Recurring Decimals and Fractions Conversion Techniques | IGCSE数学：循环小数与分数互化技巧

📚 Recurring Decimals and Fractions Conversion Techniques | IGCSE数学：循环小数与分数互化技巧

Recurring decimals, also known as repeating decimals, are decimals that have a digit or a group of digits repeating infinitely. In IGCSE Mathematics (Edexcel), converting between recurring decimals and fractions is a core skill that appears in both Foundation and Higher Tier papers. This article covers the standard methods, common traps, and quick shortcut techniques to help you score full marks.

循环小数，又称重复小数，是指小数部分有一个数字或一组数字无限重复的小数。在 Edexcel 的 IGCSE 数学考试中，循环小数与分数的互化是基础卷和高级卷都会考查的核心技能。本文将系统讲解标准方法、常见陷阱以及快速技巧，帮助你在考试中拿到满分。

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1. What Is a Recurring Decimal? | 什么是循环小数？

A recurring decimal is written using a dot above the first and last digit of the repeating block. For example, 0.3333... is written as 0.\dot{3} in standard notation, though in this article we use the notation 0.3̇ (with a dot above 3). The repeating block is called the repetend.

循环小数用在小数循环节的首位和末位数字上方加点来表示。例如 0.3333... 写作 0.\dot{3}，而本文中我们使用 0.3̇（在 3 上方加点）的记法。循环出现的数字块称为循环节。

Key examples of notation:

关键记法示例：

• 0.6666... = 0.\dot{6} (single digit repeats)
• 0.6666... = 0.\dot{6}（单个数字循环）
• 0.272727... = 0.\dot{2}\dot{7} (two digits repeat)
• 0.272727... = 0.\dot{2}\dot{7}（两个数字循环）
• 0.123123123... = 0.\dot{1}2\dot{3} (three digits repeat)
• 0.123123123... = 0.\dot{1}2\dot{3}（三个数字循环）

In Edexcel IGCSE, you are expected to recognize the notation and convert these decimals into fractions in their simplest form.

在 Edexcel IGCSE 考试中，你不仅要能识别这种记法，还要能将循环小数化为最简分数。

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2. Converting a Pure Recurring Decimal (One Repeating Digit) | 纯循环小数（单个循环数字）的互化

Let us start with the simplest case: a decimal with exactly one repeating digit, such as 0.\dot{3} or 0.\dot{7}.

我们先从最简单的情形开始：只有一个循环数字的小数，例如 0.\dot{3} 或 0.\dot{7}。

Method (standard algebraic method): Let x equal the decimal. Multiply both sides of the equation by 10 for every digit in the repeating block. Since there is one repeating digit, multiply by 10. Then subtract the original equation to eliminate the repeating part.

方法（标准代数法）：设 x 等于这个小数。根据循环节的位数，对方程两边乘以对应的 10 的幂。因为这里只有一个循环数字，所以乘以 10。然后用新方程减去原方程，消去循环部分。

Example: Convert 0.\dot{6} to a fraction.

例题：将 0.\dot{6} 化为分数。

Let x = 0.\dot{6}
10x = 6.\dot{6}
10x − x = 6.\dot{6} − 0.\dot{6}
9x = 6
x = 6/9 = 2/3

Therefore 0.\dot{6} = 2/3.

因此 0.\dot{6} = 2/3。

Another example: Convert 0.\dot{1} to a fraction.

另一个例子：将 0.\dot{1} 化为分数。

Let x = 0.\dot{1}
10x = 1.\dot{1}
9x = 1
x = 1/9

This shows a quick rule: a single-digit recurring decimal 0.\dot{n} equals n/9. For example 0.\dot{8} = 8/9.

这揭示了一个快速规律：单个循环数字的小数 0.\dot{n} 等于 n/9。例如 0.\dot{8} = 8/9。

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3. Converting Two Repeating Digits | 两个循环数字的互化

When two digits repeat, such as 0.\dot{2}\dot{7}, multiply by 100 (10²) instead of 10. This is because the repeating block has two digits.

当有两个数字循环时，例如 0.\dot{2}\dot{7}，应乘以 100（即 10²）而不是 10。因为循环节有两位数字。

Example: Convert 0.\dot{2}\dot{7} to a fraction.

