1, 2, 3, 4, and their negatives.
Of course not. The square root of 2 is less than 3, and (pi) is less than 4 .
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.
Yes. All negative integers are also less than 0.
The set of positive odd integers.
This is the 'null' or 'empty' set.There are no numbers greater than '-3' and less than '-9'.
The positive integers whose squares are less than 30 are 1, 2, 3, 4, and 5. This is because the squares of these numbers are 1, 4, 9, 16, and 25, respectively, all of which are less than 30. The next integer, 6, has a square of 36, which exceeds 30.
The product of four consecutive integers is always one less than a perfect square. The product of four consecutive integers starting with n will be one less than the square of n2 + 3n + 1
Of course not. The square root of 2 is less than 3, and (pi) is less than 4 .
Try squaring different integers (hint: in this case, the integers will be fairly small). If you find that the square of one integer is less than 27, and the square of the next integer is more than 27, you have your answer.
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.
22
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.
There are 20.
All but the square numbers - 53 of them.
16
Integers which are the squares of integers are called perfect squares or square numbers. Perfect squares less than 101 are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100.
To have a whole number square root, the number is a perfect square. Thus the numbers will be the squares of multiples of 11. Thus the first number will be (1×11) × (1×11) = 11² = 121 The next candidate will be (2×11) × (2×11) = 22² = 484 The next possible candidate will be (3×11) × (3×11) = 33² = 1089 which is too large. Thus there are two multiples of 11 less than 1000 whose square roots are whole numbers, namely 121 (11²) and 484 (22²)