Approximation and error is the skill that sits underneath every other topic in IB Mathematics Applications & Interpretation HL. AI is a modelling subject built on real, messy measurements, so almost every answer you write is an approximation — rounded, quoted in standard form, or read straight off a calculator — and every approximation carries an error. Because a GDC is always allowed, the marks are never in the arithmetic; they are in rounding correctly, quoting the right number of figures, and knowing how far the true value could really lie from the one you wrote down.
This cheatsheet condenses the whole of SL 1.6 — decimal places and significant figures, standard form, percentage error, and the upper and lower bounds of rounded data — then pushes it to HL depth by showing how those bounds propagate through sums, products and quotients, right up to judging a regression model or an Euler estimate by its percentage error. The traps that quietly cost accuracy marks are flagged throughout, and the print-ready PDF is at the bottom, free to download.
§1 — Rounding to decimal places SL 1.6
Approximation replaces an exact quantity with a nearby, easier value — by rounding to a set number of decimal places or significant figures, or by writing it in standard form. Rounding to decimal places is the most basic version: keep a fixed number of digits after the point.
Round to $n$ d.p.:keep $n$ digits after the decimal point; the very next digit decides — $\ge 5$ rounds up, $<5$ rounds down
Example:$5.6449\to 5.64$ (2 d.p.), because the third decimal is $4$
TrapRound in one step from the exact value using the single deciding digit. Do not chain-round from the far right ($5.6449\to5.645\to5.65$ is wrong — the deciding digit is the $4$, so it is $5.64$).
NoteOn a GDC, keep the full stored value through every intermediate step and round only the final answer. Rounding halfway and then reusing the rounded number is the classic way to lose an accuracy mark.
§2 — Significant figures SL 1.6
Significant figures count from the first non-zero digit, and IB reports final answers to 3 s.f. unless a question says otherwise. Leading zeros are placeholders — they are never significant.
First s.f.:the first non-zero digit; in $0.00408$ the first significant figure is the $4$
Round to $n$ s.f.:keep $n$ significant digits, use the next digit to round, then fill remaining whole-number places with zeros
Handy identity:for any value between $1$ and $10$, quoting 3 s.f. is identical to 2 d.p. — e.g. $6.7194\to 6.72$
TrapRounding to s.f. must preserve the size of the number. $48\,700$ to 2 s.f. is $49\,000$ — replace the dropped whole-number digits with zeros, do not delete them (writing $49$ changes the number by a factor of a thousand).
NoteFix the last kept digit by rounding with the next figure, never by chopping the number off. $6.798$ to 3 s.f. is $6.80$, not $6.79$.
§3 — Standard (scientific) form SL 1.1
Standard form writes any number as a single digit before the point times a power of ten — the compact way AI handles very large populations or very small tolerances.
Form:$a\times 10^{k}$ with $1\le a<10$ and $k$ an integer
Sign of $k$:$k$ counts the places the decimal point moves; large numbers give $k>0$, numbers below $1$ give $k<0$
Examples:$6\,720\,000=6.72\times10^{6}$; $\;0.00048=4.8\times10^{-4}$
TrickMove the point until exactly one non-zero digit sits in front of it, then count the moves for $k$: point moved left $\Rightarrow$ positive $k$; a small number (point moved right) $\Rightarrow$ negative $k$.
TrapThe coefficient must satisfy $1\le a<10$. $72\times10^{5}$ is not standard form — adjust it to $7.2\times10^{6}$. Significant figures of a small number start at the first non-zero digit, not at the leading zeros.
§4 — Percentage error SL 1.6
Percentage error measures how far an approximation $v_A$ falls from the exact (accepted) value $v_E$, as a percentage of the exact value.
$$\varepsilon=\left|\frac{v_A-v_E}{v_E}\right|\times100\%$$
Symbols:$v_A$ = approximate / measured value; $v_E$ = exact / accepted value
TrapAlways divide by the EXACT (accepted) value — it sits in the denominator, never the approximation. Dividing by the measured value instead is the single most common slip, and never forget the $\times100$.
NoteThe modulus makes percentage error always positive, so it does not matter whether your approximation is an over- or under-estimate — you report the size of the gap.
§5 — Upper & lower bounds from rounding SL 1.6
A rounded measurement does not fix a single true value — it fixes an interval of possible true values. Step half a unit each way to reach the bounds.
Half-unit rule:a value rounded to a unit $u$ could really be anything within $\pm\tfrac{u}{2}$
Interval:$v_A-\tfrac{u}{2}\le v_{\text{true}}
Examples:nearest $0.1\Rightarrow\pm0.05$; nearest whole cm $\Rightarrow\pm0.5$; nearest $10\text{ g}\Rightarrow\pm5\text{ g}$
TrickAdd the half-unit for the upper bound, subtract it for the lower bound. The true value can equal the lower bound but stays strictly below the upper bound.
