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Math 8 · Mathematics
Rational numbers · Numbers

Approximation and Error

Before you start, make sure you know: Approximation and rounding (Math 7)

A video with exactly 3,894,215 views shows up in your app as "3.9M". Your bathroom scale says 128.4 lb, even though your true weight has more decimal places. And the gas pump says $\$3.459$ per gallon, but nobody can pay a tenth of a cent. We use approximate numbers all day long. The important question is: how far off are we?

Truncating and Rounding

There are two ways to keep fewer digits. Both start by choosing a place value: the nearest whole number, tenths, hundredths, thousandths, and so on.

  • Truncate: just cut the number off at that place. Every digit after it is thrown away. $\pi = 3.14159\ldots$ truncated to the hundredths is $3.14$.

  • Round: keep the value that is closest to the original. Look at the first digit you are removing. If it is 5 or more, add 1 to the last digit you keep. If it is less than 5, leave the last digit alone.

Why does the "5 or more" rule work? The halfway point between two neighbors is the 5. Anything at or past halfway is closer to the larger neighbor.

Number line from 2.7 to 2.8 with the midpoint 2.75 marked, showing that 2.7846 is closer to 2.8 and so rounds to 2.8. 2.72.82.75 rounds to 2.7rounds to 2.8 2.7846 → 2.8

For large numbers it works the same way. To round $2{,}468{,}310$ to the nearest hundred thousand, look at the ten-thousands digit (a 6), so round up to $2{,}500{,}000$. To the nearest million, look at the hundred-thousands digit (a 4), so round down to $2{,}000{,}000$.

Use the slider to compare truncating and rounding on the same number. Notice that a rounding error is never more than half a unit of the place you round to.

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The irrational number $\sqrt{10} = 3.16227766...$ appears on your calculator screen. Copy the following table into your notebook and complete it by making the indicated…

The irrational number $\sqrt{10} = 3.16227766...$ appears on your calculator screen.

Copy the following table into your notebook and complete it by making the indicated approximations for that number:

Type of approximation To the nearest unit To the tenths To the hundredths To the thousandths
By truncating
By rounding
Solve
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You are helping to organize a charity race at your school. When calculating the exact distance of a section of the route with the GPS, you get 2.3458 miles. To make the…

You are helping to organize a charity race at your school. When calculating the exact distance of a section of the route with the GPS, you get 2.3458 miles.

To make the information posters, you need to simplify this figure. Write what this distance would be if we approximate by rounding:

  1. a) To the nearest unit (miles).
  2. b) To the tenths.
  3. c) To the hundredths.

Use the following number line to help you visualize where the number falls:

2,34 2,36 2,35 2,345 2,355 Halfway 2,3458 Rounding
Solve
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You have gone to the supermarket to buy ingredients for a dinner. The true prices of the products are: Pasta: $\$1.85$ Tomato sauce: $\$2.15$ Grated cheese: $\$3.48…

You have gone to the supermarket to buy ingredients for a dinner. The true prices of the products are:

  • Pasta: $\$1.85$

  • Tomato sauce: $\$2.15$

  • Grated cheese: $\$3.48$

  • Soft drink: $\$1.25$

  1. a) Make a quick estimate of the total cost by rounding each price to the nearest dollar before adding.
  2. b) Calculate the exact sum of the prices.
  3. c) Calculate the difference (absolute error) between your estimate and the true price. Do you think it was a good estimate to know if you had $\$10$ in your pocket?
Solve

Significant Figures and Scientific Notation

Significant figures are the digits that carry information about precision. When we write $3{,}894{,}215$ as "3.9 million," we use 2 significant figures.