例题：将 0.\dot{2}\dot{7} 化为分数。

Let x = 0.\dot{2}\dot{7} = 0.272727...
100x = 27.\dot{2}\dot{7} = 27.272727...
100x − x = 27.\dot{2}\dot{7} − 0.\dot{2}\dot{7}
99x = 27
x = 27/99 = 3/11

Therefore 0.\dot{2}\dot{7} = 3/11.

因此 0.\dot{2}\dot{7} = 3/11。

Quick rule: For a two-digit repetend ab repeating (0.\dot{a}\dot{b}), the fraction is (ab)/99. Simplify if possible.

快速规则：对于两位循环节 ab（即 0.\dot{a}\dot{b}），分数等于 ab/99。记得要约分。

Caution: Always check whether the fraction can be simplified. In the example above, 27/99 simplifies to 3/11.

注意：务必检查分数是否可以约分。上例中 27/99 化简为 3/11。

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4. Converting Three Repeating Digits | 三个循环数字的互化

For a three-digit recurring block, multiply by 1000 (10³).

对于三位循环节，需要乘以 1000（10³）。

Example: Convert 0.\dot{1}2\dot{3} (meaning 0.123123123...) to a fraction.

例题：将 0.\dot{1}2\dot{3}（即 0.123123123...）化为分数。

Let x = 0.123123123...
1000x = 123.123123123...
1000x − x = 123.
999x = 123
x = 123/999 = 41/333

Wait! so 123/999 = 41/333. Since 41 is prime and does not divide 333, this is the simplest form.

注意！123/999 可以约分：123 和 999 有公因数 3，因此 123/999 = 41/333。因为 41 是质数且不能整除 333，所以这是最简形式。

Quick rule: For a three-digit repetend abc, the fraction is (abc)/999, simplified.

快速规则：对于三位循环节 abc，分数等于 (abc)/999，并需要化简。

In general, for a repeating block of n digits, the denominator after subtracting will be 10ⁿ − 1, which can also be written as a number consisting of n nines (e.g. 9, 99, 999).

一般来说，循环节有 n 位时，相减后分母为 10ⁿ − 1，即 n 个 9 组成的数（如 9、99、999）。

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5. Converting a Mixed Recurring Decimal (Non-Repeating + Repeating) | 混循环小数（非循环部分 + 循环部分）的互化

A mixed recurring decimal has digits after the decimal point that do not all repeat. For example, 0.12\dot{3} means 0.123333... where "12" is non-repeating and "3" repeats. These are common in Edexcel papers.

混循环小数是指小数点后的数字并非全部循环。例如 0.12\dot{3} 表示 0.123333...，其中 "12" 不循环，而 "3" 循环。这类题目在 Edexcel 考试中很常见。

Method:

方法：

• First, write x with the decimal expansion.
• 先写出 x 的小数展开式。
• Multiply by 10 to shift the decimal point so that the non-repeating part is moved to the left.
• 乘以 10 的幂，使小数点的位置移动到非循环部分之后。
• Then multiply by a further power of 10 to align the repeating parts.
• 再乘以另一个 10 的幂，使循环部分对齐。
• Subtract and solve.
• 相减并求解。

Example: Convert 0.12\dot{3} to a fraction.

例题：将 0.12\dot{3} 化为分数。

Let x = 0.123333...
Multiply by 100: 100x = 12.333...
Multiply by 1000: 1000x = 123.333...
1000x − 100x = 123.333... − 12.333...
900x = 111
x = 111/900 = 37/300

Check: 0.123333... = 37/300 ≈ 0.123333..., which confirms the result.

验证：0.123333... = 37/300 ≈ 0.123333...，结果正确。

Alternative clean method: Write the decimal as a sum of a terminating part and a recurring part:

另一种简洁方法：将这个小数拆成有限小数与循环小数之和：

0.12\dot{3} = 0.12 + 0.003333... = 12/100 + (1/100)(0.\dot{3}) = 12/100 + (1/100)(1/3) = 36/300 + 1/300 = 37/300

Both methods yield the same fraction.

两种方法得到相同的结果。

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6. Fraction to Recurring Decimal Conversion | 分数化循环小数

Converting a fraction to a decimal is often easier: divide the numerator by the denominator using long division. If the decimal terminates, the denominator in simplest form has only prime factors 2 and 5. If the denominator has any other prime factor, the decimal will recur.

