Upper and Lower Bounds | 上下界

📚 Upper and Lower Bounds | 上下界

In real life, measurements are never exact. When we round a number or give a measurement to a certain degree of accuracy, the true value could be slightly larger or slightly smaller. Understanding upper and lower bounds allows us to find the range in which the true value must lie, and to calculate the greatest or least possible results of calculations.

在现实生活中，测量永远不可能是完全精确的。当我们把一个数四舍五入到某个精度，或者给一个带单位的测量结果时，真实值可能比记录值稍大或稍小。理解上界和下界，能帮助我们找到真实值必定落入的范围，并计算一个计算结果可能出现的最大值或最小值。

1. What Are Bounds? | 什么是边界？

If a value x is given as a correct to a certain accuracy, then the lower bound (LB) is the smallest possible value that x could be, and the upper bound (UB) is the largest possible value that x could be before it would round to the next number.

如果某个值 x 以一定精度表示为 a，那么下界（LB）就是 x 可能取到的最小值，而上界（UB）则是 x 在四舍五入跳到下一个数之前可能取到的最大值。

For example, if a length is measured as 7.3 cm correct to 1 decimal place, then the true length L satisfies:

7.25 ≤ L < 7.35

Here the lower bound is 7.25 cm and the upper bound is 7.35 cm. The upper bound is not included because if the true length were exactly 7.35 cm, it would round to 7.4 cm, not 7.3 cm.

例如，一根长度测量为 7.3 cm，精确到 1 位小数，那么真实长度 L 满足：

7.25 ≤ L < 7.35

这里下界是 7.25 cm，上界是 7.35 cm。上界不被包含，因为如果真实长度恰好是 7.35 cm，它应该四舍五入为 7.4 cm，而不是 7.3 cm。

2. Rounding and Truncation | 四舍五入与截断

There is an important difference between rounding and truncating. When a number is truncated, the discarded digits are simply cut off, so the truncated value is always smaller than or equal to the original number (for positive numbers).

在数学中，四舍五入与截断有重要区别。截断是指直接去掉后面的数字，因此对于正数来说，截断后的值总是小于或等于原数。

For truncation to 1 decimal place, if a number is recorded as 7.3, then its true value lies in the interval:

7.3 ≤ x < 7.4

The lower bound is exactly 7.3, not 7.25, because any number from 7.3 up to (but not including) 7.4 would truncate to 7.3.

对于截断到 1 位小数的情况，如果一个数记录为 7.3，那么它的真实值落在区间：

7.3 ≤ x < 7.4

此时下界正好是 7.3，而不是 7.25，因为从 7.3 到小于 7.4 的任何数截断后都会是 7.3。

In IGCSE Edexcel questions, you must check whether the phrase “correct to” means rounding or whether the question explicitly says “truncated”.

在 IGCSE Edexcel 考试中，你需要注意题目说的是 “correct to”（四舍五入）还是明确给出 “truncated”（截断）。

3. Finding the Lower and Upper Bounds | 找出上下界

To find the bounds of a number rounded to a given place value, take half of the place value unit. Add this half-unit to the rounded value to get the upper bound, and subtract it from the rounded value to get the lower bound.

要找到一个四舍五入数的上下界，取该数位单位的二分之一。把这个半单位加上去得到上界，减下去得到下界。

Here are some common cases:

Given value	Accuracy	Lower Bound	Upper Bound
80 (to nearest 10)	nearest 10	75	85
4.5 (to 1 d.p.)	0.1	4.45	4.55
0.06 (to 2 d.p.)	0.01	0.055	0.065
2600 (to 2 s.f.)	100	2550	2650

Notice that for significant figures, the place value of the last significant digit determines the half-unit. For example, 2600 to 2 significant figures means the last significant digit is in the hundreds place, so the half-unit is 50.

注意，对于有效数字，最后一个有效数字所在的数位决定了半单位。例如，2600 精确到 2 位有效数字，意味着最后一个有效数字在百位，因此半单位是 50。

4. Bounds in Addition and Subtraction | 加减法中的边界

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

当两个测量值相加时，最大可能的和等于两个上界相加；最小可能的和等于两个下界相加。

For subtraction, the situation is reversed: the greatest possible difference is found by taking the upper bound of the first number minus the lower bound of the second number. The least possible difference is the lower bound of the first number minus the upper bound of the second.

