Upper and Lower Bounds | 上界与下界

📚 Upper and Lower Bounds | 上界与下界

When a measurement is rounded, the exact value lies within a range of possible values. The smallest value that would round up to the given measurement is called the lower bound, and the largest value that would round down to it is called the upper bound. Understanding these bounds is essential for calculating the greatest and least possible results when performing operations on rounded numbers.

当一个测量值经过四舍五入后，真实值必然落在某一个可能区间内。能四舍五入为该测量值的最小值称为下界，而能四舍五入为该测量值的最大值称为上界。理解界限对于计算近似数进行运算后的最大和最小可能结果至关重要。

1. What Are Bounds? | 什么是界限？

Suppose a length is given as 5 cm to the nearest centimetre. The actual length could be as low as 4.5 cm and as high as 5.5 cm (but not including 5.5 cm). Here, 4.5 cm is the lower bound and 5.5 cm is the upper bound. The notation for this range is 4.5 ≤ x < 5.5.

假设一个长度被四舍五入到最近的整数厘米，结果为 5 cm。实际长度可能低至 4.5 cm，高至 5.5 cm（但不包括 5.5 cm）。这里 4.5 cm 是下界，5.5 cm 是上界。该区间的记法为 4.5 ≤ x < 5.5。

The difference between the upper bound and the lower bound is called the length of the interval. For 5 cm rounded to the nearest cm, the interval length is 1 cm, which is exactly the rounding accuracy.

上界与下界之差称为区间长度。对于四舍五入到最近厘米的 5 cm，区间长度为 1 cm，正好等于舍入精度。

2. Error Intervals | 误差区间

An error interval is the range of possible values within which the exact value must lie. It is written as an inequality. For a value rounded to the nearest 10, e.g., 80, the error interval is 75 ≤ x < 85.

误差区间是指真实值必须落在的可能值范围，用不等式表示。例如，一个数四舍五入到最近的 10 得到 80，则其误差区间为 75 ≤ x < 85。

Rounded to the nearest whole number: the lower bound is half a unit below, the upper bound is half a unit above.
Rounded to a given number of decimal places: the lower and upper bounds are ± half of the place value.
Rounded to a given number of significant figures: similar to decimal places, using the place value of the last retained digit.

四舍五入到最近的整数：下界为数值下方半个单位，上界为上方半个单位。
四舍五入到指定小数位数：上下界为 ± 该位值的一半。
四舍五入到指定有效数字：与小数位数类似，以最后保留数字的位值为基准。

3. Finding Bounds from Rounded Values | 从舍入值求上下界

To determine the bounds, identify the degree of accuracy first. If a number is rounded to 1 decimal place, the accuracy is 0.1. The lower bound is value – 0.05 and the upper bound is value + 0.05. For example, 7.3 cm rounded to 1 d.p. has lower bound 7.25 cm and upper bound 7.35 cm.

求上下界时，首先要确定精确度。如果一个数保留 1 位小数，其精度为 0.1，下界为数值减去 0.05，上界为数值加上 0.05。例如，7.3 cm 保留 1 位小数时，下界为 7.25 cm，上界为 7.35 cm。

When rounding to significant figures, the same principle applies. For example, 6400 rounded to 2 significant figures has accuracy 100 (since the last retained digit is in the hundreds place). Thus lower bound is 6350 and upper bound is 6450.

当按有效数字舍入时，原理相同。例如，6400 保留 2 位有效数字，其精度为 100（因为最后保留的数字位于百位）。因此下界为 6350，上界为 6450。

4. Bounds in Addition and Subtraction | 加减法中的界限

When adding two quantities, the greatest possible sum is obtained by adding their upper bounds. The least possible sum is obtained by adding their lower bounds.

两个量相加时，最大可能和等于上界相加；最小可能和等于下界相加。

For subtraction, the greatest possible difference is upper bound of the first number minus lower bound of the second number. The least possible difference is lower bound of the first number minus upper bound of the second number.

对于减法，最大可能差为第一个数的上界减去第二个数的下界；最小可能差为第一个数的下界减去第二个数的上界。

example: a = 5.2 (to 1 d.p.), b = 3.7 (to 1 d.p.)
a + b maximum = 5.25 + 3.75 = 9.00
a − b maximum = 5.25 − 3.65 = 1.60

例子：a = 5.2（保留1位小数），b = 3.7（保留1位小数）
a + b 最大值 = 5.25 + 3.75 = 9.00
a − b 最大值 = 5.25 − 3.65 = 1.60

5. Bounds in Multiplication and Division | 乘除法中的界限

For multiplication, the maximum product is the product of the upper bounds, while the minimum product is the product of the lower bounds. This is true for positive numbers, which is the usual case in IGCSE problems.

乘法中，最大乘积等于上界相乘，最小乘积等于下界相乘。对于正数而言（IGCSE 中通常如此），这一结论成立。

For division, the maximum quotient is obtained by dividing the upper bound of the numerator by the lower bound of the denominator. The minimum quotient is obtained by dividing the lower bound of the numerator by the upper bound of the denominator.

除法中，最大商等于分子上界除以分母下界；最小商等于分子下界除以上界。

Operation 运算	Maximum 最大值	Minimum 最小值
a × b	U(a) × U(b)	L(a) × L(b)
a ÷ b	U(a) ÷ L(b)	L(a) ÷ U(b)

Where U(a) and L(a) represent the upper and lower bounds of a.

其中 U(a) 与 L(a) 分别表示 a 的上界与下界。

6. Bounds for Powers and Roots | 幂与根中的界限

For a positive number raised to a power n, the maximum value is (upper bound)ⁿ and the minimum value is (lower bound)ⁿ. For square roots, the maximum root is the square root of the upper bound, and the minimum root is the square root of the lower bound.

