Upper and Lower Bounds (AI SL)
Whenever a measurement has been rounded, the true value could be anywhere in a small window either side of the number you're given. This page covers how to find that window - the lower bound and upper bound - and how to carry those bounds through a calculation, with worked examples and the mistakes that lose the most marks. It's part of the broader Approximation & Error topic.
13 questions on this sub-topic.
The bounds rule
Covered under IB syllabus reference SL1.6: upper and lower bounds of rounded numbers - e.g. if \(x = 4.1\) to one decimal place, then \(4.05 \le x < 4.15\). This isn't in the formula booklet, but it follows directly from what rounding means, so it's worth knowing cold.
Lower bound
lower bound \(= x - \tfrac12 \times\) (place value)
The smallest value that would still round to \(x\). E.g. \(12.5\) cm to the nearest \(0.1\) cm has lower bound \(12.5 - 0.05 = 12.45\) cm.
Upper bound
upper bound \(= x + \tfrac12 \times\) (place value)
The largest value that would still round to \(x\), never actually reached. So the true value \(\ell\) satisfies \(x - \tfrac12(\text{unit}) \le \ell < x + \tfrac12(\text{unit})\).
Need the full syllabus wording and formula-booklet reference table? See Approximation & Error. For GDC keystrokes on bounds questions, see the parent topic's GDC guidance.
Worked examples
A length is given as \(12.5\) cm, rounded to the nearest 0.1 cm.
(a) Find the lower bound.
(b) Find the upper bound.
Worked solution
(a) ‘Nearest \(0.1\) cm’ means the true value lies within half a unit, i.e. \(\pm0.05\) cm, of \(12.5.\) M1
Lower bound \(= 12.5 - 0.05 = 12.45\) cm. A1
(b) Upper bound \(= 12.5 + 0.05 = 12.55\) cm. A1 So the true length \(\ell\) satisfies \(12.45 \le \ell < 12.55.\) A1
Two readings are \(A = 50\) and \(B = 30\), each to the nearest whole number.
(a)(i) State the bounds for \(A\).
(a)(ii) State the bounds for \(B\).
(b) Find the maximum possible value of \(A - B\).
Worked solution
(a) Step 1 - Half-unit is \(0.5\) (nearest whole number). \(A \in [49.5, 50.5)\), \(B \in [29.5, 30.5).\) M1 A1
(b) Step 1 - Maximise \(A-B\). A difference is largest when the first term is as big as possible and the subtracted term as small as possible: use \(\max A\) and \(\min B\). R1
\(50.5 - 29.5 = 21.\) A1
Common mistakes
- Using the wrong half-unit. The half-unit comes from the place value you rounded to, not from the number itself - a value given "to the nearest 10" has half-unit \(5\), not \(0.5\).
- Writing the upper bound with \(\le\) instead of \(<\). A value exactly at the upper bound would have rounded up to the next unit, so the bounds are \(x - \tfrac12(\text{unit}) \le v < x + \tfrac12(\text{unit})\) - strict on the top, not strict on the bottom.
- Pairing the wrong bounds in a compound calculation. To maximise a sum or product, pair the upper bounds together; to maximise a difference \(A - B\), pair the upper bound of \(A\) with the lower bound of \(B\) - swapping this pairing gives the minimum instead.
Ready to practise properly?
13 upper-and-lower-bounds questions, marked instantly like the real exam.
Quick answers
How do you find the upper and lower bound of a rounded number?
Take half of the rounding unit and subtract it for the lower bound, add it for the upper bound. A value of \(4.1\) rounded to 1 decimal place has bounds \(4.05 \le x < 4.15\).
Why is the upper bound written with a strict inequality?
Because a value exactly at the upper bound would round up to the next unit instead, so the true value can get arbitrarily close to the upper bound but never quite reach it: \(x - \tfrac12(\text{unit}) \le x < x + \tfrac12(\text{unit})\).