Secondary Mathematics: Approximation, Significant Figures, Bounds and Error Intervals

Secondary Mathematics · Worked Repair Guide 32

Rounded numbers are not exact points. They represent intervals of possible original values. Once that is understood, significant figures, upper and lower bounds, maximum error and guaranteed conclusions become one connected topic rather than several isolated rules.

This guide develops one central habit: identify the rounding unit before constructing the interval. A value rounded to the nearest 0.1 has a half-step of0.05. A value rounded to the nearest10 has a half-step of5. Those half-steps determine the lower and upper boundaries.

All measurements and numerical contexts below are invented teaching examples. This worked guide complements Singapore School Mathematics: Rounded Data, Thresholds and Guaranteed Conclusions, whose job is broader reasoning about decisions under approximate inputs.

1. Approximation should be marked as approximation

The fraction 1/3 is exact. The decimal0.333 is a finite approximation to 1/3.

Therefore write 1/3≈0.333 when0.333 has been rounded to three decimal places.

Entry check: π≈3.142 to three decimal places. The equals sign would falsely claim π is exactly the terminating decimal3.142.

Exact forms should usually be kept through intermediate work when practical, then rounded at the requested final stage.

2. Decimal places count positions after the decimal point

7.486 rounded to two decimal places is7.49 because the third decimal digit is6.

12.0436 rounded to three decimal places is12.044.

0.00678 rounded to two decimal places is0.01, which is very coarse relative to the original value.

Decimal places describe location relative to the decimal point, not the number of important digits.

3. Significant figures begin at the first non-zero digit

0.004728 to three significant figures is0.00473. The significant digits are4,7,2 before rounding.

48372 to three significant figures is48400.

Trailing zeros after a decimal point can be significant when written deliberately: 2.50 has three significant figures.

Scientific notation can make intended significant figures clearer: 1.20×10³ visibly has three significant figures.

4. Rounding intervals come from half the rounding unit

A value recorded as7.3 to the nearest0.1 represents true values from7.25 up to but not including7.35:

7.25≤x<7.35.

The lower boundary is included because7.25 rounds to7.3 under the usual half-up school convention for positive values. The upper boundary7.35 rounds to7.4, so it is excluded from the7.3 interval.

The half-step is0.05 because the rounding unit is0.1.

5. Nearest integer intervals have half-unit boundaries

If a positive measurement is recorded as24 to the nearest whole unit, its interval is:

23.5≤x<24.5.

The maximum possible absolute rounding error is less than0.5 unit; the error bound is0.5.

Do not write 23≤x≤25. That interval uses a whole-unit step instead of a half-step.

6. Nearest ten, hundred and thousand use larger half-steps

730 to the nearest10 represents725≤x<735.

2400 to the nearest100 represents2350≤x<2450.

65000 to the nearest1000 represents64500≤x<65500.

Read the place value of the last retained digit to identify the rounding unit.

7. Significant-figure bounds depend on magnitude

Suppose 4800 is stated to two significant figures. The last retained significant digit is the hundreds digit, so the rounding unit is100.

Therefore 4750≤x<4850.

If0.048 is stated to two significant figures, the last retained digit is the thousandths place, so the rounding unit is0.001 and the interval is0.0475≤x<0.0485.

Significant figures do not have one fixed decimal place; the place depends on the number’s scale.

8. Absolute error is the size of the difference

If the true value is12.37 and an approximation is12.4, absolute error=|12.4−12.37|=0.03.

Absolute error is non-negative.

It has the same unit as the measured quantity.

Do not confuse actual absolute error, which needs the true value, with the maximum possible error implied by a rounding statement.

9. Percentage error compares error with a reference value

Using true value12.37 and approximation12.4, percentage error=0.03/12.37×100%≈0.243%.

The denominator matters. When the true value is known, percentage error is conventionally measured relative to that true value unless the problem defines another reference.

A small absolute error can be a large percentage error when the true quantity is small.

Always state which value is being used as the base if the context is not explicit.

10. Bounds for a sum use corresponding extremes

Suppose positive quantities a and b satisfy2.5≤a<2.7 and4.1≤b<4.3.

Then lower bound of a+b is2.5+4.1=6.6.

Upper boundary is2.7+4.3=7.0, so a+b<7.0.

Thus 6.6≤a+b<7.0.

11. Bounds for a difference use opposite extremes

With the same intervals, lower bound of b−a uses smallest b minus largest possible a:4.1−2.7=1.4 as the lower boundary.

Upper bound uses largest possible b minus smallest a:4.3−2.5=1.8 as the upper boundary.

Under the half-open source intervals, b−a lies strictly above1.4 and strictly below1.8 in this particular case, because the relevant upper endpoints are not attained.

For school bound calculations, the key structural idea is “lower difference = lower first − upper second; upper difference = upper first − lower second”, with endpoint inclusion handled from the original intervals where required.

12. Positive products use low×low and high×high

Suppose length l is12.4 cm to nearest0.1 and width w is8.2 cm to nearest0.1.

Then12.35≤l<12.45 and8.15≤w<8.25.

For positive quantities, area A=lw has lower bound12.35×8.15=100.6525 cm².

Its upper boundary is12.45×8.25=102.7125 cm², not attained under these half-open intervals.

13. Positive quotients use low numerator/high denominator and high numerator/low denominator

Suppose distance d is100 m to nearest1 m, so99.5≤d<100.5. Time t is12.0 s to nearest0.1 s, so11.95≤t<12.05.

For positive speed v=d/t, lower bound=99.5/12.05≈8.257 m/s.

Upper boundary=100.5/11.95≈8.410 m/s.

