IB Maths AI SL Topic 1 โ€” Number & Algebra Paper 1 & 2 Half-unit rule ~7 min read

Upper & Lower Bounds

Bounds are rounding in reverse. If someone tells you a length is 5.7 cm (to 1 d.p.), what could the TRUE value have been? Anything from 5.65 cm up to (but not quite) 5.75 cm โ€” those are the lower bound (LB) and upper bound (UB). Same idea as rounding: just go HALF the rounding unit either side of the given value.

๐Ÿ“˜ What you need to know

What are upper and lower bounds?

When a number has been rounded, you only know the rounded answer โ€” not the original. The original could have been any value that would round to what you were told. The full range of possible “originals” is the interval from the lower bound to the upper bound.

The half-unit rule LB = value โˆ’ unit2   ยท   UB = value + unit2
Bounds of 5.7 cm (given to 1 d.p.) 5.6 [ 5.65 LB (included) 5.7 GIVEN VALUE ) 5.75 UB (not included) 5.8 half-unit = 0.05 half-unit = 0.05 5.65 โ‰ค true value < 5.75
Any true value in the teal interval would round to 5.7. The LB (5.65) is included โ€” anything exactly 5.65 would round UP to 5.7. The UB (5.75) is NOT included โ€” anything 5.75 or above rounds UP to 5.8.

Same idea for any accuracy. Just identify the rounding unit first, then take half of it. A few quick examples:

Combining bounds โ€” sums, differences, products

When two rounded quantities are combined, the answer has its own bounds. The trick is figuring out which combination gives you the biggest possible answer and which gives the smallest. For sums and products it’s intuitive; for differences and quotients it’s not โ€” bounds work AGAINST each other.

OperationMax possibleMin possible
Sum (a + b)UBa + UBbLBa + LBb
Difference (a โˆ’ b)UBa โˆ’ LBbLBa โˆ’ UBb
Product (a ร— b)UBa ร— UBbLBa ร— LBb
Quotient (a รท b)UBa รท LBbLBa รท UBb
Why differences flip: to make a โˆ’ b as BIG as possible, you want a as big as it can be AND b as small as it can be. So max difference uses UBa and LBb. Same logic for quotient: to make a รท b biggest, make the top big AND the bottom small. (All assuming positive numbers โ€” which AI SL problems usually are.)

๐Ÿงญ Recipe โ€” finding bounds of any combined quantity

  1. Find the rounding unit for each given value: look at the last accurate digit. 1 d.p. โ†’ unit 0.1; 2 d.p. โ†’ 0.01; “nearest 10” โ†’ 10; 3 s.f. on 470 โ†’ 1.
  2. Calculate the LB and UB for each value: value ยฑ half-unit. Write them out clearly.
  3. Identify the operation: is it a sum, difference, product, or quotient? This decides which bounds you combine.
  4. Apply the right combination: from the table above. For max difference take UB โˆ’ LB; for min difference take LB โˆ’ UB.
  5. Round only at the END: keep full precision in working, then round your final UB and LB to a sensible accuracy (often matching the given values’ accuracy).

Worked examples

WE 1

Basic bounds from 1 decimal place

The length of a pencil is measured as 5.7 cm to 1 decimal place. Find the lower and upper bounds, and write the bounds as an inequality.

Step 1 โ€” find the rounding unit 1 d.p. โ†’ unit = 0.1 half-unit = 0.05 Step 2 โ€” apply the half-unit rule LB = 5.7 โˆ’ 0.05 = 5.65 UB = 5.7 + 0.05 = 5.75 LB = 5.65 cm, UB = 5.75 cm inequality form: 5.65 โ‰ค L < 5.75. The lower bound IS included, the upper bound IS NOT (because 5.75 would round UP to 5.8).
WE 2

Bounds from significant figures

A mass is given as 230 g to 2 significant figures. Find the upper and lower bounds.

