Approximate numbers and errors | Grade 10

A ruler says your textbook is 24.4 cm wide. A calculator says $\sqrt{2} = 1.414213562\ldots$ and the display just stops. In real life almost every number you meet, from a length to a population, is approximate. The real question is not "is it exact?" but "how far off can it be?". This lesson gives you the tools to answer that.

Approximate numbers and absolute error

Sometimes the exact value is impossible to write down ( $\pi$ , $\sqrt{2}$ ) and sometimes it is impossible to know (the exact mass of a bag of rice). Then we use a nearby number instead.

Let $a$ be the exact value and $\bar{a}$ an approximation of it.

The smaller $d$ is, the better the approximation. For $\pi \approx 3.14$ we can take $d = 0.002$ , so $\pi$ lies between $3.138$ and $3.142$ .

The number

\pi

lies inside the interval

3.14 \pm 0.002

Relative error

Is an error of 1 cm serious? For the length of a pencil, yes. For the length of a football pitch, not at all. We compare the error with the size of the number.

Rounding and significant figures

To round a number to a chosen place, look at the digit just to the right of that place.

If that digit is less than 5, keep the digits on the left and replace the others by zeros (or drop them after the decimal point).
If that digit is 5 or more, add 1 to the last kept digit and then do the same.

For example, $7\,459$ rounded to hundreds is $7\,500$ ; $3.14159$ rounded to 2 decimal places is $3.14$ .

When you round to a place, the error is at most half a unit of that place. So $3.14$ (rounded to hundredths) has $d = 0.005$ , and $7\,500$ (rounded to hundreds) has $d = 50$ .

The significant figures of a number are its digits starting from the first non-zero digit on the left. So $0.0305$ has 3 significant figures ( $3, 0, 5$ ), and $2.40$ has 3 ( $2, 4, 0$ ). To round to $k$ significant figures, keep the first $k$ of them and round as above. For example, $0.0071849$ to 3 significant figures is $0.00718$ , and $12\,345$ to 3 significant figures is $12\,300$ .

Reliable digits

Suppose $a = \bar{a} \pm d$ . A digit of $\bar{a}$ is reliable (a "correct digit") if $d$ is at most half a unit of the place where that digit stands. All digits to the left of a reliable digit are reliable too. When we report a result, we keep only reliable digits and round off the rest.

Example Approximating

\sqrt{2}

with a calculator

The calculator shows $\sqrt{2} = 1.41421356\ldots$ . Round it to 2 and to 3 decimal places. Give the error bound, the relative error bound and the reliable digits in each case.

Solution.

To 2 decimal places: $\bar{a} = 1.41$ . The error is $|\sqrt{2} - 1.41| = 0.0042\ldots \leq 0.005$ , so $d = 0.005$ . Half a unit of the hundredths place is $0.005$ , so all digits $1, 4, 1$ are reliable. The relative error is $\delta \leq \dfrac{0.005}{1.41} \approx 0.35\%$ .
To 3 decimal places: $\bar{a} = 1.414$ . The error is $|\sqrt{2} - 1.414| = 0.00021\ldots \leq 0.0005$ , so $d = 0.0005$ . All four digits $1, 4, 1, 4$ are reliable. The relative error is $\delta \leq \dfrac{0.0005}{1.414} \approx 0.035\%$ .
More decimal places give a smaller error bound and more reliable digits.

Key terms

English	Tiếng Việt
approximate number	số gần đúng
absolute error	sai số tuyệt đối
error bound	độ chính xác (của số gần đúng)
relative error	sai số tương đối
rounding	quy tròn
significant figure	chữ số có nghĩa
reliable digit	chữ số chắc

Summary

An approximation $\bar{a}$ of the exact value $a$ has absolute error $\Delta = |a - \bar{a}|$ . An error bound $d$ satisfies $\Delta \leq d$ , and we write $a = \bar{a} \pm d$ .
The relative error is $\delta = \Delta/|\bar{a}|$ , bounded by $d/|\bar{a}|$ . Use it to compare the accuracy of different measurements.
Rounding to a place gives an error of at most half a unit of that place.
A digit is reliable if $d$ is at most half a unit of its place. Report only reliable digits.

