📚 Example 1.7.2: Proving that √2 is Irrational | 例 1.7.2：证明根号 2 是无理数
In this AQA A-level Pure Mathematics worked example, we prove one of the most famous results in elementary number theory: the square root of 2 cannot be written as a fraction a/b, where a and b are integers with no common factor.
在这个 AQA A-level 纯数学例题中，我们证明初等数论中最著名的结论之一：根号 2 不能写成分数 a/b 的形式，其中 a 和 b 是没有公因数的整数。
1. The Problem and Its Importance | 问题及其重要性
The statement to prove is: √2 is irrational. In AQA exam language, this means √2 cannot be expressed in the form p/q, where p and q are integers and q ≠ 0.
要证明的命题是：√2 是无理数。在 AQA 考试的表述中，这意味着 √2 不能表示成 p/q 的形式，其中 p 和 q 是整数且 q ≠ 0。
This example appears in the proof section of the AQA specification because it introduces contradiction, a core method for Pure Mathematics.
这个例子出现在 AQA 考纲的证明部分，因为它引入了反证法，这是纯数学的核心方法。
2. What Proof by Contradiction Means | 反证法的含义
s assumption leads to something impossible, such as 1 = 0 or two contradictory statements.
要用反证法证明命题 S，我们先假设 S 为假，然后证明这个假设会导致不可能的情况，例如 1 = 0 或两个相互矛盾的结论。
Here, S is '√2 is irrational'. Its negation is '√2 is rational', so we begin by assuming √2 can be written as a fraction in lowest terms.
这里 S 是 '√2 是无理数'。它的否定是 '√2 是有理数'，所以我们先假设 √2 可以写成最简分数。
3. Step 1: Assume √2 Is Rational | 步骤1：假设根号2是有理数
Assume, for contradiction, that √2 is rational. Then there exist integers a and b with b ≠ 0 such that:
为了引出矛盾，假设 √2 是有理数。那么存在整数 a 和 b，且 b ≠ 0，使得：
√2 = a/b
We can also choose a and b to have no common factor other than 1, so the fraction a/b is in its simplest form. This 'lowest terms' assumption will be important later.
我们还可以选取 a 和 b 除 1 以外没有公因数，因此分数 a/b 是最简形式。这个 '最简形式' 的假设在后面很重要。
4. Step 2: Square Both Sides | 步骤2：两边平方
Square both sides of the equation to remove the square root:
将等式两边平方以消去根号：
2 = a² / b²
Multiplying both sides by b² gives:
两边同乘 b² 得到：
a² = 2b²
This equation is central to the proof. It tells us that a² is exactly twice an integer, so a² must be an even number.
这个等式是证明的核心。它告诉我们 a² 是某个整数的两倍，所以 a² 一定是偶数。
5. Step 3: Deduce That a Is Even | 步骤3：推出 a 是偶数
even, so a² is even. A standard result states that if the square of an integer is even, then the integer itself must be even.
由于 a² = 2b²，右边是偶数，所以 a² 是偶数。一个标准结论是：如果一个整数的平方是偶数，那么这个整数本身也一定是偶数。
² = 4n² + 4n + 1 = 2(2n² + 2n) + 1, which is odd. Therefore a cannot be odd, so a is even.
我们可以通过考虑奇数来验证：奇数具有 2n + 1 的形式，其平方为 (2n + 1)² = 4n² + 4n + 1 = 2(2n² + 2n) + 1，这是一个奇数。因此 a 不可能是奇数，所以 a 是偶数。
Thus we can write a = 2k for some integer k.
因此我们可以将 a 写成 a = 2k，其中 k 是某个整数。
6. Step 4: Substitute a = 2k | 步骤4：代入 a = 2k
Substitute a = 2k into the equation a² = 2b²:
将 a = 2k 代入方程 a² = 2b²：
(2k)² = 2b²
Expanding the left-hand side gives 4k² = 2b². Dividing both sides by 2 gives:
展开左边得到 4k² = 2b²。两边除以 2 得到：
2k² = b²
Now the equation shows that b² is twice an integer, so b² is even.