TrapThe unit $u$ is the last place quoted. A length given as $1.5$ m (1 d.p.) has $u=0.1$, so the swing is $\pm0.05$ — not $\pm0.5$. Read the precision off the last digit, not the first.
§6 — Propagating bounds through sums & products SL 1.6
A quantity built from rounded measurements — an area, a perimeter, a total — has its own bounds. Find each measurement's bounds first, then push the extremes through the formula.
Sum:maximum = sum of the upper bounds; minimum = sum of the lower bounds. Perimeter $P=2(L+W)$ is largest at $2(L_{\max}+W_{\max})$
Product:for positive quantities, maximum $=$ largest $\times$ largest; minimum $=$ smallest $\times$ smallest. Area $=L\times W$ is largest at $L_{\max}\times W_{\max}$
TrickGet each side's upper and lower bound first, then combine — never round back to the given figures partway through, or the compounded error disappears.
NoteA difference flips one bound: for $a-b$ the maximum uses $a$'s upper bound and $b$'s lower bound, $\ (a-b)_{\max}=a_{\max}-b_{\min}$. Any quantity you subtract contributes its opposite bound.
§7 — Propagating bounds through quotients SL 1.6
Rates and densities are quotients — speed $=\dfrac{d}{t}$, density $\rho=\dfrac{m}{V}$ — and a bound in the denominator flips its role, because dividing by a smaller number makes the result larger.
Maximum quotient:largest numerator $\div$ smallest denominator — max over min
Minimum quotient:smallest numerator $\div$ largest denominator
TrapTo make a fraction large, make the bottom small. Feeding both quantities' upper bounds into a quotient is wrong — the denominator's bounds swap, so the maximum density uses $m_{\max}$ over $V_{\min}$.
NoteA mass quoted "to the nearest $10$ g" carries bounds of $\pm5$ g and a volume "to the nearest cm³" carries $\pm0.5$ cm³ — build each interval first, then combine max-over-min.
§8 — Application: judging a model by percentage error SL 1.6
An AI regression or growth model returns a prediction, never the exact truth. To judge how good the model is, treat its output as the approximate value and the measured data as the exact value, then report the percentage error.
Exponential model:for $V=P\,b^{\,t}$ the estimate at time $t$ is $P b^{\,t}$
$$\varepsilon=\left|\frac{P\,b^{\,t}-V_{\text{actual}}}{V_{\text{actual}}}\right|\times100\%$$
TrapThe model's prediction is the approximation; the measured value is treated as exact, so it goes in the denominator. Evaluate $b^{\,t}$ with the full power $t$, not $t-1$.
NoteA good model keeps the percentage error small across all the data points. A percentage error that is consistently one-signed (always over- or always under-shooting) is a sign the wrong type of model was fitted.
§9 — Euler's method: estimate & shrinking error AHL 5.16
Euler's method steps a differential equation forward in fixed steps of size $h$, producing an approximate solution whose error you can quantify against the exact curve.
Euler step:$y_{i+1}=y_i+h\,f(x_i,y_i)$ — follow the tangent for one step of width $h$
Error vs exact:for growth $\dfrac{dy}{dt}=ky$ compare with the exact solution $y_0e^{kt}$, then $\varepsilon=\left|\dfrac{y_{\text{Euler}}-y_0e^{kt}}{y_0e^{kt}}\right|\times100\%$
TrapFor exponential growth Euler undershoots the true curve — the tangent lines fall below the accelerating curve. Divide the error by the exact solution $y_0e^{kt}$, the value in the denominator.
TrickReducing the step $h$ shrinks the percentage error: roughly halving $h$ roughly halves the error. Smaller steps mean a closer estimate at the cost of more calculation.
§10 — Exam attack plan All sections
| Question cue | What to do | Watch for |
|---|
| "Round to $n$ d.p. / s.f." | Keep $n$ digits, round with the next digit | Don't chain-round; fill zeros to keep the size |
| "Write in standard form" | $a\times10^{k}$ with $1\le a<10$ | Numbers below $1$ give $k<0$; fix $a$ into range |
| "Percentage error" | $\dfrac{|v_A-v_E|}{|v_E|}\times100$ | Exact value in the denominator; keep the $\times100$ |
| "Upper / lower bound of a measurement" | Add / subtract $\tfrac12$ of the last-place unit | Unit is the last place quoted; upper is exclusive |
| "Max / min area (product)" | $L_{\max}\times W_{\max}$ and $L_{\min}\times W_{\min}$ | Find each side's bounds first, then multiply |
| "Max / min perimeter or total (sum)" | Sum the uppers / sum the lowers | Don't round back to the given figures midway |
| "Max / min speed or density (quotient)" | Biggest top $\div$ smallest bottom (max over min) | The denominator's bounds flip |
| "Percentage error of a model / Euler" | Prediction = approx, measured data = exact | Measured value in the denominator; Euler undershoots growth |
Worked Example — AI HL-Style Bounds & Percentage Error
Question (AI HL Paper 2 style — 6 marks)
A rectangular solar panel is measured as length $L=1.6$ m and width $W=0.9$ m, each rounded to the nearest $0.1$ m. An engineer models the daily energy output as proportional to the panel's area.