分数化小数通常更简单：直接用分子除以分母做长除法。如果小数能有限终止，最简分数的分母只能含有质因数 2 和 5。如果分母含有其他质因数，则小数必定会循环。

Example: Convert 5/6 to a decimal.

例题：将 5/6 化为小数。

5 ÷ 6 = 0.8333... = 0.8\dot{3}

Since the denominator 6 = 2 × 3 contains a factor 3, the decimal repeats.

因为分母 6 = 2 × 3 含有因数 3，所以小数会循环。

Example: Convert 7/12 to a decimal.

例题：将 7/12 化为小数。

7 ÷ 12 = 0.58333... = 0.58\dot{3}

Again, denominator 12 = 2² × 3 has a factor 3, so the decimal recurs.

同样，分母 12 = 2² × 3 含有因数 3，所以小数循环。

To determine which pattern emerges, always simplify the fraction first and examine the denominator's prime factors.

要判断小数是否会循环，务必先将分数化简，再检查分母的质因数。

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7. Shortcut Formula for Recurring Decimals | 循环小数互化的快捷公式

For an exam, the algebraic method is reliable, but a shortcut can save time. The general formula for a pure recurring decimal is:

在考试中，代数法虽然可靠，但快捷公式可以节省时间。纯循环小数的一般公式为：

0.\dot{a₁}a₂...aₙ̇ = (a₁a₂...aₙ) / (99...9) （n 个 9）

where a₁a₂...aₙ is the integer formed by the repeating block.

其中 a₁a₂...aₙ 是循环节形成的整数。

For mixed recurring decimals, e.g. 0.m\dot{r} where m is a finite block of k digits and r is a repeating block of n digits:

对于混循环小数，例如 0.m\dot{r}，其中 m 是包含 k 位数字的有限部分，r 是包含 n 位数字的循环部分：

0.m\dot{r} = (mr − m) / (10ᵏ(10ⁿ − 1))

This formula is derived from the algebraic method but can be applied directly. However, always simplify the final fraction.

这个公式由代数法推导而来，可以直接使用。不过最终分数一定要化简。

Example: 0.12\dot{3} = (123 − 12) / (100 × 9) = 111/900 = 37/300.

例如：0.12\dot{3} = (123 − 12)/(100 × 9) = 111/900 = 37/300。

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8. Worked Exam-Style Examples | 考试经典例题

Let us work through several questions that mirror the style of Edexcel IGCSE exam questions.

我们来做几道与 Edexcel IGCSE 真题风格一致的练习题。

Example 1: Convert 0.\dot{4}5\dot{4} to a fraction. Here the repeating block is 454, so n = 3.

例题1：将 0.\dot{4}5\dot{4} 化为分数。这里循环节是 454，所以 n = 3。

x = 0.454454454...
1000x = 454.454454...
999x = 454
x = 454/999

454/999 is already in simplest form (454 = 2 × 227; 999 = 3³ × 37, no common factors).

454/999 已是最简分数（454 = 2 × 227；999 = 3³ × 37，无公因数）。

Example 2: Convert 0.2\dot{3}5\dot{6} to a fraction. This means the block "56" repeats after the "23".

例题2：将 0.2\dot{3}5\dot{6} 化为分数。这意味着在 "23" 之后循环块 "56" 重复出现。

Let x = 0.23565656...
100x = 23.565656...
100x = 23.\dot{5}\dot{6}
Now let y = 23.\dot{5}\dot{6}
100y = 2356.\dot{5}\dot{6}
99y = 2356 − 23 = 2333
y = 2333/99
So 100x = 2333/99
x = 2333/9900

Check if it simplifies: 2333 = 2333 (prime?), 9900 = 2² × 3² × 5² × 11. No common factors, so x = 2333/9900.

检查能否约分：2333 是质数，9900 = 2² × 3² × 5² × 11，没有公因数，所以 x = 2333/9900。

Example 3 (non-calculator manipulation): Show that 0.\dot{5} = 5/9 and use this to find 0.5\dot{5} as a fraction.

