📚 Proof by Contradiction | 反证法
Proof by contradiction is a fundamental technique in A-Level Edexcel Mathematics that allows you to demonstrate the truth of a statement by showing that assuming the opposite leads to an impossible situation. This method, also known as indirect proof, is particularly useful when a direct proof is difficult to construct, and it frequently appears in exam questions on irrational numbers, number theory, and logical arguments.
反证法是A-Level Edexcel数学中的一种基本证明技巧，通过展示假设命题的反面会导致不可能的情况，从而证明原命题为真。这种方法也称为间接证明，在直接证明难以构建时尤其有用，并且经常出现在关于无理数、数论和逻辑论证的考试题中。
1. What is Proof by Contradiction? | 什么是反证法？
At its core, a proof by contradiction aims to establish a statement P by temporarily supposing that P is false. From this assumption, we use valid logical deductions to derive a contradiction — something that violates known facts or the original hypothesis. Since logical reasoning cannot produce a contradiction from true premises, the only resolution is that our initial assumption must be incorrect; thus P is true.
反证法的核心思想是，为了证明命题P成立，我们暂时假设P不成立。从这个假设出发，通过有效的逻辑推导，得出了一个矛盾——即与已知事实或原假设相违背的结论。由于正确的逻辑推理不可能从真实的前提出发推出矛盾，唯一的解释就是最初的假设是错误的，因此P为真。
2. Logical Foundation | 逻辑基础
he negation of P yields a logical impossibility, then the negation cannot be true. Mathematically, if we want to prove an implication 'if p then q' then show this leads to a contradiction. This is because p → q is logically equivalent to ¬(p ∧ ¬q).
该方法基于排中律：任何一个明确表述的数学命题要么为真，要么为假，不存在第三种可能。如果假设P的否定会导致逻辑上的不可能，那么该否定就不可能是真的。从数学上讲，如果我们想证明蕴涵式“若p则q”，我们可以假设p真且q假，然后证明这将导致矛盾。这是因为p → q在逻辑上等价于¬(p ∧ ¬q)。
3. Step-by-Step Method | 分步方法
Step 1: Clearly state the proposition you intend to prove. Write down its negation precisely — this is the assumption you will work with. Step 2: Use algebraic manipulation, properties of numbers, or previously established theorems to derive consequences of the negated assumption. Step 3: en number equals an odd number, or a number both has and does not have a certain property. Step 4: Conclude that the negated assumption is impossible, and therefore the original proposition must be true.
步骤1：清晰陈述你想要证明的命题。精确写出该命题的否定——这将是你将要处理的假设。步骤2：运用代数运算、数的性质或先前已证明的定理，从否定假设出发推导出一系列结论。步骤3：持续推导，直至出现不一致的情况，例如1=0、一个数既是偶数又是奇数，或者一个数同时具有和不具有某个性质。步骤4：得出否定假设不可能成立，因此原命题必定为真。
4. Example 1: √2 is Irrational | 例1：根号2是无理数
This is the classic proof that √2 cannot be expressed as a fraction. Assume, for contradiction, that √2 is rational. Then √2 ≠ 0, and the fraction is in its lowest terms (a and b are coprime, having no common factor greater than 1).
这是证明√2不能表示为分数的经典例子。假设√2是有理数，从而引出矛盾。那么√2 = a/b，其中a和b为整数且b ≠ 0，并且该分数已化为最简形式（a与b互质，没有大于1的公因数）。
t a = 2k for some integer k. Substituting back, we get (2k)² = 2b² → 4k² = 2b² → b² = 2k². Hence b² is also even, so b must be even. But this contradicts our assumption that a and b have no common factor, since both are now divisible by 2. The contradiction shows our original assumption is false; therefore √2 is irrational.
两边平方得2 = a²/b²，所以a² = 2b²。这意味着a²是偶数，从而a必须是偶数（因为奇数的平方为奇数）。设a = 2k，其中k为整数。代入后得到(2k)² = 2b² → 4k² = 2b² → b² = 2k²。因此b²也是偶数，所以b必定为偶数。但这与我们假设a和b没有公因数相矛盾，因为它们现在都能被2整除。该矛盾表明最初假设为假，因此√2是无理数。
5. Example 2: Infinite Number of Primes | 例2：素数有无穷多个
Euclid' imes, say p₁, p₂, ..., pₙ. Consider the number N = p₁ × p₂ × ... × pₙ + 1. Clearly N is greater than 1, so it must be divisible by some prime (by the Fundamental Theorem of Arithmetic).
欧几里得关于素数无穷多的证明是反证法的另一个优雅范例。假设情况相反：只有有限个素数，设为p₁, p₂, ..., pₙ。考虑数字N = p₁ × p₂ × ... × pₙ + 1。显然N大于1，根据算术基本定理，它必能被某个素数整除。
ivides N. Therefore p divides the difference N - (p₁ × p₂ × ... × pₙ) = 1. This means p divides 1, which is impossible for any prime. The contradiction implies our initial supposition of a finite list of primes is false — there must be infinitely many primes.
设该素数为p。由于我们的列表包含了所有素数，p必然是p₁, p₂, ..., pₙ中的某一个。因此p能整除乘积p₁ × p₂ × ... × pₙ，同时显然p也能整除N。那么p就能整除差值N - (p₁ × p₂ × ... × pₙ) = 1。这意味着p整除1，这对任何素数来说都是不可能的。这一矛盾意味着我们关于素数有限的初始假设是错误的——素数必有无穷多个。
6. Example 3: If n² is Even then n is Even | 例3：若n²为偶，则n为偶
ow suppose, for contradiction, that n is odd. An odd integer can be written as n = 2k + 1 for some integer k. Squaring gives n² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1, which is clearly odd. But we were given that n² is even. This is a contradiction, so our assumption that n is odd must be false. Hence n is even.