Two numbers with their significant figures highlighted: 0.0045 has 2 because leading zeros do not count, and 3.050 has 4 because the final zero shows precision. 0.0045 0.00: don't count 45: 2 significant figures 3.050 4 significant figures (the last 0 shows precision)

Very large and very small numbers are easiest to write in scientific notation, $a \times 10^n$ with $1 \le a < 10$. Every digit of $a$ is significant, so it also shows exactly how many significant figures you kept. In $4{,}500$ the zeros might or might not be significant, but $4.5 \times 10^3$ clearly has 2.

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A famous streamer has uploaded a video that has exactly 3,894,215 views. a) Write this amount rounded to the nearest million. b) Write the amount rounded to the nearest…

A famous streamer has uploaded a video that has exactly 3,894,215 views.

  1. a) Write this amount rounded to the nearest million.
  2. b) Write the amount rounded to the nearest hundred thousand.
  3. c) How many significant figures have you used in part b?
Solve
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In the scientific and technological field, it is crucial to know with what precision we are working. Round the following numbers to three significant figures: a) The…

In the scientific and technological field, it is crucial to know with what precision we are working. Round the following numbers to three significant figures:

  1. a) The number of views of a viral video: $14\,592\,301$.
  2. b) The mass of a dust particle: $0,00040961$ grams.
  3. c) The number $\pi$: $3,141592...$
  4. d) The capacity of a hard drive in bytes: $1\,024\,576\,000$.

Then, express each result using scientific notation.

Solve

Absolute Error

How wrong is an approximation? The simplest answer is to subtract. The result is the absolute error.

Number line from 116 to 122 inches showing a measured value of 118 and a true value of 120, with the gap between them labeled as the absolute error of 2 inches. 116118120122 measuredtrue value Eₐ = 2 in.

Because of the absolute value, it doesn't matter whether you guess too high or too low. The absolute error is never negative.

Notice that the absolute error alone can be misleading. An error of 2 inches is huge when measuring a pencil, but tiny when measuring a football field. To compare measurements of different sizes, we need a second kind of error.

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When measuring the length of your study desk, the true factory value is 120 inches. You use a tape measure and get a measurement of 119 inches. a) Calculate the absolute…

When measuring the length of your study desk, the true factory value is 120 inches. You use a tape measure and get a measurement of 119 inches.

  1. a) Calculate the absolute error ($E_a$) made in your measurement. Remember that the absolute error is the positive difference between the true value and the approximate value: $E_a = |V_{real} - V_{aprox}|$.
  2. b) If your classmate measures the school hallway, which is actually 5,000 inches long, and gets 4,999 inches, his absolute error is 1 inch, the same as yours. Does this mean that both measurements have the same precision? Explain your answer (you can rely on the concept of relative error).
Solve
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We want to calculate the area of a circle whose radius we have measured with a ruler. The measurement of the radius is $r = 5$ in. We know that the area of a circle is…

We want to calculate the area of a circle whose radius we have measured with a ruler.

The measurement of the radius is $r = 5$ in.

We know that the area of a circle is $A = \pi \cdot r^2$.

  1. a) If we use the approximation of $\pi \approx 3$, what is the calculated area?
  2. b) If we use the approximation of $\pi \approx 3,14$, what is the calculated area?
  3. c) Knowing that the "true" value (using the $\pi$ key on the calculator) is approximately 78.54 sq in., calculate the absolute error made in part a and in part b. Which approximation of $\pi$ is more appropriate to use in 8th grade math?
Solve

Relative Error

Mistaking 1 inch on a desk and mistaking 1 inch on a phone screen are not the same. On the desk it barely matters. On the phone it is a disaster. The relative error compares the error with the size of the true value.

Two bars drawn to the same scale: an error of 1 inch on a 48-inch desk is small, while an error of 1 inch on a 6-inch phone screen is large. Desk: 48 in., you read 49 in. Eₐ = 1 in. · Eᵣ = 1/48 ≈ 0.021 = 2.1% → pretty good Phone screen: 6 in., you read 7 in. Eₐ = 1 in. · Eᵣ = 1/6 ≈ 0.167 = 16.7% → poor

A smaller relative error means a better-quality measurement.

