it depends on the number, for example the square root of nine is a natural number. it's 3. the square root of 3 is not a natural number. It is 1.73... and it goes on and on.
When it can be square rooted
No, the square root of 21 is not a natural number. Natural numbers are positive integers (1, 2, 3, ...), and the square root of 21 is approximately 4.58, which is not an integer. Therefore, it does not qualify as a natural number.
It is an odd number It is a prime number It is a rational number It is an irrational number when square rooted It is V as a Roman numeral
√225 = 15 so it is a natural number.
28 does not have a square root.
When it can be square rooted
No, it is not.
No, it is not.
yes they can if i know what you are asking
- square rooted by 6
infinity is not considered a number it is considered a theory therefore it can not be square rooted
No because it can't be squared rooted evenly
No - a natural number is a whole number. Therefore, the square root of 49 is a natural number, but the square root of 50 is not.
No, it is not even a real number. It is an imaginary number.
The square of number returns to its original value when square rooted as for example 3 squared is 9 and its square root is 3
Natural numbers are those numbers used for counting. The square root of 14 is the irrational number 3.74165... . Therefore, the square root of 14 is not a natural number.
Not in the case of Rational Numbers -- only in Imaginary Numbers, typically referring to the imaginary component.