Why a⁰ = 1 and a⁻³ = 1/a³
Tonight's idea
\(a^5\) means five copies of \(a\) multiplied together. The small number, the exponent, is a count, and every law below is just counting. Here \(a\) is any number except 0, because we are about to divide by it.
Multiply \(a^3 \cdot a^4\): three copies times four copies is seven copies. Divide \(a^7 / a^3\): three cancel, four remain. Raise a power to a power, \((a^3)^4\): four groups of three, twelve in all.
\(a^m \cdot a^n = a^{m+n}\), \(\quad \dfrac{a^m}{a^n} = a^{m-n}\), \(\quad (a^m)^n = a^{mn}\).
Now the strange cases. Nobody can multiply zero copies of \(a\), but the division law does not care. \(a^3 / a^3\) is 1, and the law says it is \(a^{3-3} = a^0\). If the law is to keep working, \(a^0\) must equal 1.
Same for negatives. \(a^2 / a^5\): two copies cancel, three are left on the bottom, so the answer is \(1/a^3\). The law says \(a^{2-5} = a^{-3}\). So \(a^{-3}\) must mean \(1/a^3\).
The picture says the same thing. Going right, each step divides by 2. Keep going past \(2^1 = 2\): \(2^0 = 1\), then \(2^{-1} = \tfrac12\), then \(2^{-2} = \tfrac14\). Nothing else lets the pattern continue.
One more law for free: \((ab)^n = a^n b^n\), since \(n\) copies of \(ab\) regroup into \(n\) copies of \(a\) and \(n\) copies of \(b\).
Watch one: \((2^3)^4\), \(2^{-5}\), and \(10^6 \div 10^9\)
Power of a power: \((2^3)^4 = 2^{12} = 4096\). Not \(2^7\); four groups of three is twelve.
Negative exponent: \(2^{-5} = \dfrac{1}{2^5} = \dfrac{1}{32}\). The minus sign means "one over", nothing about the answer being negative.
Division: \(10^6 \div 10^9 = 10^{6-9} = 10^{-3} = \dfrac{1}{1000}\). Check by cancelling: six zeros on top, nine on the bottom, three left on the bottom.
Scientific notation uses the same idea, with \(10^{-3}\) for thousandths: see Big and tiny numbers.
Your turn
- Write each as a single power, then as a number: (a) \(5^7 \cdot 5^{-3}\); (b) \((3^2)^5 \div 3^7\); (c) \(10^{-2} \cdot 10^5\).
- Evaluate: (a) \(2^{10} / 2^{13}\); (b) \((-3)^0\); (c) \(-3^0\); (d) \(\left(\tfrac12\right)^{-3}\). Two of these are traps. Say which.
- Find \(n\) in each case by writing both sides as powers of the same base: (a) \(4^n = 2^{10}\); (b) \(9^n \cdot 3 = 3^7\); (c) \(2^n = \tfrac{1}{32}\).
- (a) Which is bigger, \(2^{300}\) or \(3^{200}\)? No calculator is needed. Rewrite each as a power with exponent 100. (b) Show that \(4^{50} \cdot 25^{50}\) is a power of 10, and say how many digits it has.
Stretch: what the laws force \(x^{1/2}\) to mean
Suppose fractional exponents obey the same laws. Then \((x^{1/2})^2 = x^{1} = x\). So \(x^{1/2}\) is a number whose square is \(x\): it has to be \(\sqrt{x}\). Use the same reasoning to evaluate \(8^{1/3}\), \(8^{2/3}\) and \(16^{-3/4}\).
Answers
- (a) \(5^{4} = 625\). (b) \(3^{10} \div 3^7 = 3^3 = 27\). (c) \(10^3 = 1000\).
- (a) \(2^{-3} = \tfrac18\). (b) \(1\): anything non-zero to the power 0 is 1. (c) \(-1\): the exponent applies to 3 only, then the minus sign. (d) \(8\): one over \(\left(\tfrac12\right)^3 = \tfrac18\). The traps are (c) and (d).
- (a) \(4^n = (2^2)^n = 2^{2n}\), so \(2n = 10\), \(n = 5\). (b) \(3^{2n} \cdot 3^1 = 3^{2n+1} = 3^7\), so \(n = 3\). (c) \(\tfrac{1}{32} = 2^{-5}\), so \(n = -5\).
- (a) \(2^{300} = (2^3)^{100} = 8^{100}\) and \(3^{200} = (3^2)^{100} = 9^{100}\). Since \(9 > 8\), \(3^{200}\) is bigger. (b) \(4^{50} \cdot 25^{50} = (4 \cdot 25)^{50} = 100^{50} = 10^{100}\): a 1 followed by 100 zeros, so 101 digits.
Stretch. \((8^{1/3})^3 = 8\), so \(8^{1/3} = 2\). \(8^{2/3} = (8^{1/3})^2 = 2^2 = 4\). \(16^{-3/4} = 1 / 16^{3/4} = 1/(16^{1/4})^3 = 1/2^3 = \tfrac18\).