对于正数的 n 次幂，最大值为（上界）ⁿ，最小值为（下界）ⁿ。对于平方根，最大根为对应上界的平方根，最小根为对应下界的平方根。

Be careful when n is even and the value can be negative: the analysis changes. However, in IGCSE bounds questions, quantities such as length, area or volume are always positive.

当指数 n 为偶数且数值可能为负时需小心，分析会有所不同。然而，IGCSE 界限问题中的长度、面积或体积等量始终为正。

7. Applying Bounds to Real-World Problems | 界限在实际问题中的应用

Bounds are often used to find the maximum and minimum possible values of formulas involving measurements. For example, a rectangular field measures 12 m by 8 m, both to the nearest metre. The maximum area is obtained from the upper bounds of both sides: 12.5 × 8.5 = 106.25 m². The minimum area is 11.5 × 7.5 = 86.25 m².

界限常用于求包含测量值的公式的最大和最小可能结果。例如，一块矩形田地测得长 12 m、宽 8 m（均精确到米）。最大面积由上界计算：12.5 × 8.5 = 106.25 m²；最小面积由下界计算：11.5 × 7.5 = 86.25 m²。

Another typical problem involves speed. If a car travels 200 km to the nearest 10 km in 4 hours to the nearest hour, the maximum speed is 205 km ÷ 3.5 h = 58.6 km/h, and the minimum speed is 195 km ÷ 4.5 h = 43.3 km/h.

另一个典型问题是速度。若一辆汽车行驶 200 km（精确到 10 km）用时 4 小时（精确到小时），则最大速度为 205 km ÷ 3.5 h = 58.6 km/h，最小速度为 195 km ÷ 4.5 h = 43.3 km/h。

8. Common Mistakes and Tips | 常见错误与技巧

Mistake 1: Using the given rounded value directly in subsequent calculations. Always use bounds when the question asks for maximum or minimum.
Mistake 2: Forgetting that the upper bound is not included in the error interval. Write ≤ for the lower bound and < for the upper bound.
Mistake 3: In division, switching the upper and lower bounds. Remember: upper ÷ lower gives maximum, lower ÷ upper gives minimum.

错误 1：在后续计算中直接使用舍入值。当问题要求最大或最小时，必须使用上下界。
错误 2：忘记误差区间的上界不包括实际值。下界用 ≤，上界用 <。
错误 3：除法中混淆上下界。记住：上界 ÷ 下界得到最大值，下界 ÷ 上界得到最小值。

Always write down the error interval before starting a calculation. This helps avoid sign errors and ensures you know which bound to use.

在计算之前先写出误差区间，这有助于避免符号错误，并确保你知道该使用哪个边界。

9. Worked Exam-Style Example | 考试风格例题详解

A cylinder has a height of 10.0 cm and a radius of 4.0 cm, both measured to 1 decimal place. Calculate the lower and upper bounds of the volume. (Use π = 3.14)

一个圆柱体的高为 10.0 cm，半径为 4.0 cm，均保留 1 位小数。计算体积的下界与上界。（取 π = 3.14）

Step 1: Identify bounds. Height: 9.95 ≤ h < 10.05. Radius: 3.95 ≤ r < 4.05.

步骤 1：确定上下界。高：9.95 ≤ h < 10.05。半径：3.95 ≤ r < 4.05。

Step 2: Volume formula is V = πr²h. For the lower bound, use the smallest radius and height: V_min = 3.14 × 3.95² × 9.95 = 3.14 × 15.6025 × 9.95 ≈ 487.9 cm³.

步骤 2：体积公式为 V = πr²h。计算下界时使用最小半径和高：V_min = 3.14 × 3.95² × 9.95 = 3.14 × 15.6025 × 9.95 ≈ 487.9 cm³。

Step 3: For the upper bound, use the largest radius and height: V_max = 3.14 × 4.05² × 10.05 = 3.14 × 16.4025 × 10.05 ≈ 517.6 cm³.

步骤 3：计算上界时使用最大半径和高：V_max = 3.14 × 4.05² × 10.05 = 3.14 × 16.4025 × 10.05 ≈ 517.6 cm³。

Therefore, 487.9 cm³ ≤ V < 517.6 cm³ (using π = 3.14).

因此，487.9 cm³ ≤ V < 517.6 cm³（取 π = 3.14）。

10. Practice Questions | 练习题

A quantity x is measured as 6.4, correct to 2 significant figures. Write down the error interval.
p = 3.0 and q = 2.0, both to 1 decimal place. Find the maximum value of p/q.
The side of a square is 5 cm, measured to the nearest centimetre. Find the minimum possible perimeter.
A force F is given by F = ma, where m = 2.5 kg (2 s.f.) and a = 1.2 m/s² (2 s.f.). Find the maximum force.

量 x 测量得 6.4，精确到 2 位有效数字。写出其误差区间。
p = 3.0 与 q = 2.0，均保留 1 位小数。求 p/q 的最大值。
正方形的边长为 5 cm，精确到最近厘米。求最小可能周长。
力 F = ma，其中 m = 2.5 kg（2 位有效数字），a = 1.2 m/s²（2 位有效数字）。求最大力。

Answers: 1. 6.35 ≤ x < 6.45 2. 3.05 ÷ 1.95 ≈ 1.56 3. 4.5 × 4 = 18 cm 4. 2.55 × 1.25 = 3.1875 N

答案：1. 6.35 ≤ x < 6.45 2. 3.05 ÷ 1.95 ≈ 1.56 3. 4.5 × 4 = 18 cm 4. 2.55 × 1.25 = 3.1875 N

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Upper and Lower Bounds | 上界与下界