(a) Write down the lower and upper bounds of $L$ and of $W$.
(b) Hence find the lower and upper bounds of the panel's area.
(c) In fact the panel's true area is its upper bound. Find the percentage error in using the nominal area $1.6\times0.9=1.44\text{ m}^2$ as an approximation.
Solution
- (a) Each side is rounded to the nearest $0.1$ m, so the half-unit is $0.05$ m. Stepping half a unit each way: $1.55\le L<1.65$ and $0.85\le W<0.95$. (A1)(A1) (the unit is the last place quoted, $0.1$, so the swing is $\pm0.05$.)
- (b) Area is a product, so the maximum uses each side's upper bound and the minimum uses each lower bound: $A_{\max}=1.65\times0.95=1.5675\text{ m}^2$ and $A_{\min}=1.55\times0.85=1.3175\text{ m}^2$. (M1)(A1) So $1.3175\le A<1.5675$ (m²).
- (c) Here the exact value is the upper bound $v_E=1.5675$ and the approximation is $v_A=1.44$: $\varepsilon=\left|\dfrac{1.44-1.5675}{1.5675}\right|\times100=\dfrac{0.1275}{1.5675}\times100=8.13\%$ (3 s.f.). (M1)(A1)
Examiner's note: for the area, push each side to its own bound before multiplying — the maximum area is $1.65\times0.95$, never the nominal $1.44$ nudged up. In part (c) the exact value ($1.5675$) goes in the denominator, not the approximation, and the answer keeps its units and 3 s.f. Do not forget the $\times100$, and quote area in $\text{m}^2$.
Common Student Questions
What is the difference between decimal places and significant figures?
Decimal places are counted from the decimal point; significant figures are counted from the first non-zero digit. So $0.00408$ has $5$ decimal places but only $3$ significant figures (the $4$, $0$ and $8$). The one place they coincide: for a number between $1$ and $10$, quoting 3 s.f. is exactly the same as 2 d.p. — for example $6.7194$ becomes $6.72$ either way.
In percentage error, do I divide by the exact value or the approximation?
Always by the exact (accepted) value. The formula is $\dfrac{|v_A-v_E|}{|v_E|}\times100$, and the accepted value $v_E$ sits in the denominator every time. Dividing by the measured or approximate value instead is the classic error that loses the mark. If the question gives a "true" or "accepted" figure, that is the one on the bottom.
A length is "12.0 cm to 1 decimal place" — what are its bounds?
The unit is the last place quoted, here $0.1$ cm, so the swing is half of that, $\pm0.05$ cm. The bounds are therefore $11.95\le x<12.05$. Note the upper bound is strict: the true value can equal the lower bound ($11.95$) but must stay below the upper bound ($12.05$), because a value of exactly $12.05$ would round up to $12.1$.
For the area of a rectangle from rounded sides, why isn't the maximum just the nominal area rounded up?
Because the error compounds through the product. You must push each side to its own upper bound first, then multiply: $A_{\max}=L_{\max}\times W_{\max}$. For sides $1.6$ and $0.9$ (to $0.1$ m) that is $1.65\times0.95=1.5675$, which is larger than the nominal $1.44$ by more than either side alone would suggest. Rounding the sides back to their given values partway through throws the compounded uncertainty away.
Does Euler's method over- or under-estimate the true solution?
For exponential growth it under-estimates: each step follows the tangent, which lies below the accelerating curve, so Euler's estimate falls short of $y_0e^{kt}$. To measure how far short, use percentage error and divide by the exact solution. Reducing the step size $h$ shrinks the error — roughly halving $h$ roughly halves the error — at the cost of doing more steps.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
- Notes PDF — every rounding, standard-form and bounds method in full, with worked edge cases and the exact GDC keystrokes for keeping full accuracy.
- Tutorial booklet — 25+ AI HL-style questions from basic rounding to full bounds-propagation and model-error applications.
- Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations for every percentage-error and bounds method.
- Predicted-paper questions — the exact approximation, bounds and percentage-error question types most likely in the next session.
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