对于减法，情况相反：最大差等于第一个数的上界减去第二个数的下界；最小差等于第一个数的下界减去第二个数的上界。

Example: Two lengths are 7.3 cm and 2.4 cm, each correct to 1 decimal place. The maximum possible perimeter of a rectangle using these lengths is:

2 × (7.35 + 2.45) = 2 × 9.80 = 19.60 cm

The minimum perimeter is:

2 × (7.25 + 2.35) = 2 × 9.60 = 19.20 cm

在加减法中，误差是线性相加的，所以不需要使用百分误差。

5. Bounds in Multiplication and Division | 乘除法中的边界

For multiplication of positive numbers, the maximum product is found by multiplying the two upper bounds. The minimum product is found by multiplying the two lower bounds.

对于正数的乘法，最大乘积等于两个上界相乘；最小乘积等于两个下界相乘。

For division, to maximise the quotient, divide the upper bound of the numerator by the lower bound of the denominator. To minimise the quotient, divide the lower bound of the numerator by the upper bound of the denominator.

对于除法，要得到最大商，需要用分子的上界除以分母的下界；要得到最小商，需要用分子的下界除以分母的上界。

Suppose a rectangle has length L = 5.2 cm and width W = 3.4 cm, both to 1 decimal place. Then:

Maximum area = 5.25 × 3.45 = 18.1125 cm²

最大面积 = 5.25 × 3.45 = 18.1125 cm²
Minimum area = 5.15 × 3.35 = 17.2525 cm²

最小面积 = 5.15 × 3.35 = 17.2525 cm²

For division, if a car travels a distance D = 200 km to the nearest 10 km in a time T = 4 hours to the nearest 0.1 hour, the maximum speed is:

(200 + 5) ÷ (4 − 0.05) = 205 ÷ 3.95 ≈ 51.9 km/h

而最小速度是：

(200 − 5) ÷ (4 + 0.05) = 195 ÷ 4.05 ≈ 48.1 km/h

6. Composite Calculations | 复合计算

Many exam questions involve more than one operation. In such cases, you must work step by step, deciding each intermediate result using the appropriate upper or lower bounds.

许多考试题目涉及不止一种运算。在这种情况下，你必须逐步计算，每一步都依据题目要求选择合适的上界或下界。

Example: The formula \( s = \frac{1}{2}at^2 \) is often used, but here we avoid LaTeX and write it as:

s = ½ × a × t²

If a = 9.8 m/s² (to 1 d.p.) and t = 2.1 s (to 1 d.p.), then the maximum value of s is:

½ × 9.85 × 2.15² = ½ × 9.85 × 4.6225 ≈ 22.8 m

The minimum value is:

½ × 9.75 × 2.05² = ½ × 9.75 × 4.2025 ≈ 20.5 m

Notice that for \( t^2 \), the upper bound of \( t^2 \) is obtained by squaring the upper bound of t, not by multiplying two different bounds.

注意，对于 t²，其上界是 t 的上界的平方，而不是两个不同边界的乘积。

7. Applied Problems: Speed, Area, Density | 应用问题：速度、面积、密度

Edexcel IGCSE often asks real-life bound questions. The key is to ask yourself: which bound makes the result bigger or smaller?

Edexcel IGCSE 经常考查结合生活实际的边界问题。关键在于问自己：取哪个边界的组合会让结果变得更大或更小？

Speed = distance ÷ time. Maximum speed = max distance ÷ min time.

速度 = 距离 ÷ 时间。最大速度 = 最大距离 ÷ 最小时间。
Area of rectangle = length × width. Maximum area = max length × max width.

矩形面积 = 长 × 宽。最大面积 = 最大长 × 最大宽。
Density = mass ÷ volume. Maximum density = max mass ÷ min volume.

密度 = 质量 ÷ 体积。最大密度 = 最大质量 ÷ 最小体积。

Always write down the interval at the start and label the bounds clearly.