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We are going to calculate the relative error ($E_r$) to compare two different weighings. Remember that the formula is $E_r = \frac{E_a}{|V_{real}|}$. 1. Case A: You…

We are going to calculate the relative error ($E_r$) to compare two different weighings. Remember that the formula is $E_r = \frac{E_a}{|V_{real}|}$.

  1. Case A: You weigh an apple whose true weight is 200 g and your scale shows 205 g.

  2. Case B: You weigh your dog whose true weight is 20,000 g (20 kg) and the scale shows 20,005 g.

Calculate the relative error in both cases (you can express it in decimal form or as a percentage) and state which measurement was more precise (of higher quality).

Solve
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Two students make different measurements for a science project. Lucy measures the length of a soccer field and gets $105$ ft, with an absolute error of $0,5$ ft. Mark…

Two students make different measurements for a science project.

  • Lucy measures the length of a soccer field and gets $105$ ft, with an absolute error of $0,5$ ft.

  • Mark measures the length of a cell phone screen and gets $15$ in, with an absolute error of $0,1$ in.

Which of the two measurements is more precise (has higher quality)? Justify your answer by calculating the relative errors of both measurements.

Lucy Soccer field L = 105 ft Eₐ = 0.5 ft Mark Cell phone screen L = 15 in Eₐ = 0.1 in
Solve
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In a nanotechnology laboratory and an astronomical observatory, two different measurements have been taken with their respective precision instruments: 1. The length of…

In a nanotechnology laboratory and an astronomical observatory, two different measurements have been taken with their respective precision instruments:

  1. The length of a microscopic chip: $0,00045 \text{ m}$ (with a possible error of $\pm 0,00001 \text{ m}$).

  2. The distance between two satellites: $45000 \text{ km}$ (with a possible error of $\pm 100 \text{ km}$).

Without performing complex calculations, one might think that the chip measurement is "finer" because it is the smaller number, but which of the two measurements has higher scientific quality or precision?

Justify your answer by calculating the relative error of both measurements and expressing it as a percentage.

Solve

Error Bounds and Intervals

Often you don't know the true value, but you do know how the number was rounded. Suppose a digital thermometer rounds to the nearest tenth of a degree and shows 98.6 °F. The true temperature was anywhere that would round to 98.6.

Rounded to the nearest Unit $u$ Error bound $u/2$
whole number 1 0.5
tenth 0.1 0.05
hundredth 0.01 0.005
thousandth 0.001 0.0005
thousand 1,000 500
Number line from 98.5 to 98.7 with the interval from 98.55 to 98.65 highlighted around the displayed value 98.6. 98.598.698.7 98.5598.65 the true temperature is somewhere in here 98.6 ± 0.05

The same idea works for headlines. "About 25,000 people" rounded to the nearest thousand means the true crowd was between $24{,}500$ and $25{,}500$, and the error is at most 500.

Error bounds also tell you how many decimal places you need. Rounding to $n$ decimal places gives an error of at most $0.5 \times 10^{-n}$. If you want the error below $0.001$, rounding to the thousandths gives a bound of $0.0005$, which works.

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Digital bathroom scales usually have a precision of $100$ grams ($0,1$ kg). This means that if the scale reads $54,2$ kg, the true weight could be slightly above or…

Digital bathroom scales usually have a precision of $100$ grams ($0,1$ kg). This means that if the scale reads $54,2$ kg, the true weight could be slightly above or slightly below, but rounded to that decimal.

  1. a) Write the interval of possible true values for a measured weight of $54,2$ kg.
  2. b) What is the absolute error bound of this scale?
Solve
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In the news, they announced that "approximately 25,000 people"* attended a demonstration. a) If we know that this figure was obtained by rounding to the nearest…

In the news, they announced that "approximately 25,000 people" attended a demonstration.

