📚 Number Concepts and Operations | 数的概念与运算
Welcome to the first topic of your IGCSE Edexcel Mathematics journey. In this article, we cover the essential building blocks of the number system – from types of numbers to fractions, powers, standard form, and the order of operations. Mastering these foundational skills is critical because they appear in every subsequent topic, whether you are solving equations, working with graphs, or tackling trigonometry. Let's build your confidence step by step, with clear explanations and plenty of examples that match the style of Edexcel exam questions.
欢迎开启 IGCSE Edexcel 数学的学习之旅，这里是复习系列的第一篇。我们将从数系的基本模块入手，涵盖数的种类、分数、幂、标准形式和运算顺序等核心内容。扎实掌握这些基础非常重要，因为后续无论是解方程、画图像还是处理三角学问题，都会反复用到。本文将逐步带你建立信心，所有讲解和示例都与 Edexcel 考试风格对齐，帮助你稳扎稳打地起步。
1. Types of Numbers | 数的种类
The number system you will use throughout your IGCSE is the set of real numbers. Real numbers can be divided into natural numbers (1, 2, 3, …), whole numbers (0, 1, 2, …), integers (…, −2, −1, 0, 1, 2, …), rational numbers (any number that can be written as a fraction a/b where a and b are integers and b ≠ 0), and irrational numbers (numbers that cannot be written as a simple fraction, such as π and √2). All of these together form the real number line.
你在 IGCSE 阶段使用的数系是实数集。实数可以进一步分为自然数（1, 2, 3, …）、整数（含 0, 1, 2, … 及 …−2, −1, 0, 1, 2, …）、有理数（任何可写成 a/b 形式的数，其中 a 和 b 是整数且 b ≠ 0）和无理数（不能写成简单分数的数，例如 π 和 √2）。所有这些数共同构成了实数轴。
- Natural numbers: 1, 2, 3, …
- Whole numbers: 0, 1, 2, 3, …
- Integers: … −2, −1, 0, 1, 2, …
- Rational examples: 1/2, −3, 0.75, 0.3̇ (recurring decimals belong here)
- Irrational examples: π, √2, √3, e
Rational numbers include integers, fractions, terminating decimals and recurring decimals. An irrational number's decimal expansion goes on forever without repeating. Recognising the type of number helps you decide how to simplify expressions and justify answers in exams.
有理数包括整数、分数、有限小数和循环小数。无理数的小数展开无限且不循环。能够识别数的种类，有助于你在化简表达式时做出正确选择，并在考试中为答案提供理由。
2. Prime Numbers, Factors and Multiples | 质数、因数与倍数
A prime number is a positive integer greater than 1 that has exactly two distinct factors: 1 and itself. The first few primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Note that 2 is the only even prime. A factor of a number divides into it exactly without leaving a remainder, while a multiple is the product of that number and any integer. Understanding prime factorisation – expressing a number as a product of prime factors – is the key to finding highest common factors (HCF) and lowest common multiples (LCM).
质数是大于 1 且恰好有两个不同因数的正整数，即 1 和它本身。最初几个质数是 2, 3, 5, 7, 11, 13, 17, 19, 23, 29。注意 2 是唯一的偶质数。一个数的因数是能整除它的数，倍数则是该数与任意整数的乘积。理解质因数分解——将合数写成质因数乘积的形式——是求最大公因数（HCF）和最小公倍数（LCM）的关键。
Prime factorisation example: 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5
质因数分解示例：60 = 2 × 2 × 3 × 5 = 2² × 3 × 5
3. Highest Common Factor (HCF) and Lowest Common Multiple (LCM) | 最大公因数与最小公倍数
The HCF of two or more numbers is the largest number that divides into all of them. The LCM is the smallest number that is a multiple of all of them. Using prime factorisation, you can find the HCF by multiplying the lowest powers of common prime factors, and the LCM by multiplying the highest powers of all prime factors present.