该命题可以通过逆否命题直接证明，但用反证法同样富有启发性。假设前提条件：n²为偶数。现在，为了引出矛盾，我们假设n为奇数。一个奇数可以写成n = 2k + 1，其中k为整数。平方后得到n² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1，显然为奇数。但已知条件说n²是偶数。这就产生了矛盾，因此我们关于n为奇数的假设必定是假的。所以n是偶数。
This proof is frequently extended in exams to cases like 'if n² is divisible by 3, then n is divisible by 3', where you assume n is not divisible by 3 and write n = 3k ± 1 to reach a contradiction.
这个证明在考试中经常被推广到类似“若n²能被3整除，则n能被3整除”的情形，此时可以假设n不能被3整除，并设n = 3k ± 1，从而推出矛盾。
7. Example 4: Sum of Rational and Irrational is Irrational | 例4：有理数与无理数之和为无理数
A very common A-Level question asks: 'Prove that if a is rational and b is irrational, then a + b is irrational.' under subtraction, the difference of two rationals is rational. Then b = (a + b) - a must be rational. However, this contradicts the given fact that b is irrational. Hence our assumption that a + b is rational cannot be true; a + b is irrational.
一个非常常见的A-Level考题是：“证明若a为有理数且b为无理数，则a + b为无理数。”我们用反证法来证明。假设其否定：a为有理数，b为无理数，但a + b为有理数。由于有理数集对减法封闭，两个有理数的差仍为有理数。那么b = (a + b) - a必然是有理数。但这与已知b为无理数相矛盾。因此，我们关于a + b为有理数的假设不可能成立；a + b是无理数。
This elegant proof hinges on nothing more than the definition of irrational numbers and the closure properties of the rationals. It can be readily adapted to products and quotients under appropriate non-zero conditions.
这个简洁的证明仅依赖于无理数的定义以及有理数的封闭性。它很容易推广到在适当的非零条件下的乘积与商的情况。
8. Example 5: log₂5 is Irrational | 例5：log₂5是无理数
Suppose, for contradiction, that log₂5 is rational. Then it can be expressed as a fraction p/q where p and q are positive integers. By the definition of a logarithm, this means 2p/q = 5. Raising both sides to the power q gives 2p = 5q. Now examine the parity: 2p is an even integer (since p ≥ 1), while 5q is odd (any power of 5 ends in 5 or 25, and is never divisible by 2). An even number cannot equal an odd number. This contradiction proves that log₂5 cannot be rational; it is irrational.
假设log₂5为有理数，以导出矛盾。那么它可以表示为分数p/q，其中p和q为正整数。根据对数的定义，这意味着2p/q = 5。两边同时q次方，得到2p = 5q。现在考察奇偶性：2p是偶数（因为p ≥ 1），而5q是奇数（5的任何次幂都以5或25结尾，永远不能被2整除）。一个偶数不可能等于一个奇数。这一矛盾证明log₂5不可能是无理数；它是无理数。
9. Common Mistakes to Avoid | 常见错误避免
Incorrect negation: The most frequent error is to state the negation of the proposition inaccurately. For an implication 'if P then Q', the negation is 'P is true and Q is false', not 'if P then not Q'. Circular reasoning: Avoid using the result you are trying to prove within your derivation. Weak contradiction: Ensure the contradiction is genuine and unambiguous, not just an unexpected outcome. Muddy conclusion: After reaching the contradiction, explicitly state that the original assumption is false and therefore the original proposition is true.
否定错误：最常见的错误是未能准确地陈述命题的否定。对于“若P则Q”的蕴涵式，其否定是“P真且Q假”，而不是“若P则非Q”。循环论证：避免在推导中使用你正在试图证明的结论。矛盾不充分：确保矛盾是真实且不含糊的，而不仅仅是一个意外的结果。结论模糊：在得出矛盾后，必须明确陈述原假设为假，因此原命题为真。
10. Exam Strategy and Practice | 考试策略与练习
In Edexcel A-Level exams, proof by contradiction questions are often worth 3–5 marks and require a clear, logical structure. Start by writing 'Assume, for contradiction, that...' and finish with 'This is a contradiction; therefore the statement is true.' Practise spotting the type of statement suited for contradiction: ones involving irrationality, infinity, 'there is no', or uniqueness. Below are some typical exercises you can attempt:
在Edexcel A-Level考试中，反证法题目通常分值在3到5分，要求清晰、逻辑严谨的结构。开头可写“为引出矛盾，假设……”，结尾写“这是矛盾的；因此该命题为真。” 练习时要善于识别适合用反证法的命题类型：涉及无理性、无穷性、“不存在”或唯一性的命题。以下是一些你可以尝试的典型练习题：
♦ Prove that √3 is irrational. ♦ Prove there is no smallest positive rational number. ♦ Prove that if a and b are integers and ab is odd, then both a and b are odd. ♦ Prove that the product of a non-zero rational and an irrational is irrational. ♦ Prove that log₃7 is irrational. (Hint: adapt the method for log₂5.)
♦ 证明√3是无理数。 ♦ 证明不存在最小的正有理数。 ♦ 证明若a和b为整数且ab为奇数，则a和b均为奇数。 ♦ 证明一个非零有理数与一个无理数的乘积是无理数。 ♦ 证明log₃7是无理数。（提示：仿照log₂5的方法。）
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