√2 is irrational.
How to prove if that √2 is irrational.
√2 can be expressed as a fraction in the form (a/b), where a and b are integers with no common factors (other than 1) and b is not equal to 0.
So, we have:
√2 = a/b
Square both sides of the equation to get rid of the square root:
2 = a²/b²
2b² = a²
Now, consider the Rational Root Theorem. If √2 is rational, then a and b have no common factors other than 1. However, on the right side of the equation, a² is even, which means a must also be even. Let a = 2k for some integer k.
Substitute a = 2k into the equation:
2b² = (2k)²
2b² = 4k²
b² = 2k²
Now, notice that b² is also even, which implies b must also be even.
However, if both a and b are even, then they have the common factor of 2, which contradicts our assumption that a/b is in its simplest form.
Therefore, our assumption that √2 is rational leads to a contradiction, and we conclude that √2 is irrational.