Answer:
To prove that none of the terms of the sequence 2, 5, 8 are perfect squares, we need to show that no integer exists such that its square equals 2, 5, or 8.
First, let's consider 2. We can start by assuming that there exists an integer n such that n^2 = 2. However, this leads to a contradiction, because the square of any integer is always non-negative, so n^2 cannot be equal to a negative number like 2. Therefore, there is no integer n such that n^2 = 2, and 2 is not a perfect square.
Similarly, let's consider 5. Again, assume that there exists an integer n such that n^2 = 5. However, this leads to another contradiction, because the square of any integer is always either a perfect square or a number that is slightly larger than a perfect square (for example, 4^2 = 16 and 6^2 = 36). Since 5 is not a perfect square, it cannot be equal to the square of any integer, and there is no integer n such that n^2 = 5.
Finally, let's consider 8. We can again assume that there exists an integer n such that n^2 = 8. However, this leads to a contradiction, because the square of any even integer is always a multiple of 4 (for example, (2n)^2 = 4n^2). Since 8 is not a multiple of 4, it cannot be equal to the square of any integer, and there is no integer n such that n^2 = 8.
Therefore, we have shown that none of the terms of the sequence 2, 5, 8 are perfect squares.