Powers with negative exponents

So far, the exponents of powers were natural numbers, or 00. But we can also have negative exponents, such as

2−32^{-3}

But how do we define such a power? Let's have another look at our progression of decreasing exponents:

23=8:222=4:221=2:220=1:2\begin{array}{lll} 2^3&=8 \quad\quad :2\\ 2^2&=4 \quad\quad :2\\ 2^1&=2 \quad\quad :2\\ 2^0&=1 \quad\quad :2\\ \end{array}

Clearly, if we want to continue the pattern of "dividing by 2" if we increase the exponent by 11, we get

2−1=12:22−2=14:22−3=18\begin{array}{lll} 2^{-1} &= \frac{1}{2}\quad\quad :2\\ 2^{-2} &= \frac{1}{4}\quad\quad :2\\ 2^{-3} &= \frac{1}{8}\\ \end{array}

But note that we can write these fractions also as powers:

2−1=12=1212−2=14=1222−3=18=123\begin{array}{lll} 2^{-1} &= \frac{1}{2}=\frac{1}{2^1}\\ 2^{-2} &= \frac{1}{4}=\frac{1}{2^2}\\ 2^{-3} &= \frac{1}{8}=\frac{1}{2^3} \end{array}

And a simple rule emerges:

2−n=12n2^{-n}=\frac{1}{2^n}

and more generally we have that

a−n=1an\boxed{a^{-n}=\frac{1}{a^n}}

and also

ca−n=can\boxed{ca^{-n}=\frac{c}{a^n}}

(This last equation follows from c⋅a−n=c⋅1an=canc\cdot a^{-n}=c\cdot \frac{1}{a^n} = \frac{c}{a^n}). We can extend this definition to brackets as well. e.g.

(x+y)−1=1x+y(x+y)^{-1}=\frac{1}{x+y} 3(a2−1+b)−3=3(a2−1+b)33(a^2-1+b)^{-3}=\frac{3}{(a^2-1+b)^3}
Note 1

Key to a good understanding of negative exponents is to know your rules about fractions. In particular the multiplication of fractions. Recall that

abcd=ab⋅dc\dfrac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b} \cdot \frac{d}{c}

and also that

c⋅ab=cabc \cdot \frac{a}{b}=\frac{ca}{b}

because we can write c=c1c=\frac{c}{1}. Another useful rule is how to evaluate double fractions:

abcd=ab⋅dc\dfrac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b} \cdot \frac{d}{c}

and therefore

acd=a⋅dc\dfrac{a}{\frac{c}{d}}=a \cdot \frac{d}{c}

because we can write a=a1a=\frac{a}{1}.

Exercise 1

Are these statements correct? Argue with the definitions of powers and the properties of fractions.

