📚 Example 1.8.2: Proof by Contradiction – Irrationality of √2 | 例题 1.8.2：反证法证明根号2是无理数
In AQA A Level Mathematics, Example 1.8.2 is a standard proof by contradiction: show that √2 is irrational. This article walks through the logic, the standard exam wording, and the common mistakes that students make.
在 AQA A Level 数学中，例题 1.8.2 是一个经典的反证法证明：证明 √2 是无理数。本文将详细讲解逻辑、考试标准写法以及学生常见的错误。
1. Proof by Contradiction | 反证法
Proof by contradiction assumes the opposite of what you want to prove. If that assumption leads to an impossible or contradictory result, then the assumption must be false, so the original claim is true.
反证法先假设要证明的结论的反面成立。如果这个假设推出了一个不可能或矛盾的结果，那么该假设一定为假，因此原命题成立。
In this example, the statement is: '√2 is irrational'. Its opposite is: '√2 is rational'. We start by assuming that √2 is rational and show that this leads to a contradiction.
在本例中，命题是：“√2 是无理数”。它的反面是：“√2 是有理数”。我们先假设 √2 是有理数，然后说明这个假设会推出矛盾。
2. Rational and Irrational Numbers | 有理数与无理数
A rational number can be written in the form p/q, where p and q are integers and q ≠ 0. An irrational number cannot be expressed in this way.
有理数可以写成 p/q 的形式，其中 p 和 q 是整数且 q ≠ 0。无理数不能这样表示。
Examples of rational numbers: 1/2, -3, 0.75, 4/9
Examples of irrational numbers: √2, √3, π, e
有理数的例子：1/2、-3、0.75、4/9。
无理数的例子：√2、√3、π、e。
The proof will use the fact that a rational number can always be written as a fraction in its lowest terms, meaning p and q have no common factors other than 1.
证明将用到以下事实：有理数总能写成最简分数，即 p 和 q 除了 1 以外没有其他公因数。
3. Setting Up the Contradiction | 建立反证假设
We begin by assuming that √2 is rational. This means:
我们先假设 √2 是有理数。这意味着：
√2 = p/q, where p, q ∈ ℤ, q ≠ 0 and gcd(p, q) = 1
its lowest terms. This is crucial because later we will show that p and q have a common factor of 2, which contradicts this statement.
这里 gcd(p, q) = 1 表示 p/q 是最简分数。这一点非常关键，因为后面我们将证明 p 和 q 有公因数 2，这与 gcd(p, q) = 1 矛盾。
At this stage, we have not assumed anything unusual. We are simply using the definition of a rational number.
在这一步，我们并没有假设任何异常的东西。我们只是使用了有理数的定义。
4. Squaring Both Sides | 两边平方
We square both sides of the equation to eliminate the square root on the left.
我们对等式两边平方，以消去左边的根号。
(√2)² = (p/q)²
This gives:
由此得到：
2 = p² / q²
Multiplying both sides by q² gives the key equation:
两边乘以 q²，得到关键方程：
p² = 2q²
This equation is the starting point of the contradiction. It tells us that p² is twice another integer q², so p² must be an even number.
这个方程是矛盾的起点。它告诉我们 p² 是另一个整数 q² 的两倍，所以 p² 一定是偶数。
5. Deducing That p Is Even | 推出 p 是偶数
From p² = 2q², we know p² is even because it equals 2 multiplied by an integer.
由 p² = 2q² 可知 p² 是偶数，因为它等于 2 乘以一个整数。
We also need the fact: if p² is even, then p must be even. This is because the square of an odd number is always odd. Therefore p cannot be odd if p² is even.
我们还需要一个事实：如果 p² 是偶数，那么 p 一定是偶数。这是因为奇数的平方总是奇数。因此如果 p² 是偶数，p 就不可能是奇数。
So we can write p as twice an integer:
因此我们可以把 p 写成某个整数的两倍：
p = 2k, where k ∈ ℤ
This substitution is the next step in exposing the contradiction.
这个代入是暴露矛盾的下一步。
6. Substituting p = 2k | 代入 p = 2k
We replace p with 2k in the equation p² = 2q².
我们在方程 p² = 2q² 中用 2k 替换 p。
(2k)² = 2q²
Expanding the left side gives:
展开左边得到：
4k² = 2q²
Dividing both sides by 2 simplifies this to:
两边除以 2，简化为：
q² = 2k²
This new equation has the same form as p² = 2q². It shows that q² is also an even number.
这个新方程与 p² = 2q² 具有相同的形式。它表明 q² 也是偶数。
7. Deducing That q Is Also Even | 推出 q 也是偶数
From q² = 2k², the same reasoning applies: q² is even, so q must also be even.
由 q² = 2k²，同样的推理适用：q² 是偶数，所以 q 也一定是偶数。
We have now shown that both p and q are even. This means that both p and q have a common factor of 2.
我们现在已经证明 p 和 q 都是偶数。这意味着 p 和 q 都有公因数 2。
. If both p and q are even, then gcd(p, q) is at least 2, which contradicts gcd(p, q) = 1.
但一开始我们假设 p/q 是最简分数，即 gcd(p, q) = 1。如果 p 和 q 都是偶数，那么 gcd(p, q) 至少为 2，这与 gcd(p, q) = 1 矛盾。
8. The Contradiction | 矛盾点
The contradiction is clear: our assumption that √2 can be written as a fraction in lowest terms forces p and q to have a common factor of 2. This is impossible if the fraction is truly in lowest terms.
矛盾已经十分清楚：我们假设 √2 可以写成最简分数，结果却推出 p 和 q 有公因数 2。如果这个分数真的是最简分数，这是不可能的。
Therefore, the assumption that √2 is rational must be false.
因此，√2 是有理数这一假设一定是错误的。
Since a number is either rational or irrational, and √2 is not rational, we conclude that √2 is irrational.
由于一个数要么是有理数，要么是无理数，而 √2 不是有理数，所以我们得出结论：√2 是无理数。
9. Standard Exam Wording | 标准考试表述
In AQA exams, a typical question may be worded as:
在 AQA 考试中，典型题目可能这样表述：
Prove by contradiction that √2 is irrational.
Use proof by contradiction to show that √2 cannot be expressed as a fraction in its lowest terms.
用反证法证明 √2 是无理数。
使用反证法证明 √2 不能表示为最简分数。
You should always state the assumption clearly, derive the contradiction, and finish with a conclusion that cancels the assumption.
你应始终清楚地写出假设，推导出矛盾，
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
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