两个或多个数的最大公因数（HCF）是能够整除这些数的最大数。最小公倍数（LCM）则是这些数的倍数中最小的一个。利用质因数分解，你可以用公共质因数的最低次幂相乘得到 HCF，用所有出现质因数的最高次幂相乘得到 LCM。
Example: Find HCF and LCM of 48 and 180. 48 = 2⁴ × 3, 180 = 2² × 3² × 5. HCF = 2² × 3 = 12; LCM = 2⁴ × 3² × 5 = 720.
示例：求 48 和 180 的 HCF 与 LCM。48 = 2⁴ × 3，180 = 2² × 3² × 5。HCF = 2² × 3 = 12；LCM = 2⁴ × 3² × 5 = 720。
4. Fractions and Decimals | 分数和小数
Fractions represent parts of a whole. Edexcel questions will test your ability to add, subtract, multiply, and divide fractions, as well as convert between mixed numbers and improper fractions. Always simplify your answers to their lowest terms. To add or subtract fractions, write them with a common denominator; to multiply, multiply the numerators and denominators; to divide, multiply by the reciprocal.
分数表示整体的一部分。Edexcel 题目会考察你对分数的加、减、乘、除运算，以及带分数与假分数转换的能力。所有结果必须化为最简形式。加减分数时要先通分；乘法直接分母乘分母、分子乘分子；除法则是乘以其倒数。
Key operations (fraction examples): + 1/4 = 8/12 + 3/12 = 11/12; 2/3 × 1/4 = 2/12 = 1/6; 2/3 ÷ 1/4 = 2/3 × 4/1 = 8/3 = 2 2/3.
关键运算（分数示例）：2/3 + 1/4 = 8/12 + 3/12 = 11/12；2/3 × 1/4 = 2/12 = 1/6；2/3 ÷ 1/4 = 2/3 × 4/1 = 8/3 = 2 2/3。
Recognise terminating decimals (e.g., 1/4 = 0.25) and recurring decimals (e.g., 1/3 = 0.3̇).
小数遵循相同的运算法则。分数化小数用分子除以分母即可。要能区分有限小数（如 1/4 = 0.25）和循环小数（如 1/3 = 0.3̇）。
5. Recurring Decimals to Fractions | 循环小数化分数
A common IGCSE skill is to express a recurring decimal as an exact fraction. For a decimal with a single repeating digit, multiply by 10, subtract the original, and solve. For repeating cycles of length two, multiply by 100, and so on. Always eliminate the recurring part through subtraction.
IGCSE 常见考点之一是将循环小数表示为精确分数。对于只有一位循环节的小数，乘以 10，减去原数，然后求解。若循环节长度为两位，则乘以 100，依此类推。核心是通过减法消去循环部分。
Example 1: Let x = 0.6̇. Then 10x = 6.6̇. Subtract: 10x − x = 6.6̇ − 0.6̇ → 9x = 6 → x = 6/9 = 2/3.
示例 1：设 x = 0.6̇，则 10x = 6.6̇。相减得 10x − x = 6.6̇ − 0.6̇ → 9x = 6 → x = 6/9 = 2/3。
Example 2: x = 0.1̇8̇. 100x = 18.1̇8̇. Subtract: 99x = 18 → x = 18/99 = 2/11.
示例 2：x = 0.1̇8̇。100x = 18.1̇8̇。相减 99x = 18 → x = 18/99 = 2/11。
6. Powers and Roots | 幂与根
aᵐ)ⁿ = aᵐⁿ, a⁰ = 1, and a⁻ⁿ = 1/aⁿ. Roots are the inverse of powers. The square root of a, √a, is the positive number that gives a when squared. The cube root ∛a is the number that gives a when cubed.