Step 1 โ€” find the rounding unit 230 to 2 s.f. means rounded to the nearest 10 (the last “significant” place is the tens column) unit = 10 โ†’ half-unit = 5 Step 2 โ€” apply the half-unit rule LB = 230 โˆ’ 5 = 225 UB = 230 + 5 = 235 LB = 225 g, UB = 235 g the trick for s.f. is identifying WHICH place the rounding happened to. For 230 (2 s.f.), the 3 is the last significant digit, in the tens column โ€” so the rounding is to the nearest 10. Not the nearest 1, not the nearest 100.
WE 3

Bounds from “to the nearest”

The population of a town is given as 4500 to the nearest 100. Find the lower and upper bounds.

Step 1 โ€” find the rounding unit (already stated) unit = 100 โ†’ half-unit = 50 Step 2 โ€” apply the half-unit rule LB = 4500 โˆ’ 50 = 4450 UB = 4500 + 50 = 4550 LB = 4450, UB = 4550 when the question says “to the nearest X”, you don’t have to figure out the unit โ€” they’ve told you. X is the unit. Just halve it and go.
WE 4

Sum bounds โ€” combined length

Two planks are measured as 24.6 cm and 18.3 cm, both to 1 decimal place. They are laid end to end. Find the maximum and minimum possible total length.

Step 1 โ€” find bounds for each plank (half-unit = 0.05) plank A: LB = 24.55, UB = 24.65 plank B: LB = 18.25, UB = 18.35 Step 2 โ€” combine for sum (max = UB + UB, min = LB + LB) MAX total = 24.65 + 18.35 = 43.00 cm MIN total = 24.55 + 18.25 = 42.80 cm max = 43.00 cm, min = 42.80 cm for sums, both bounds work the same way โ€” biggest + biggest = biggest, smallest + smallest = smallest. Easy to remember because it matches your intuition.
WE 5

Difference bounds โ€” race times

Mariam runs 100 m in 12.4 s. Daniyal runs 100 m in 13.1 s. Both times are recorded to 1 d.p. Find the maximum and minimum possible difference between their times (Daniyal โˆ’ Mariam).

Step 1 โ€” find bounds for each time (half-unit = 0.05) Mariam: LB = 12.35, UB = 12.45 Daniyal: LB = 13.05, UB = 13.15 Step 2 โ€” biggest difference: Daniyal as SLOW as possible, Mariam as FAST as possible MAX diff = UB_Daniyal โˆ’ LB_Mariam = 13.15 โˆ’ 12.35 = 0.80 s Step 3 โ€” smallest difference: Daniyal as FAST as possible, Mariam as SLOW as possible MIN diff = LB_Daniyal โˆ’ UB_Mariam = 13.05 โˆ’ 12.45 = 0.60 s max difference = 0.80 s, min difference = 0.60 s for differences, the bounds FLIP โ€” biggest of A minus SMALLEST of B gives the largest gap. If you used UB โˆ’ UB by mistake, you’d get 0.70 โ€” which is the difference of the rounded values, not a bound. Always pair UB with LB and vice versa.
WE 6

Product bounds โ€” rectangle area

A rectangle has length 8 cm and width 5 cm, each measured to the nearest centimetre. Find the upper and lower bounds for the area of the rectangle.

Step 1 โ€” find bounds for each side (half-unit = 0.5) length: LB = 7.5, UB = 8.5 width: LB = 4.5, UB = 5.5 Step 2 โ€” for products (positive values), max = UB ร— UB, min = LB ร— LB MAX area = UB_L ร— UB_W = 8.5 ร— 5.5 = 46.75 cmยฒ MIN area = LB_L ร— LB_W = 7.5 ร— 4.5 = 33.75 cmยฒ UB of area = 46.75 cmยฒ, LB of area = 33.75 cmยฒ notice how wide the gap is โ€” the “true” area could be anywhere from 33.75 to 46.75 cmยฒ, a 13 cmยฒ range, even though the side lengths were only 1 cm uncertain each. Errors COMPOUND when you multiply: a small uncertainty in each side becomes a much bigger uncertainty in the area. This is why precise measurements matter.

๐Ÿ’ก Top tips

โš  Common mistakes

Up next: Percentage Error. Once you know what the true value could be (bounds) and what the rounded answer actually is, the natural next question is “how far off is the rounded answer from the true value, as a percentage?” That’s percentage error โ€” a single number summarising the size of any rounding or measurement error.

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