The denominator works oppositely: a larger positive denominator makes the quotient smaller.

14. Bounds for powers depend on sign and domain

If positive r satisfies2.95≤r<3.05, then r² ranges from2.95²=8.7025 up to but not including3.05²=9.3025.

This monotonic reasoning works because r is positive.

If an interval crosses zero, squaring needs more care: the minimum may be0 even when neither endpoint is closest in ordinary numerical order.

Do not mechanically square “lower” and “upper” endpoints without checking how the function behaves over the interval.

15. Maximum error from rounding is half the rounding unit

If a length is measured as8.6 cm to nearest0.1 cm, the true value lies in8.55≤L<8.65.

The error bound is0.05 cm.

The actual absolute error could be smaller or even zero if the true value happens to equal8.6.

“Maximum possible error” and “actual error” therefore answer different questions.

16. Relative error bounds help compare measurements of different sizes

A quantity reported as100 cm to nearest1 cm has error bound0.5 cm, at most about0.5% relative to the reported magnitude.

A quantity reported as2 cm to nearest1 cm has the same absolute error bound0.5 cm but a much larger relative scale.

Using the reported values for a quick comparison gives0.5/100=0.5% versus0.5/2=25%.

These are approximate relative-error comparisons based on the recorded magnitudes, not exact actual percentage errors unless true values are known.

17. Rounded inputs can support guaranteed threshold decisions

Suppose a positive mass is recorded as9.8 kg to nearest0.1 kg. Then9.75≤m<9.85.

Can we guarantee m>9.7? Yes, because even the lower bound9.75 exceeds9.7.

Can we guarantee m≥9.8? No. Values such as9.76 round to9.8 but are below9.8.

This is the bridge from numerical bounds to decision logic.

18. Capstone: bound a calculated density

An invented object has mass120 g to nearest1 g and volume50 cm³ to nearest1 cm³.

119.5≤m<120.5 and49.5≤V<50.5.

Density ρ=m/V is positive. Lower bound=119.5/50.5≈2.366 g/cm³.

Upper boundary=120.5/49.5≈2.434 g/cm³.

Therefore the rounded inputs do not justify reporting density as though it were known to many exact decimal places.

19. Independent practice

1. Round7.486 to2 decimal places.
2. Round0.004728 to3 significant figures.
3. Round48372 to3 significant figures.
4. Write the error interval for7.3 to nearest0.1.
5. Write the error interval for24 to nearest integer.
6. Write the error interval for730 to nearest10.
7. Write the error interval for2400 to nearest100.
8. Write the error interval for4800 to2 significant figures.
9. Write the error interval for0.048 to2 significant figures.
10. True value12.37 is approximated by12.4. Find absolute error.
11. Find percentage error for Question10 relative to the true value.
12. If2.5≤a<2.7 and4.1≤b<4.3, find bounds for a+b.
13. Length12.4 cm and width8.2 cm are each correct to nearest0.1 cm. Find the lower bound for area.
14. Find the corresponding upper boundary for area in Question13.
15. Distance100 m to nearest1 m and time12.0 s to nearest0.1 s. Find a lower bound for speed.
16. Find an upper boundary for speed in Question15.
17. A radius is3.0 cm to nearest0.1 cm. Find lower and upper boundaries for r².
18. A measurement is8.6 cm to nearest0.1 cm. State its error bound.
19. A mass is9.8 kg to nearest0.1 kg. Is m>9.7 guaranteed?
20. For the same mass, is m≥9.8 guaranteed?

20. Worked answers

1. 7.49.

2. 0.00473.

3. 48400.

4. 7.25≤x<7.35.

5. 23.5≤x<24.5.

6. 725≤x<735.

7. 2350≤x<2450.

8. 4750≤x<4850.

9. 0.0475≤x<0.0485.

10. 0.03.

11. Approximately0.243%. 0.03/12.37×100.

12. 6.6≤a+b<7.0.

13. 100.6525 cm². 12.35×8.15.

14. 102.7125 cm² as the upper boundary. The exact product remains below this value for the stated half-open input intervals.

15. Approximately8.257 m/s. 99.5/12.05.

16. Approximately8.410 m/s as the upper boundary. 100.5/11.95.

17. 8.7025≤r²<9.3025. From2.95≤r<3.05.

18. 0.05 cm.

19. Yes. The lower bound9.75 is already above9.7.

20. No. Allowed values below9.8 still round to9.8.

21. Diagnose bounds errors by finding the rounding unit first

Common failures include using a full rounding step instead of a half-step, confusing decimal places with significant figures, treating the upper boundary as included automatically, using high/high for a lower quotient, or reporting many decimals from rounded inputs as exact information.

A useful repair note says “find the last retained place”, “half that unit gives the error bound”, “positive quotient lower=low/high”, or “rounded value represents an interval”.

Then change only the rounding precision. A secure learner should be able to predict how the interval width changes before calculating the new endpoints.

22. Continue through the BTT learning routes

Return to the BTT Mathematics Hub or BTT Mathematical Lab. Use Units, Scale and Measurement for measurement foundations and Rounded Data, Thresholds and Guaranteed Conclusions for decision-level reasoning under approximate inputs.

Within Batch08, continue to Probability Trees, Conditional Events and Without-Replacement Reasoning, Histograms, Frequency Density and Grouped Data, or Distance–Time, Speed–Time and Motion Graphs.

23. Sources and scope

The rounded measurements, bounds and practice questions are original teaching material. Error intervals use the standard half-step interpretation for positive values under ordinary school rounding conventions.

For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match compound bounds, percentage-error and significant-figure depth to the learner’s actual subject level and school programme.