  1. a) If we know that this figure was obtained by rounding to the nearest thousand, what is the maximum and minimum number of people who could have actually attended?
  2. b) Represent this interval on the number line.
  3. c) What is the maximum absolute error that could be made with this headline?
24.000 ? 25.000 ? 26.000 Minimum Approximation Upper limit ? ? Maximum absolute error = ?
Solve
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A digital bicycle speedometer shows the speed to one decimal place. At a given moment, it shows $24,5 \text{ mph}$. We know that this device approximates by rounding to…

A digital bicycle speedometer shows the speed to one decimal place. At a given moment, it shows $24,5 \text{ mph}$.

We know that this device approximates by rounding to the nearest tenth.

  1. a) Write the interval of possible actual values in which the exact speed of the bicycle lies. That is, what is the minimum and maximum speed the cyclist could be traveling for the device to show 24.5?
  2. b) If the cyclist maintains this speed for exactly 2 hours, what is the maximum difference in miles (absolute error bound) between the distance calculated with the value 24.5 and the possible actual distance?
Solve

Approximating in Calculations

When a calculation uses approximate numbers, the answer is approximate too. The golden rule: keep all the digits while you calculate and round only at the end. Every early rounding adds error.

Circles are the classic example. Suppose a circle has radius $r = 10$ ft, so $A = \pi r^2 = 100\pi \approx 314.159$ ft².

A circle with a radius of 10 feet; its area is pi times the radius squared. r = 10 ft A = π · r²
Value used for $\pi$ Area (ft²) Absolute error (ft²)
3 300 about 14.16
3.14 314 about 0.16
3.1416 314.16 about 0.0007

More digits of $\pi$ means less error. In Grade 8 math, $3.14$ is usually good enough. When you want the best answer, use the $\pi$ key on your calculator, or leave $\pi$ in the answer for an exact result.

Roots work the same way. An answer with radicals is exact, and a decimal is an approximation.

Rounding early also matters in money. Suppose gas costs $\$3.459$ per gallon and you buy 12 gallons.

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In many gas stations, the price of fuel is shown with three decimal places (thousandths of a dollar), although we pay with coins that only go up to cents (hundredths).…

In many gas stations, the price of fuel is shown with three decimal places (thousandths of a dollar), although we pay with coins that only go up to cents (hundredths).

The price of diesel today is $\$1,489$/gallon.

You are going to fill up with exactly $35$ gallons.

  1. a) Calculate the exact true price by multiplying the $35$ gallons by the price with three decimal places.
  2. b) Calculate the price if we first round the price per gallon to two decimal places ($\$1,49$/gal).
  3. c) What is the difference in dollars between both methods? Who benefits from the prior rounding, the customer or the gas station?
Solve
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We want to calculate the area of a rectangle whose base measures $\sqrt{5}$ in and its height measures $\sqrt{3}$ in. a) Calculate the exact area using radicals. b)…

We want to calculate the area of a rectangle whose base measures $\sqrt{5}$ in and its height measures $\sqrt{3}$ in.

  1. a) Calculate the exact area using radicals.
  2. b) Approximate $\sqrt{5}$ and $\sqrt{3}$ to the hundredths and calculate the approximate area with those values.
  3. c) Determine the absolute error made when using the approximate values with respect to the exact value from part a.
√5 in √3 in AREA = ?
Solve
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We want to paint a helicopter landing pad (helipad) which has a circular shape. To calculate the amount of paint needed, we measure the radius and obtain…

We want to paint a helicopter landing pad (helipad) which has a circular shape. To calculate the amount of paint needed, we measure the radius and obtain $r = 12,5 \text{ m}$. We know that the formula for the area is $A = \pi \cdot r^2$.