现在这个等式表明 b² 是某个整数的两倍，所以 b² 是偶数。
7. Step 5: Deduce That b Is Even | 步骤5：推出 b 是偶数
b² is even, then b must be even. Therefore b can be written as b = 2m for some integer m.
使用与前面相同的推理，如果 b² 是偶数，那么 b 一定是偶数。因此 b 可以写成 b = 2m，其中 m 是某个整数。
We have now shown that both a and b are even.
我们现在已经证明了 a 和 b 都是偶数。
8. The Contradiction | 矛盾的出现
At the start, we assumed that a/b was in its lowest terms, meaning a and b have no common factor other than 1. But we have just proved that both a and b are even, so they share a common factor of 2.
一开始，我们假设 a/b 是最简分数，即 a 和 b 除 1 以外没有公因数。但我们刚刚证明了 a 和 b 都是偶数，所以它们有公因数 2。
This contradicts the assumption that a/b is in lowest terms. Since the only assumption we made was that √2 is rational, that assumption must be false.
这与 a/b 是最简分数的假设相矛盾。由于我们唯一做出的假设是 √2 是有理数，因此该假设一定是错误的。
Therefore √2 cannot be rational, so it is irrational.
因此 √2 不可能是有理数，所以它是无理数。
9. Writing the Proof Concisely | 简洁书写证明
In an AQA exam, a concise version of this proof is usually sufficient. A model answer might look like this:
在 AQA 考试中，这个证明的简洁版本通常就足够了。一个标准答案可能如下：
Assume √2 b², so b² = 2k². Hence b² is even, so b is even. This contradicts gcd(a, b) = 1. Therefore √2 is irrational.
假设 √2 = a/b，其中 a、b 为整数，b ≠ 0，且 gcd(a, b) = 1。则 2 = a²/b²，所以 a² = 2b²。因此 a² 是偶数，所以 a 是偶数；设 a = 2k。则 4k² = 2b²，所以 b² = 2k²。因此 b² 是偶数，所以 b 是偶数。这与 gcd(a, b) = 1 相矛盾。因此 √2 是无理数。
10. Common Mistakes | 常见错误
One common mistake is forgetting to state that a/b is in lowest terms. Without this, proving that a and b are even does not necessarily create a contradiction, because a fraction like 2/4 can be reduced to 1/2.
一个常见错误是忘记说明 a/b 是最简分数。如果没有这一点，证明 a 和 b 都是偶数并不一定会产生矛盾，因为像 2/4 这样的分数可以约分为 1/2。
Another mistake is assuming that if a² is even, then a is even without justification. AQA examiners expect you to either quote this as a known result or justify it briefly using the odd-integer form 2n + 1.
另一个错误是在没有证明的情况下直接认为若 a² 是偶数则 a 是偶数。AQA 考官希望考生要么引用这一已知结论，要么用奇数形式 2n + 1 进行简要证明。
Finally, avoid writing √2 = a/b and then immediately treating √2 as a decimal. This is a proof about exact representation, not an approximation exercise.
最后，避免写出 √2 = a/b 后立即把 √2 当作小数处理。这是一个关于精确表示的证明，而不是近似计算练习。
11. Exam-Style Extension: Proving √3 Is Irrational | 考试扩展：证明根号3是无理数
AQA often asks students to adapt this proof to other square roots, such as √3 or √5. The structure is identical, but the parity argument needs to be replaced by divisibility by 3 or 5.
AQA 经常要求学生将此证明改编到其他平方根，如 √3 或 √5。结构完全相同，但奇偶性论证需要替换为被 3 或 5 整除的论证。
For √3, assume √3 a remainder of 1 modulo 3. Therefore a must be divisible by 3. Write a = 3k, then 3b² = 9
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