's pretty evident that it can't have an integer square root.
The square of 5 is 25 ... an integer, rational.
They are +5 and -5, which are both rational.
Square root of 25 = 251/2 = 5, which is an integer. So the square root of 25 is integer.
You cannot. The square root of 5 is irrational.
The square root of 5 is an irrational number