Determine if √2 is rational or irrational and give a reason for your answer.
The square root of 2 (√2) is an irrational number.
We can prove this using proof by contradiction. Assume √2 is rational. This means it can be expressed as the ratio of two integers, a/b, where a and b have no common factors (excluding 1 and -1). We can express this as √2 = a/b.
If we square both sides, we get 2 = (a^2)/(b^2). Rearranging the equation, we have (a^2) = 2*(b^2).
This implies that (a^2) is an even number. From this, we can conclude that a must also be even. So let's write a as 2n, where n is an integer.
Substituting this value back into the equation, we get (2n)^2 = 2*(b^2). Simplifying, we have 4*(n^2) = 2*(b^2), which further simplifies to 2*(n^2) = (b^2).
Similarly, this implies that (b^2) is an even number, and consequently, b must also be even. However, this contradicts our initial assumption that a/b has no common factors.
Therefore, we have reached a contradiction, showing that our assumption that √2 is rational must be false. Thus, √2 is irrational.