Big and tiny numbers
Tonight's idea
\(2^3\) means \(2 \times 2 \times 2\). The exponent counts the twos. That one sentence gives you every rule. \(2^3 \times 2^4\) is three twos times four twos, seven twos in all: \(2^7\). So \(a^m \times a^n = a^{m+n}\). Dividing cancels twos: \(2^7 \div 2^3 = 2^4\). And \((2^3)^2\) is three twos, twice: \(2^6\).
What is \(2^0\)? Use the rule: \(2^3 \div 2^3 = 2^0\), and \(8 \div 8 = 1\). So \(2^0 = 1\). What about \(2^{-1}\)? \(2^2 \div 2^3 = 2^{-1}\), and \(4 \div 8 = \tfrac12\). Negative exponents mean "divide instead".
Powers of 10 are the most useful of all, because \(10^n\) is a 1 followed by \(n\) zeros, and \(10^{-3} = 0.001\): a decimal point, two zeros, then a 1. Scientific notation writes any number as (something between 1 and 10) \(\times 10^n\): the distance to the Sun is \(1.5 \times 10^8\) km, the width of an atom is about \(1 \times 10^{-10}\) m. The exponent tells you the size at a glance, and sizes multiply by adding exponents.
Here is a number worth knowing. Light travels \(3 \times 10^5\) km every second. A year has about \(3.16 \times 10^7\) seconds. So a light-year, the distance light covers in a year, is \(3 \times 10^5 \times 3.16 \times 10^7 \approx 9.5 \times 10^{12}\) km. Nearly ten trillion kilometres, and it is the small unit of astronomy.
Watch one
How long does sunlight take to reach Earth? Distance \(1.5 \times 10^8\) km, speed \(3 \times 10^5\) km/s. Divide: \(\tfrac{1.5}{3} \times 10^{8-5} = 0.5 \times 10^3 = 500\) seconds, which is 8 minutes 20 seconds. The Sun you see is 8 minutes old.
Simplify \(\dfrac{5^6}{5^4}\): six fives over four fives leaves two fives, \(5^2 = 25\).
Your turn
Tonight's core: the example and exercises 1 to 3. The rest can wait for another visit.
- Simplify, giving both the power and the number: \(3^2 \times 3^3\); \(\;(2^3)^2\); \(\;\dfrac{7^5}{7^3}\); \(\;10^0\); \(\;10^{-2}\).
- Scientific notation: write 7 500 000 and 0.000 42 in it; write \(3.2 \times 10^5\) and \(6 \times 10^{-3}\) as ordinary numbers.
- The Moon is about 384 000 km away. How long does its light take to reach you? Express 384 000 in scientific notation first.
- Proxima Centauri, the nearest star after the Sun, is 4.2 light-years away. Write that in kilometres in scientific notation. A fast rocket travels 50 000 km/h. Roughly how many years would the trip take? (There are about 8800 hours in a year.)
- Work out roughly: seconds in a year; heartbeats in an 80-year life at 70 per minute. Write each as \(a \times 10^n\) with \(a\) to one decimal place.
- An atom is about \(10^{-10}\) m across. How many atoms, side by side, make a line 1 mm long? (1 mm \(= 10^{-3}\) m.)
Stretch: folding paper to the Moon
A sheet of paper is 0.1 mm thick, which is \(10^{-4}\) m. Fold it in half and it is twice as thick; after \(n\) folds it is \(2^n \times 10^{-4}\) m. The Moon is \(3.84 \times 10^8\) m away. How many folds would reach it?
You need \(2^n\) to be at least \(3.84 \times 10^{12}\). A helpful shortcut: \(2^{10} = 1024 \approx 10^3\), so \(2^{40} \approx 10^{12}\). Start there and work up. (Nobody has folded a sheet of paper more than about a dozen times. Why do you think that is? Think about area as well as thickness.)
Answers
- \(3^5 = 243\); \(2^6 = 64\); \(7^2 = 49\); 1; \(\tfrac1{100} = 0.01\).
- \(7.5 \times 10^6\); \(4.2 \times 10^{-4}\); 320 000; 0.006.
- \(3.84 \times 10^5\) km \(\div\, 3 \times 10^5\) km/s \(= 1.28\) s. Moonlight is 1.3 seconds old.
- \(4.2 \times 9.5 \times 10^{12} \approx 4.0 \times 10^{13}\) km. Hours: \(4.0 \times 10^{13} \div (5 \times 10^4) = 8 \times 10^8\) hours. Years: \(8 \times 10^8 \div 8800 \approx 9 \times 10^4\), about 90 000 years. The nearest star is very far away.
- Seconds in a year: \(60 \times 60 \times 24 \times 365 = 31\,536\,000 \approx 3.2 \times 10^7\). Heartbeats: \(70 \times 60 \times 24 \times 365 \times 80 \approx 2.9 \times 10^9\), about three billion.
- \(10^{-3} \div 10^{-10} = 10^7\): ten million atoms.
Stretch. \(2^{40} \approx 1.1 \times 10^{12}\), \(2^{41} \approx 2.2 \times 10^{12}\), \(2^{42} \approx 4.4 \times 10^{12}\). So 42 folds: thickness \(4.4 \times 10^8\) m, past the Moon. The catch is area. Each fold halves the area, and if you alternate directions (21 folds one way, 21 the other) each width is halved 21 times. A sheet the size of a football field, 100 m by 70 m, would end up about 48 by 33 micrometres across, thinner than a hair but still hundreds of thousands of atoms wide. Meanwhile its thickness is \(4.4 \times 10^8\) m. A column far taller than it is wide cannot be folded again, which is why real folding stops long before 42.