Bounds & Truncation
Bounds & Truncation (Higher)
When a measurement is rounded, the true value lies within a range. Bounds (also called error intervals) describe the smallest and largest values the true measurement could be.
Upper and Lower Bounds
When a value is rounded to a given degree of accuracy, the lower bound is the smallest value that rounds up to the given number, and the upper bound is the smallest value that would round up to the next number.
Rule: The bounds extend half a unit either side of the last rounded digit.
| Measurement | Accuracy | Lower Bound | Upper Bound |
|---|---|---|---|
| 7 cm | Nearest cm | 6.5 cm | 7.5 cm |
| 3.2 kg | 1 d.p. | 3.15 kg | 3.25 kg |
| 150 m | Nearest 10 | 145 m | 155 m |
| 2.46 s | 2 d.p. | 2.455 s | 2.465 s |
The true value v satisfies: lower bound ≤ v < upper bound
Note the strict inequality (<) on the upper bound — if the value equalled the upper bound, it would round up to the next number.
Truncation
Truncation means cutting off digits without rounding. For example, 3.78 truncated to 1 d.p. gives 3.7 (not 3.8).
For a truncated value, the error interval is different:
- 3.7 (truncated to 1 d.p.): the true value lies in 3.7 ≤ v < 3.8
The lower bound equals the truncated value itself, and the upper bound is one unit of the last digit higher.
Calculations with Bounds
When combining measurements, you must consider which bounds give the maximum or minimum result.
| Operation | Maximum Result | Minimum Result |
|---|---|---|
| A + B | Upper A + Upper B | Lower A + Lower B |
| A − B | Upper A − Lower B | Lower A − Upper B |
| A × B | Upper A × Upper B | Lower A × Lower B |
| A ÷ B | Upper A ÷ Lower B | Lower A ÷ Upper B |
Key insight: For subtraction and division, to get the biggest result you use the upper bound of the first value and the lower bound of the second (subtracting less or dividing by less gives more).
Worked Example: A rectangle measures 12.4 cm by 7.8 cm, both to 1 d.p. Find the upper and lower bounds of the area.
- Length: LB = 12.35, UB = 12.45
- Width: LB = 7.75, UB = 7.85
- Maximum area = 12.45 × 7.85 = 97.7325 cm²
- Minimum area = 12.35 × 7.75 = 95.7125 cm²
Worked Example: A car travels 180 miles (nearest 10) in 2.5 hours (nearest 0.1). Find the bounds of its average speed.
- Distance: LB = 175, UB = 185
- Time: LB = 2.45, UB = 2.55
- Maximum speed = 185 ÷ 2.45 = 75.5 mph (to 1 d.p.)
- Minimum speed = 175 ÷ 2.55 = 68.6 mph (to 1 d.p.)
Suitable Degree of Accuracy
A common exam question asks you to use bounds to determine how many decimal places or significant figures you can state a result to with certainty.
Method: Calculate the upper and lower bounds, then find where they first disagree.
Worked Example: Using the car example above (speed between 68.6 and 75.5), can we state the speed to 1 s.f.?
- Lower bound rounds to 70 (1 s.f.)
- Upper bound rounds to 80 (1 s.f.)
- They disagree at 1 s.f., so we cannot state the speed to any degree of accuracy with certainty.
Exam Tips
- Write error intervals using correct notation: 6.5 ≤ x < 7.5 (not 6.5 ≤ x ≤ 7.5)
- For division, remember the counterintuitive rule: divide by the LOWER bound to get the MAXIMUM
- Questions about truncation have a different error interval from rounding — read carefully which one is stated