Explanation: There are several ways to prove that √2 is an irrational number, but one common method is using proof by contradiction. Here is one possible proof:
Assume the opposite, that √2 is a rational number and can be written as a fraction a/b, where a and b are integers and b ≠ 0. Then, we can square both sides of the equation:
(a/b)^2 = 2
Expanding the left side, we get:
a^2/b^2 = 2
Multiplying both sides by b^2, we get:
a^2 = 2b^2
This means that a^2 is even, which implies that a must be even. Let a = 2c, where c is also an integer. Then:
(2c)^2 = 2b^2
Expanding the left side again, we get:
4c^2 = 2b^2
Dividing both sides by 2, we get:
2c^2 = b^2
This means that b^2 is also even, which implies that b must be even. But this contradicts the assumption that a and b have no common factors except for 1. Therefore, our assumption that √2 is a rational number is false, and √2 must be an irrational number.
This proof shows that √2 cannot be expressed as a fraction of integers, and therefore it is an irrational number.