求证2^(0.5)为无理数：反证法的应用逻辑问询

Let's walk through this proof by contradiction step by step—no overly complex jargon, just straight-forward logical reasoning:

• First, we start with the core of contradiction proof: assume the opposite of our target conclusion is true. That means we suppose √2 is a rational number.
• By definition, any rational number can be expressed as a fraction of two coprime integers (integers that share no common divisor other than 1). So let's write:
  √2 = a/b
  where a and b are positive integers, and gcd(a, b) = 1 (they're coprime, no shared factors beyond 1).
• Square both sides to eliminate the square root:
  (√2)² = (a/b)²
  Simplify this to:
  2 = a²/b²
  Multiply both sides by b² to rearrange:
  a² = 2b²
• Now, a² equals twice an integer (2b²), which means a² is even. A key fact: if a squared number is even, the original number must also be even (since odd × odd = odd, even × even = even). So we can rewrite a as 2k, where k is some integer.
• Substitute a = 2k back into the equation a² = 2b²:
  (2k)² = 2b²
  Calculate the left-hand side:
  4k² = 2b²
  Divide both sides by 2 to simplify:
  2k² = b²
• Following the same logic as before: b² is twice an integer, so b² is even, which means b itself must be even.
• Here's the contradiction: we initially assumed a and b are coprime (no common divisors other than 1), but we just proved both a and b are even—so they share a common divisor of 2. This directly contradicts our starting assumption.

Since our assumption that √2 is rational leads to an impossible logical conflict, that assumption must be false. Therefore, √2 is irrational.

内容的提问来源于stack exchange，提问作者Yasmina Anand