Multiply by 1
Tonight's idea
Here is a fraction that equals 1: \(\dfrac{1\ \text{km}}{1000\ \text{m}}\). Top and bottom are the same length. Its flip, \(\dfrac{1000\ \text{m}}{1\ \text{km}}\), equals 1 as well.
Multiplying by 1 leaves a quantity alone. So multiplying by one of these fractions cannot change a distance. It can only change how the distance is written: the units. Choose the one whose bottom cancels the unit you have. Units cancel like letters in algebra: metres on top, metres on the bottom, gone.
\[3.2\ \cancel{\text{km}} \times \frac{1000\ \text{m}}{1\ \cancel{\text{km}}} = 3200\ \text{m}.\]
That is the whole method. Everything else is chaining. A rate such as metres per second has a unit on top and a unit on the bottom, so it needs two chains at once, one for each. A squared unit needs the same factor twice, because cm² means cm × cm.
The method has a second use. If a formula hands you seconds when you asked for metres, do not check the arithmetic. The formula is wrong.
The grade 6 sheet The 125-million-dollar unit mistake tells the story of the spacecraft lost to exactly this kind of slip.
Watch one: a sprinter in km/h
Usain Bolt ran 100 m in 9.58 s, an average of \(100 / 9.58 = 10.44\) m/s. What is that in km/h? Two chains at once: one fraction turns metres into kilometres, the other turns seconds into hours. Each equals 1.
\[\frac{10.44\ \cancel{\text{m}}}{1\ \cancel{\text{s}}} \times \frac{1\ \text{km}}{1000\ \cancel{\text{m}}} \times \frac{3600\ \cancel{\text{s}}}{1\ \text{h}} = \frac{10.44 \times 3600}{1000}\ \frac{\text{km}}{\text{h}} \approx 37.6\ \text{km/h}.\]
Metres cancel, seconds cancel, kilometres over hours remain. Had the hour fraction been upside down, the seconds would have piled up as \(\text{s}^2\) instead of cancelling. That is the signal to flip it. Finally, check the size: 37.6 km/h is a fast bicycle, which is about right for the fastest human.
Your turn
- Write each as a chain of fractions equal to 1, cancel the units, then compute. (a) 3.25 hours in seconds. (b) 72 inches in metres, given 1 inch = 2.54 cm.
- Two chains at once. (a) 108 km/h in m/s. (b) Cheese at $18.00 per kilogram, in cents per gram.
- (a) A floor is 4.2 m by 3.6 m. Tiles are 30 cm squares. Convert the floor's area to cm², using the factor twice, and find how many tiles are needed. (b) Aluminium has density 2.7 g/cm³. Convert to kg/m³. How many times does the length factor appear?
- A dropped stone falls \(d\) metres in \(t\) seconds, where \(g = 9.8\ \text{m/s}^2\) (metres per second, per second). Three candidate formulas: \(d = gt\), \(d = \tfrac12 g t^2\), \(d = g/t\). (a) Work out the units each one produces. Which can be right? (b) Use the survivor: how far does a stone fall in 3 s?
Stretch: a pendulum
A pendulum of length \(L\) metres swings back and forth in time \(T = 2\pi\sqrt{L/g}\), with \(g = 9.8\ \text{m/s}^2\). Show that the units under the root simplify to \(\text{s}^2\), so \(T\) comes out in seconds. Find \(T\) for \(L = 1\) m. Then explain what goes wrong if someone puts \(L\) in centimetres.
Answers
- (a) \(3.25\ \cancel{\text{h}} \times \frac{60\ \cancel{\text{min}}}{1\ \cancel{\text{h}}} \times \frac{60\ \text{s}}{1\ \cancel{\text{min}}} = 11{,}700\) s. (b) \(72\ \cancel{\text{in}} \times \frac{2.54\ \cancel{\text{cm}}}{1\ \cancel{\text{in}}} \times \frac{1\ \text{m}}{100\ \cancel{\text{cm}}} = 1.8288\) m, about 1.83 m.
- (a) \(\frac{108\ \cancel{\text{km}}}{1\ \cancel{\text{h}}} \times \frac{1000\ \text{m}}{1\ \cancel{\text{km}}} \times \frac{1\ \cancel{\text{h}}}{3600\ \text{s}} = 30\) m/s. (b) \(\frac{18\ \cancel{\$}}{1\ \cancel{\text{kg}}} \times \frac{100\ \text{c}}{1\ \cancel{\$}} \times \frac{1\ \cancel{\text{kg}}}{1000\ \text{g}} = 1.8\) cents per gram.
- (a) \(4.2 \times 3.6 = 15.12\ \text{m}^2\). Then \(15.12\ \text{m}^2 \times \frac{100\ \text{cm}}{1\ \text{m}} \times \frac{100\ \text{cm}}{1\ \text{m}} = 151{,}200\ \text{cm}^2\). One tile is \(30 \times 30 = 900\ \text{cm}^2\), so \(151{,}200 / 900 = 168\) tiles. (b) \(\frac{2.7\ \text{g}}{\text{cm}^3} \times \frac{1\ \text{kg}}{1000\ \text{g}} \times \left(\frac{100\ \text{cm}}{1\ \text{m}}\right)^3 = 2.7 \times \frac{1{,}000{,}000}{1000} = 2700\ \text{kg/m}^3\). The length factor appears three times, because cm³ is cm × cm × cm.
- (a) \(gt\) gives \(\frac{\text{m}}{\text{s}^2} \times \text{s} = \text{m/s}\), a speed, not a distance. \(\tfrac12 g t^2\) gives \(\frac{\text{m}}{\text{s}^2} \times \text{s}^2 = \text{m}\). \(g/t\) gives \(\text{m/s}^3\). Only \(d = \tfrac12 g t^2\) can be right. (b) \(\tfrac12 \times 9.8 \times 3^2 = 44.1\) m.
Stretch. \(\dfrac{\text{m}}{\text{m/s}^2} = \text{m} \times \dfrac{\text{s}^2}{\text{m}} = \text{s}^2\), and \(\sqrt{\text{s}^2} = \text{s}\). For \(L = 1\) m, \(T = 2\pi\sqrt{1/9.8} \approx 2.01\) s. With \(L = 100\) (centimetres) and \(g\) still in m/s², the formula gives \(2\pi\sqrt{100/9.8} \approx 20.1\): ten times too long, and the units inside the root no longer cancel, which is how you would catch it.