例题3（不含计算器的运算）：证明 0.\dot{5} = 5/9，并利用此结论将 0.5\dot{5} 化为分数。

0.5\dot{5} = 0.5 + 0.0\dot{5} = 1/2 + (1/10)(5/9) = 1/2 + 5/90 = 45/90 + 5/90 = 50/90 = 5/9

Interesting result: 0.5\dot{5} equals 5/9 exactly, the same as 0.\dot{5}. This is because 0.55... = 0.555..., which is indeed 5/9.

有趣的结果：0.5\dot{5} 等于 5/9，与 0.\dot{5} 相同。因为 0.55... 就是 0.555...，确实等于 5/9。

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9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Many students lose marks on recurring decimal questions due to avoidable errors. Here are the most common pitfalls.

许多学生在循环小数题目中因为可避免的错误而丢分。以下是最常见的陷阱。

• Multiplying by the wrong power of 10. The multiplier must match the number of repeating digits.
  乘以错误的 10 的幂。倍数必须与循环数字的位数匹配。
• Forgetting to subtract the original x. You must subtract exactly x, not the first shifted equation.
  忘记减去原方程 x。你必须减去 x 本身，而不是第一次移位后的方程。
• Not simplifying the final fraction. Edexcel mark schemes usually require the simplest form.
  没有化简最终分数。Edexcel 的评分标准通常要求最简形式。
• Misreading the notation. For example 0.\dot{2}3\dot{4} is not the same as 0.234 repeating; the former repeats 234 after the decimal? Actually 0.\dot{2}3\dot{4} means the block 234 repeats (because the dots are over 2 and 4, indicating the entire middle is included). Always carefully identify the repetend.
  误读循环点记法。例如 0.\dot{2}3\dot{4} 不是 0.234 循环；0.\dot{2}3\dot{4} 表示 234 循环（因为点在 2 和 4 上面，说明中间的数字也包含在内）。一定要仔细识别循环节。
• Arithmetic slip when subtracting. Write the equations neatly and align the decimal points.
  相减时出现算术错误。整齐地写出方程并对齐小数点。

Let' above the first and last digit of a block, the entire block between them repeats. For 0.2\dot{3}5\dot{6}, the dots on 3 and 6 mean the block "356" repeats? rst and last digit indicates that all digits between them form the repeating block. So 0.2\dot{3}5\dot{6} would mean the block "356" repeats: 0.2356356356... But the previous example assumed only "56" repeats. In standard Edexcel notation, 0.2\dot{3}5\dot{6} is not a standard way to write only "56" repeating; to show "56" repeating one would write 0.23\dot{5}\ the repetend. The Edexcel specification uses the dot-over-first-and-last convention. So 0.\dot{2}3\dot{4} = 0.234234234... and 0.2\dot{3}5\ not a valid standard notation in Edexcel because the dot over 3 and 6 indicates the block "356" epeats; two dots over the first and last digits mean the whole block between them repeats.

我们仔细看循环点记法规则：如果点在循环节的首位和末位数字上方，则它们之间的所有数字都构成循环节。例如 0.\dot{2}3\dot{4} 表示 234 循环，即 0.234234234...。但在 Edexcel 规范中，0.2\dot{3}5\dot{6} 并不是标准的表示 "56" 循环的写法；要表示 "56" 循环，应该写作 0.23\dot{5}\dot{6}。为避免歧义，本文采用以下约定：单点表示该数字循环；首末两位加点表示整个中间块循环。

Therefore, in Example 2 of section 8, the correct notation for 0.23565656... should be 0.23\dot{5}\dot{6}, not 0.2\dot{3}5\dot{6}. The calculation remains correct if we use 0.23\dot{5}\dot{6}. Let us redo it properly.

因此，第 8 节例题 2 中，0.23565656... 的正确记法应为 0.23\dot{5}\dot{6}，而不是 0.2\dot{3}5\dot{6}。如果我们使用 0.23\dot{5}\dot{6}，计算结果不变。我们重新正确计算一遍。

Let x = 0.23\dot{5}\dot{6} = 0.23565656...
100x = 23.\dot{5}\dot{6} = 23.565656...
Let y = 23.\dot{5}\dot{6}
100y = 2356.\dot{5}\dot{6}
100y − y = 2356.\dot{5}\dot{6} − 23.\dot{5}\dot{6}
99y = 2333
y = 2333/99
Therefore 100x = 2333/99
x = 2333/9900

This matches the earlier result.