幂（指数）表示重复相乘。3⁴ 表示 3 × 3 × 3 × 3 = 81。你需要熟练掌握指数运算法则：aᵐ × aⁿ = aᵐ⁺ⁿ，aᵐ ÷ aⁿ = aᵐ⁻ⁿ，(aᵐ)ⁿ = aᵐⁿ，a⁰ = 1，a⁻ⁿ = 1/aⁿ。方根是幂的逆运算。√a 表示平方后等于 a 的正数；立方根 ∛a 表示立方后等于 a 的数。
Key rules in words: ply indices. A negative index means the reciprocal, and fractional indices represent roots – a¹/² = √a, a¹/³ = ∛a.
用语言表述关键规则：同底数幂相乘，指数相加；相除则指数相减；幂的乘方，指数相乘。负指数表示倒数，分数指数表示根式——a¹/² = √a，a¹/³ = ∛a。
7. Standard Form | 标准形式
ass of an electron 0.000 000 000 000 000 000 000 000 000 000 91 kg becomes 9.1 × 10⁻³¹ kg.
标准形式将数字写作 A × 10ⁿ，其中 1 ≤ A < 10 且 n 为整数。这在处理科学中非常大或非常小的数字时特别有用。例如，光速 300 000 000 米/秒可以写为 3 × 10⁸，电子质量 0.000 000 000 000 000 000 000 000 000 000 91 千克则为 9.1 × 10⁻³¹ 千克。
parts and subtract the indices. Always check that your final A falls back into 1 ≤ A < 10.
标准形式的加、减法需要先将两个数写成相同的 10 的幂次。乘法：A 部分相乘，指数相加；除法：A 部分相除，指数相减。最后务必将 A 调整回 1 ≤ A < 10 的范围。
8. Order of Operations (BIDMAS) | 运算顺序
BIDMAS (or BODMAS) ensures everyone calculates complex expressions in the same order: Brackets, Indices, Division and Multiplication (left to right), Addition and Subtraction (left to right). This is one of the most frequently tested skills, often tucked inside other topics like substitution and equation solving.
运算顺序 BIDMAS（或 BODMAS）保证了每个人计算复杂表达式时都遵循相同顺序：括号（Brackets）、指数（Indices）、除法与乘法（从左到右）、加法与减法（从左到右）。这是考试中最高频的考点之一，经常隐藏在代值代入、解方程等题目中。
Example: acket: 3² − 1 = 9 − 1 = 8. Then multiplication: 2 × 8 = 16. Finally addition: 8 + 16 = 24.
示例：计算 8 + 2 × (3² − 1)。先算括号：3² − 1 = 9 − 1 = 8。然后乘法：2 × 8 = 16。最后加法：8 + 16 = 24。
| Order | Meaning | Example |
| B | Brackets (parentheses) | (4 + 2) first |
| I | Indices (powers, roots) | 3², √9 |
| DM | Division & Multiplication (L→R) | 10 ÷ 2 × 3 = 15 |
| AS | Addition & Subtraction (L→R) | 7 − 2 + 5 = 10 |
9. Rounding and Estimation | 近似与估算
Rounding means reducing the number of digits in a number while keeping its value close to the original. You can round to a given number of decimal places (d.p.) or significant figures (s.f.). Estimation uses rounding to make rough calculations that check the reasonableness of exact answers. Edexcel often asks candidates to estimate before calculating accurately to catch large errors.
近似是指减少数字的位数同时使其值接近原数。你可以将数字精确到指定的小数位数（d.p.）或有效数字（s.f.）。估算利用近似进行粗略计算，用以检验精确答案的合理性。Edexcel 题目常常要求先估算再精确计算，以帮助考生发现大的失误。
Rules for significant figures: The first non-zero digit is the first significant figure. For 0.0040602, 4 is the first s.f., and two s.f. gives 0.0041 (rounding up). For 12 345 to 3 s.f., write 12 300 (12 300).
