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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.
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
7 plus the square root of 5 is an irrational number because the square root of 5 is a never ending decimal number that can't be expressed as a fraction.