There is no terminating decimal that, when squared, equals 2. It can also be proved that there is no repeating decimal that, when squared, equals 2. Therefore, The square root of 2 is
a.) a rational number
b.) no a rational number
The square root of 2 is not a rational number, so the correct answer is option b.) "not a rational number."
To understand why, let's go through a proof by contradiction. Suppose that the square root of 2 is rational, which means it can be expressed as a fraction in the form a/b, where a and b are integers and b is not equal to zero. Let's assume that a/b is in its simplest form, meaning that a and b have no common factors.
If (a/b)^2 = 2, then we have (a^2)/(b^2) = 2. Rearranging the equation, we get a^2 = 2 * b^2. This implies that a^2 is an even number because it is twice the value of b^2.
Now, let's examine the possible cases for a:
1) If a is even, then a can be expressed as a = 2k, where k is an integer. Substituting this into the equation, we get (2k)^2 = 2 * b^2, which simplifies to 4k^2 = 2 * b^2. Dividing both sides by 2, we have 2k^2 = b^2. This means that b^2 is also an even number, making b an even number since the square of an odd number is odd. So, both a and b are even, contradicting our assumption that a/b is in its simplest form.
2) If a is odd, then a can be expressed as a = 2k + 1, where k is an integer. Substituting this into the equation, we get (2k + 1)^2 = 2 * b^2. Expanding and simplifying, we have 4k^2 + 4k + 1 = 2 * b^2. Rearranging, we get 2(2k^2 + 2k) + 1 = 2 * b^2. This implies that 2k^2 + 2k = (2 * b^2 - 1)/2. The right side of the equation is an odd number divided by 2, which is not an integer. Therefore, the left side, 2k^2 + 2k, must also be not an integer. But since k is an integer, k^2 and k are integers, and their sum is an integer. This is a contradiction.
In both cases, we have reached a contradiction, which means our initial assumption that the square root of 2 is rational must be false. Hence, the square root of 2 is not a rational number.