Round to decimal places or significant figures

Number

\(\displaystyle 5.67\)

Round the number to one decimal place

Move the point towards each end of the slider. Watch how both distances change

As the point moves from left to right, how do its distances change?

Locate the deciding digit

Decimal places count after the decimal point. Significant figures count from the first non-zero digit. Inspect the next digit once. Keep the retained digit for zero to four; increase it for five to nine, carrying when needed

1. 1Decimal places Keep hundredths and inspect thousandths \(\displaystyle 7.286\approx7.29\quad(2\text{ d.p.})\)
2. 2Significant figures Leading zeros locate the number. The first two significant digits set the rounding place \(\displaystyle 0.008472\approx0.0085\quad(2\text{ s.f.})\)
3. 3Zeros and signs Keep final decimal zeros when they record the requested precision. For negative numbers, round the magnitude and restore the sign. Exact halfway values go away from zero \(\displaystyle \begin{aligned}4.096&\approx4.10\quad(2\text{ d.p.})\\-2.65&\approx-2.7\quad(1\text{ d.p.})\end{aligned}\)

Number

\(\displaystyle -6.438\)

Round the number to two decimal places

Locate hundredths, then inspect the next digit. Keep the negative sign

Number

\(\displaystyle 0.003746\)

Round the number to two significant figures

Start counting at the first non-zero digit. Inspect the digit after the two you retain

Number

\(\displaystyle 8.304\)

Which answer gives the number to three significant figures?

The task is to round to one decimal place. What is wrong with this method?

Three equal lengths are each exactly one third of a metre. Predict what happens if each is replaced by 0.33 m before adding

Change the number of equal lengths. Compare adding exact thirds with adding 0.33 m for every length. Multiples of three make the exact total a whole number

What happens when the number of lengths doubles?

Keep exact values until the end

An approximation can be accurate enough for one use but accumulate error when reused. Compare the chosen method with exact arithmetic before deciding whether the difference matters

1. 1Exact representation Fractions can preserve a value that has no finite decimal expansion \(\displaystyle 12\times\frac13=4\)
2. 2Repeated approximation Rounding every length first makes the total smaller \(\displaystyle 12\times0.33=3.96\)
3. 3Report the difference Here the shortfall is four centimetres. Whether it matters depends on the permitted tolerance \(\displaystyle 4-3.96=0.04\text{ m}\)

Exact and approximate lengths

\(\displaystyle 9\times\frac13,\quad9\times0.33\)

Nine lengths are each exactly one third of a metre. What shortfall results from using 0.33 m for each length?

A calculation must be reported to two decimal places. Which method best avoids cumulative rounding error?