Since 7^2 is 49 and 8^2 is 64,
And the numbers must have a square root between 7 and 8,
The numbers have to be in between 49 and 64.
This is because if the numbers are under 49,
Its square root will be below 7.
And if the numbers are above 64,
Its square root will be above 8.
Of course they can. Every integer greater than zero is a square root.
greater than
The square root of 7 is approximately 2.645. The two consecutive integers that it lies between are 2 and 3, since (2^2 = 4) and (3^2 = 9). Therefore, (\sqrt{7}) is greater than 2 but less than 3.
√3 = 1.732050808 Therefore, 0.85 is less than the square root of 3.
The square root of 31 is an irrational number and so there are no integers for it.
greater than
Five even integers. Square root of 37 is a little more than 6 (but less than 7). Square root of 257 is a little more than 16 (less than 17). So the even integers are: 8, 10, 12, 14, 16. There are five of them.
The square root of a negative number cannot be expressed with integers. The square root of a negative number is always imaginary.
6 9
No there is not. If you are looking for prime factors of a number and you get to the square root of that number you can stop. Yes, there is. If an integer is not itself a prime, then one of its factors will be less or equal to its square root and the "co-factor" will be greater than or equal to the square root. But both cannot be greater than the square root so, when searching for factors, you can stop when you reach the square root.
Try it out! Take the square of several (in this case small) integers, until you find one square that is less than 15, and one that is more.
The square root of 148 is not an integer.