To prove that √2 is an irrational number, we will use a proof by contradiction. We will begin by assuming that √2 is a rational number. This means we can express √2 as a fraction of two integers a and b, where a and b have no common factors other than 1 (in other words, they are coprime) and b is not zero:
√2 = a/b
By squaring both sides of the equation, we obtain:
2 = a²/b²
This implies:
a² = 2b²
This equation tells us that a² is even, since it is equal to 2 times another integer (b²). If a² is even, then a must also be even (this is a key property of even numbers). So, we can express a as:
a = 2k
for some integer k. We can now substitute this back into our equation:
(2k)² = 2b²
4k² = 2b²
Dividing both sides by 2 gives us:
2k² = b²
This shows that b² is also even, which means that b must be even as well.
Now, we have established that both a and b are even numbers. This is a contradiction because we started with the assumption that a and b are coprime (having no common factors other than 1). Since both a and b are even, they both have at least the factor 2 in common, which contradicts our initial assumption.
Therefore, our initial assumption that √2 is a rational number must be false. Hence, we conclude that √2 is indeed an irrational number.