有效数字规则：第一个非零数字为第一个有效数字。对于 0.0040602，4 是第一个有效数字，保留两位有效数字结果为 0.0041（四舍五入）。将 12 345 精确到 3 位有效数字则写为 12 300。
10. Upper and Lower Bounds | 上界与下界
When a measurement is given to a certain unit or degree of accuracy, the true value lies within a half-unit interval. For a length given as 6.4 cm to the nearest millimetre, the lower bound is 6.35 cm and the upper bound is 6.45 cm. Bounds become crucial in calculation questions: when adding lengths, add the lower bounds for a minimum and upper bounds for a maximum. For multiplication, use the extremes to find the range of possible answers.
当一个测量值给出到特定单位或精度时，真实值位于一个半单位区间内。对于精确到毫米的长度 6.4 cm，下界是 6.35 cm，上界是 6.45 cm。在计算题中边界值至关重要：长度相加时，用下界之和求最小值，用上界之和求最大值。涉及乘法时，需要利用极值找出可能答案的范围。
Example: ea. Upper bounds: length 7.05 cm, width 4.05 cm. Maximum area = 7.05 × 4.05 = 28.5525 cm².
示例：一个矩形长为 7.0 cm，宽为 4.0 cm，均精确到一位小数。求面积的上界。上界：长 7.05 cm，宽 4.05 cm。最大面积 = 7.05 × 4.05 = 28.5525 cm²。
11. Ratio and Proportion Basics | 比与比例基础
Ratio compares the sizes of two or more quantities. It is written in the form a : b and is simplified just like a fraction. If a sum of money is divided in the ratio 3 : 5, the total parts are 8, so the shares are 3/8 and 5/8 of the total. Proportion means that two ratios are equal. Understanding direct proportion enables you to solve problems such as 'If 5 apples cost £1.20, how much do 8 apples cost?' by using the unitary method or multiplying both sides of the ratio.
比用于比较两个或多个量的大小，写成 a : b 的形式并像分数一样化简。如果一笔钱按 3 : 5 分配，总份数为 8，因此两人分别获得总数的 3/8 和 5/8。比例表示两个比相等。理解正比例关系可以帮助你解决诸如“5 个苹果 1.20 英镑，8 个苹果多少钱？”的问题，方法可以是计算出单价再相乘，或者对比两边同时缩放。
the given ratio parts, sum them, and then solve for the multiplier. For example, to divide £60 in the ratio 2:3:5, total parts = 10, one part = £6, so shares are £12, £18 and £30.
按给定比值分配总量是一项核心技能。写出各部分份数对应的倍数，求和，然后计算出每份对应的实际量。例如，将 60 英镑按 2:3:5 分配，总份数为 10，一份为 6 英镑，因此各部分为 12 英镑、18 英镑和 30 英镑。
12. Percentage Basics | 百分数基础
Percentage means 'per hundred' 100. You can convert between fractions, decimals and percentages fluently: 3/5 = 0.6 = 60%. Calculate a percentage of an amount by multiplying, e.g., 15% r decrease is found by dividing the change by the original value, then multiplying by 100.
百分数意为“每一百”，是将数量表示为 100 的分数的一种方法。你需要熟练进行分数、小数和百分数之间的转换：3/5 = 0.6 = 60%。求一个数量的百分之几用乘法，如 200 的 15% = 15/100 × 200 = 30。求百分数增加或减少，用变化量除以原值再乘 100。
Example: A jacket originally priced £80 is reduced by 25%. Find the new price. Discount = 25% of £80 = £20, so new price = £60. Alternatively, multiplier method: 0.75 × £80 = £60.
示例：一件夹克原价 80 英镑，降价 25%。求新价格。折扣 = 80 的 25% = 20 英镑，因此新价格 60 英镑。或用倍增因子法：0.75 × £80 = £60。
ing backwards to find the original amount before a percentage change. If the sale price is £60 after a 25% reduction, the original is £60 ÷ 0.75 = £80.
反向百分数则需反向推算出在百分变化之前的原数量。如果某商品降价 25% 后售价 60 英镑，则原价 = 60 ÷ 0.75 = 80 英镑。
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