  1. a) Calculate the "exact" area using the $\pi$ key on your calculator.
  2. b) Calculate the approximate area using the rounded value $\pi \approx 3,14$.
  3. c) Calculate the absolute error and the relative error made when using the approximation of $\pi$ compared to the calculator value.
  4. d) If a can of paint covers $5 \text{ m}^2$, does the error made affect the number of cans we must buy?
H r = 12.5 ft
Solve

Error in Sums, Products, and Averages

If you add, multiply, or average rounded numbers, the errors combine. You can still bound the final error.

For products, remember that every dimension could be a little too big or a little too small. A box measured as 12 in. by 8 in. by 5 in. with an error bound of $0.5$ in. on each side:

Three nested rectangles showing the smallest possible, measured, and largest possible base of a box measured as 12 inches by 8 inches. measured 12 × 8 Outer dashed: largest 12.5 × 8.5 in. Middle solid: measured 12 × 8 in. Inner dashed: smallest 11.5 × 7.5 in.

Truncating is a special case. It always throws digits away, so the error always goes in the same direction and builds up when you add many truncated numbers. Rounding errors go both ways, so they partly cancel each other out. A store that truncates every weight to the tenth of a gram, for example, always pays for a little less than it receives.

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A jewelry company buys gold based on weight. Their digital scale measures with a precision of hundredths of a gram ($0,01 \text{ g}$). However, the invoice they issue…

A jewelry company buys gold based on weight. Their digital scale measures with a precision of hundredths of a gram ($0,01 \text{ g}$). However, the invoice they issue truncates the weight to a single decimal place (tenths of a gram) for charging.

This week they have bought 4 batches of gold whose actual weights on the scale were:

  • Batch A: $12,48 \text{ g}$

  • Batch B: $25,19 \text{ g}$

  • Batch C: $8,76 \text{ g}$

  • Batch D: $15,95 \text{ g}$

  1. a) Calculate the actual total weight by adding the exact amounts from the scale.
  2. b) Calculate the total weight that appears on the invoices by adding the amounts truncated to the tenths.
  3. c) What is the total absolute error made with this billing method? Who loses out, the buyer or the seller?
Solve
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We have a water tank in the shape of a rectangular prism. We have measured its dimensions with a tape measure that only marks decimeters, so our measurements have an…

We have a water tank in the shape of a rectangular prism. We have measured its dimensions with a tape measure that only marks decimeters, so our measurements have an absolute error bound of $0,5 \text{ dm}$ (that is, the actual value could be 0.5 dm larger or smaller).

The measurements obtained are:

  • Length: $20 \text{ dm}$

  • Width: $10 \text{ dm}$

  • Height: $15 \text{ dm}$

Calculate the volume of the tank ($V = \text{largo} \times \text{ancho} \times \text{alto}$) and determine between which two values (interval) the actual volume of the tank lies, considering the error bounds of each measurement.

V = l × w × h V = ? 20 ± 0.5 dm 10 ± 0.5 dm 15 ± 0.5 dm
Solve
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Calculate step by step the absolute and relative error in the following combined operation, taking into account the order of operations. Given the numbers $A = 2,86$ and…

Calculate step by step the absolute and relative error in the following combined operation, taking into account the order of operations.

Given the numbers $A = 2,86$ and $B = 4,13$, approximations of some unknown actual values, but knowing that both have been rounded to two decimal places (hundredths).

Calculate the value of $C = \frac{A + B}{2}$.

Reason: What is the absolute error bound of the result $C$? (Hint: Think about the maximum and minimum value that A and B could have before being rounded).

Solve

Back to the start: The video with 3,894,215 views is shown as 3.9M, or $3{,}900{,}000$. The absolute error is $|3{,}894{,}215 - 3{,}900{,}000| = 5{,}785$ views, and the relative error is $\frac{5{,}785}{3{,}894{,}215} \approx 0.0015 = 0.15\%$. That is a tiny error, and the short number is much easier to read. Your 128.4 lb scale reading has an error bound of $0.05$ lb, so your true weight is between $128.35$ and $128.45$ lb.

Summary

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