这与之前的结果一致。

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10. Terminating vs Recurring: How to Tell Instantly | 有限小数与循环小数：如何快速判断

A fraction in its simplest form will produce a terminating decimal if and only if the denominator has no prime factors other than 2 and 5. Otherwise it produces a recurring decimal.

一个最简分数能化为有限小数，当且仅当分母除了 2 和 5 以外没有其他质因数。否则它会产生循环小数。

Fraction	Denominator factors	Decimal
3/8	8 = 2³	0.375 (terminating)
7/20	20 = 2² × 5	0.35 (terminating)
4/9	9 = 3²	0.\dot{4} (recurring)
2/7	7	0.\dot{2}8571\dot{4} (recurring)

are asked to convert 3/8, simply do long division: 3 ÷ 8 = 0.375. No need for any formula.

在考试中，如果要将 3/8 化为小数，直接做长除法：3 ÷ 8 = 0.375，不需要任何公式。

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11. Practice Questions with Solutions | 练习与答案

Try these questions yourself before reading the solutions.

请先尝试自己完成下面的练习，再看答案。

Practice 1: Convert 0.\dot{5} to a fraction.

练习1：将 0.\dot{5} 化为分数。

Practice 2: Convert 0.\dot{7}2\dot{7} (i.e. 0.727272...) to a fraction.

练习2：将 0.\dot{7}2\dot{7}（即 0.727272...）化为分数。

Practice 3: Convert 0.2\dot{4} to a fraction.

练习3：将 0.2\dot{4} 化为分数。

Practice 4: Convert 11/30 to a decimal.

练习4：将 11/30 化为小数。

Practice 5: Show that 0.\dot{9} = 1.

练习5：证明 0.\dot{9} = 1。

Solutions:

答案：

• 1: 0.\dot{5} = 5/9.
• 2: Let x = 0.\dot{7}2\dot{7}; 100x = 72.\dot{7}2\dot{7}; 99x = 72; x = 72/99 = 8/11.
• 3: Let x = 0.2\dot{4}; 10x = 2.\dot{4}; 100x = 24.\dot{4}; 90x = 22; x = 22/90 = 11/45.
• 4: 11 ÷ 30 = 0.3666... = 0.3\dot{6}.
• 5: Let x = 0.\dot{9}; 10x = 9.\dot{9}; 9x = 9; x = 1. This is a famous result.

• 1：0.\dot{5} = 5/9。
• 2：设 x = 0.\dot{7}2\dot{7}；100x = 72.\dot{7}2\dot{7}；99x = 72；x = 72/99 = 8/11。
• 3：设 x = 0.2\dot{4}；10x = 2.\dot{4}；100x = 24.\dot{4}；90x = 22；x = 22/90 = 11/45。
• 4：11 ÷ 30 = 0.3666... = 0.3\dot{6}。
• 5：设 x = 0.\dot{9}；10x = 9.\dot{9}；9x = 9；x = 1。这是一个著名结论。

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12. Final Tips for Exam Success | 考试成功要点总结

To summarise, here are the key takeaways for this topic:

总而言之，以下是本主题的关键要点：

• Always let x equal the recurring decimal, then multiply by 10, 100 or 1000 depending on the length of the repeating block.
• 总是设 x 等于循环小数，然后根据循环节的长度乘以 10、100 或 1000。
• Subtract the original x to eliminate the infinite repeating part.
• 减去原方程 x 以消去无限循环部分。
• Simplify the resulting fraction fully.
• 将所得分数完全化简。
• For mixed recurring decimals, two multiplications are often needed to align the repeating parts.
• 对于混循环小数，通常需要两次乘法来对齐循环部分。
• Know the difference between terminating and recurring decimals using denominator prime factors.
• 通过分母的质因数判断有限小数与循环小数。
• Practice the algebraic method until it becomes automatic, but also be ready to verify using the shortcut formula.
• 反复练习代数法直到熟练，同时也可以用快捷公式来验证。

With consistent practice, recurring decimal problems become some of the easiest marks on the IGCSE paper. Master the mechanics, double-check your arithmetic, and always present your working clearly to gain method marks even if a small error slips in.

只要坚持练习，循环小数题会成为 IGCSE 考试中最容易得分的题目之一。掌握方法步骤，反复检查运算，并清晰展示过程——即使出现小错误，也能获得方法分。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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