开始计算前先写出每个量的取值范围，并清楚标注哪个是上界、哪个是下界。

8. Percentage Error and Bounds | 百分比误差与边界

If you know the absolute error, you can calculate the percentage error:

percentage error = absolute error ÷ measured value × 100%

当你知道绝对误差时，可以计算百分比误差：

百分比误差 = 绝对误差 ÷ 测量值 × 100%

For a value rounded to a certain precision, the absolute error is at most half the precision. Example: a mass of 5.6 kg measured to 1 decimal place has an absolute error of at most 0.05 kg, so the percentage error is at most:

(0.05 ÷ 5.6) × 100% ≈ 0.89%

对于四舍五入到一定精度的值，绝对误差最多是该精度的一半。例如，一个质量 5.6 kg 精确到 1 位小数，其绝对误差最大为 0.05 kg，所以百分比误差最大为：

(0.05 ÷ 5.6) × 100% ≈ 0.89%

For multiplication and division, the approximate maximum percentage error of the result is the sum of the percentage errors of the inputs. This is a handy check.

对于乘法和除法，结果的最大百分比误差约等于各输入量百分比误差之和。这是一个有用的检验方法。

9. Common Mistakes to Avoid | 常见错误与陷阱

Students often make the following mistakes:

同学们常犯以下错误：

Using the lower bound of a denominator when trying to minimise a quotient (correct: denominator must be the upper bound for minimising a quotient).

在求最小商时错误地使用了分母的下界（正确做法：要求最小商，分母应使用上界）。
Forgetting to halve the unit when the number is given to the nearest whole number, e.g. 50 to the nearest 10 has bounds 45 and 55, not 49 and 51.

当数给定到最近的整十时，忘记了取一半单位。例如 50 精确到十位，上下界是 45 和 55，不是 49 和 51。
Treating the upper bound as included. In standard rounding, the upper bound is usually not included.

把上界当作包含在内。在标准四舍五入中，上界通常不包含。
Applying the same bounds to subtraction as to addition. For subtraction, the maximum is upper – lower, not upper – upper.

在减法中套用了加法的边界规则。减法中最大差是上界减下界，而不是上界减上界。

Always pause to think logically: what makes the answer larger?

遇到边界问题，先停下来想一想：怎样的取值组合会使结果更大？

10. Exam Strategy | 考试策略

In the Edexcel IGCSE exam, bounds questions often appear in both Paper 1 and Paper 2. A clear method is essential.

在 Edexcel IGCSE 考试中，边界问题在试卷 1 和试卷 2 中都会出现。掌握清晰的解题步骤至关重要。

Write down the original value and its accuracy.

写下原数值和它的精度。
Determine the half-unit of the accuracy.

确定该精度对应的半单位。
Calculate LB = value – half-unit and UB = value + half-unit.

计算下界 = 数值 – 半单位，上界 = 数值 + 半单位。
Read the operation or formula required and select the appropriate bounds.

阅读题目要求的运算或公式，选择正确的边界组合。
Show the substitution clearly and give your final answer with the correct degree of accuracy.

把代入过程写清楚，最后用正确的精度给出答案。

Marks are awarded for identifying bounds, for combining them correctly, and for a suitable final answer.

分数会分别给到：正确识别边界、正确组合边界、以及写出合适的最终答案。

11. Worked Example | 典型例题

Let us work through a classic exam question:

我们通过一道经典考试题来演练：

“A rectangular field has length 120 m correct to the nearest 10 m and width 80 m correct to the nearest 10 m. Calculate the maximum possible area of the field.”

“一块矩形田地的长度为 120 m（精确到十米），宽度为 80 m（精确到十米）。计算这块田地的最大可能面积。”

Step 1: Bounds for length: 115 ≤ L < 125, so UB = 125 m.

第一步：长度的边界：115 ≤ L < 125，所以上界 = 125 m。

Step 2: Bounds for width: 75 ≤ W < 85, so UB = 85 m.

第二步：宽度的边界：75 ≤ W < 85，所以上界 = 85 m。

Step 3: Maximum area = 125 × 85 = 10625 m².

第三步：最大面积 = 125 × 85 = 10625 m²。

Notice that if the question asks for the minimum area, use 115 × 75 = 8625 m².

注意，如果题目要求最小面积，则用 115 × 75 = 8625 m²。

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Upper and Lower Bounds | 上下界