Practice

Round $7\,459$ to tens and to hundreds, and round $3.14159$ to 2 and to 4 decimal places.
How many significant figures do $0.0305$ , $2.40$ and $40.007$ have? Round $0.0071849$ to 2 significant figures.
Use $\bar{a} = 1.73$ as an approximation of $\sqrt{3} = 1.7320508\ldots$ . Find $\Delta$ (to 4 decimal places, rounding up), then a relative error bound as a percentage (2 decimal places).
A tailor measures a cloth $8.5\text{ m}$ long with $d = 0.05\text{ m}$ , and a builder measures a wall $120\text{ m}$ long with $d = 0.5\text{ m}$ . Whose measurement is more accurate?
Which is the better approximation of $\pi$ : $\dfrac{22}{7}$ or $3.1416$ ? Compare their absolute errors.
Let $\bar{a} = 1\,536$ with $d = 4$ . Which digits are reliable? Write the result with reliable digits only.
Let $\bar{a} = 45.60$ with $d = 0.03$ . Which digits are reliable? How would you report it?
The sides of a rectangle are measured as $25.4\text{ cm}$ and $12.6\text{ cm}$ , each with $d = 0.1\text{ cm}$ . Find the perimeter and an error bound for it, then its relative error bound.
A student writes $\bar{a} = 2.500$ with $d = 0.001$ and claims that all four digits are reliable. Is that right? Use the idea of half a unit, and say how the number should be reported.
(Challenge) For the rectangle in exercise 8, the area is estimated as $25.4 \times 12.6$ . Find the possible smallest and largest areas, an error bound $d$ for the estimated area, and the reliable digits.

Show solutions

To tens: $7\,459 \to 7\,460$ (the units digit $9 \geq 5$ ). To hundreds: $7\,459 \to 7\,500$ (the tens digit $5 \geq 5$ ). Then $3.14159 \to 3.14$ (the next digit is $1$ ) and $3.14159 \to 3.1416$ (the next digit is $9$ ).
$0.0305$ has 3 significant figures, $2.40$ has 3, and $40.007$ has 5. For $0.0071849$ keep $7$ and $1$ ; the next digit is $8 \geq 5$ , so round up to $0.0072$ .
$\Delta = |1.7320508 - 1.73| = 0.0020508 \approx 0.0021$ (rounded up). So $d = 0.0021$ and $\delta \leq \dfrac{0.0021}{1.73} \approx 0.0012 = 0.12\%$ .
Tailor: $\dfrac{0.05}{8.5} \approx 0.0059 = 0.59\%$ . Builder: $\dfrac{0.5}{120} \approx 0.0042 = 0.42\%$ . The builder's measurement is more accurate because its relative error is smaller.
$\left|\dfrac{22}{7} - \pi\right| \approx 0.00126$ and $|3.1416 - \pi| \approx 0.0000073$ . The second error is much smaller, so $3.1416$ is the better approximation.
Half a unit of the tens place is $5 \geq 4$ , so the tens digit $3$ is reliable (and so are $1$ and $5$ ). Half a unit of the units place is $0.5 < 4$ , so the digit $6$ is not reliable. Report $1\,540$ (rounded to tens).
Half a unit of the tenths place is $0.05 \geq 0.03$ , so the digit $6$ is reliable, as are $4$ and $5$ . Half a unit of the hundredths place is $0.005 < 0.03$ , so the last digit $0$ is not reliable. Report $45.6$ .
Perimeter $\bar{P} = 2(25.4 + 12.6) = 76.0\text{ cm}$ . Each side may be off by $0.1$ , so $d = 2(0.1 + 0.1) = 0.4\text{ cm}$ and $P = 76.0 \pm 0.4$ . Relative error: $\delta \leq \dfrac{0.4}{76.0} \approx 0.0053 = 0.53\%$ .
No. The last digit $0$ is in the thousandths place. The error $d = 0.001$ is larger than half a unit of the thousandths place ( $0.0005$ ), so the last $0$ is not reliable. It is smaller than half a unit of the hundredths place ( $0.005$ ), so the digits $2, 5, 0$ are reliable. Report $2.50$ .
The smallest area is $25.3 \times 12.5 = 316.25$ and the largest is $25.5 \times 12.7 = 323.85$ . The estimate is $25.4 \times 12.6 = 320.04\text{ cm}^2$ . The distances are $|323.85 - 320.04| = 3.81$ and $|316.25 - 320.04| = 3.79$ , so $d = 3.81$ . Half a unit of the tens place is $5 \geq 3.81$ , so the digits $3$ and $2$ are reliable, but half a unit of the units place is $0.5 < 3.81$ , so $0$ is not. Report the area as about $320\text{ cm}^2$ .