  1. 4−2=1164^{-2}=\frac{1}{16}

  2. (3⋅10)−1=0.3(3\cdot 10)^{-1}= 0.3

  3. 3⋅10−1=0.33\cdot 10^{-1}= 0.3

  4. (0.25−2)−1=1/8\left(0.25^{-2}\right)^{-1}=1/8

  5. a7⋅a−3=a4a^7 \cdot a^{-3} = a^4

  6. 3a−2=3a23 a^{-2} = \frac{3}{a^2}

  7. (3a)−2=13a2\left(3a\right)^{-2}=\frac{1}{3a^2}

  8. (ab)−3=3ab\left(ab\right)^{-3}=\frac{3}{ab}

  9. (a2)−3=a−6\left(a^2\right)^{-3}=a^{-6}

  10. (2a2)−1=12a−2\left(2a^2\right)^{-1}=\frac{1}{2} a^{-2}

  11. 2a4=2a−4\frac{2}{a^4}=2 a^{-4}

  12. 1a−3=a3\frac{1}{a^{-3}}=a^3

  13. (12)−3=8\left( \frac{1}{2}\right)^{-3} = 8

  14. a−3b−3=(ab)−3a^{-3} b^{-3}=(ab)^{-3}

  15. ab−1=aba b^{-1} = \frac{a}{b}

  16. a−3b3=(ab)−9a^{-3} b^{3}=(ab)^{-9}

  17. a2a−2=1a^2 a^{-2} = 1

  18. (ab)−2=b2a2\left( \frac{a}{b}\right)^{-2} = \frac{b^2}{a^2}

  19. 1a4=a−4\frac{1}{a^4}=a^{-4}

  20. 1a−4=a4\frac{1}{a^{-4}}=a^{4}

  21. 3a4=3a−4\frac{3}{a^4}=3a^{-4}

  22. 13a4=13a−4\frac{1}{3a^4}=\frac{1}{3}a^{-4}

  23. (1a4)−2=a8\left( \frac{1}{a^4}\right)^{-2}=a^{8}

Solution
  1. 4−2=142=1164^{-2} =\frac{1}{4^2}=\frac{1}{16} correct
  2. (3⋅10)−1=13⋅10=1⋅13⋅10=13⋅110=0.3‾⋅0.1=0.03‾≠0.3(3\cdot 10)^{-1}= \frac{1}{3\cdot 10} = \frac{1\cdot 1}{3\cdot 10}=\frac{1}{3} \cdot \frac{1}{10}= 0.\overline{3}\cdot 0.1 = 0.0\overline{3}\neq 0.3 wrong
  3. 3⋅10−1=3⋅110=3⋅0.1=0.33\cdot 10^{-1}= 3\cdot \frac{1}{10}=3\cdot 0.1= 0.3 correct
  4. (0.25−2)−1=((14)−2)−1=(1(14)2)−1=(114⋅14)−1(1116)−1=(11116)−1=16−1=116≠18\left(0.25^{-2}\right)^{-1}= \left( \left(\frac{1}{4}\right)^{-2} \right)^{-1} = \left(\dfrac{1}{\left(\frac{1}{4}\right)^2}\right)^{-1} = \left(\dfrac{1}{\frac{1}{4}\cdot \frac{1}{4}}\right)^{-1} \left(\dfrac{1}{\frac{1}{16}}\right)^{-1} =\left(\frac{\frac{1}{1}}{\frac{1}{16}}\right)^{-1}=16^{-1}=\frac{1}{16} \neq \frac{1}{8} wrong
  5. a7⋅a−3=a7⋅1a3=a7a3=a4a^7 \cdot a^{-3} = a^7 \cdot \frac{1}{a^3} = \frac{a^7}{a^3}=a^4 correct
  6. 3a−2=3⋅1a2=31⋅1a2=3a23 a^{-2} = 3 \cdot \frac{1}{a^2} = \frac{3}{1} \cdot \frac{1}{a^2} = \frac{3}{a^2} correct
  7. (3a)−2=1(3a)2=13⋅a⋅3⋅a=19a2≠13a2\left(3a\right)^{-2}= \frac{1}{(3a)^2}=\frac{1}{3\cdot a\cdot 3\cdot a}=\frac{1}{9a^2} \neq \frac{1}{3a^2} wrong
  8. (ab)−3=1(ab)3≠3ab\left(ab\right)^{-3}= \frac{1}{(ab)^3} \neq \frac{3}{ab} wrong
  9. (a2)−3=1(a2)3=1a2⋅a2⋅a2=1a6=a−6\left(a^2\right)^{-3}= \frac{1}{(a^2)^3} =\frac{1}{a^2 \cdot a^2 \cdot a^2 } =\frac{1}{a^6} = a^{-6} correct
  10. (2a2)−1=12a2=1⋅12⋅a2=12⋅1a2=12a−2\left(2a^2\right)^{-1}= \frac{1}{2a^2} = \frac{1\cdot 1}{2 \cdot a^2} = \frac{1}{2} \cdot \frac{1}{a^2} = \frac{1}{2} a^{-2} correct
  11. 2a4=2⋅1a4=2a−4\frac{2}{a^4}= 2\cdot \frac{1}{a^4} = 2 a^{-4} correct
  12. 1a−3=11a3=a3\frac{1}{a^{-3}}= \dfrac{1}{\frac{1}{a^3}} = a^3 correct
  13. (12)−3=1(12)3=112⋅12⋅12=118=8\left( \frac{1}{2}\right)^{-3} = \dfrac{1}{\left(\frac{1}{2}\right)^3}= \dfrac{1}{\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}} = \dfrac{1}{\frac{1}{8}} = 8 correct
  14. a−3b−3=1a3⋅1b3=1a3b3=1(ab)3=(ab)−3a^{-3} b^{-3}= \frac{1}{a^3}\cdot \frac{1}{b^3} =\frac{1}{a^3 b^3} = \frac{1}{(ab)^3} = (ab)^{-3} correct
  15. ab−1=a1b=aba b^{-1} = a\frac{1}{b} = \frac{a}{b} correct
  16. a−3b3=1a3b3=b3a3≠(ab)−9a^{-3} b^{3}=\frac{1}{a^3} b^3 = \frac{b^3}{a^3} \neq (ab)^{-9} (which is 1(ab)9\frac{1}{(ab)^9}), wrong
  17. a2a−2=a2a2a2=1⋅̸a21⋅̸a2=11=1a^2 a^{-2} = a^2\frac{a^2}{a^2}=\frac{1\cdot \not{a^2}}{1\cdot \not{a^2}} =\frac{1}{1} =1 correct
  18. (ab)−2=1(ab)2=1ab⋅ab=1a2b2=b2a2\left( \frac{a}{b}\right)^{-2} = \dfrac{1}{\left(\frac{a}{b}\right)^2} = \dfrac{1}{ \frac{a}{b}\cdot\frac{a}{b}} = \dfrac{1}{\frac{a^2}{b^2}} = \frac{b^2}{a^2} correct
  19. 1a4=a−4\frac{1}{a^4}=a^{-4} correct by definition
  20. 1a−4=11a4=a4\frac{1}{a^{-4}}= \dfrac{1}{\frac{1}{a^4}} = a^{4} correct
  21. 3a4=3⋅1a4=3a−4\frac{3}{a^4}= 3\cdot \frac{1}{a^4} = 3 a^{-4} correct
  22. 13a4=1⋅13⋅a4=131a4=13a−4\frac{1}{3a^4}= \frac{1\cdot 1}{3\cdot a^4} =\frac{1}{3}\frac{1}{a^4} = \frac{1}{3}a^{-4} correct
  23. (1a4)−2=1(1a4)2=11a41a4=11a4a4=11a8=a8\left( \frac{1}{a^4}\right)^{-2}= \dfrac{1}{\left(\frac{1}{a^4}\right)^2} =\dfrac{1}{\frac{1}{a^4} \frac{1}{a^4}} = \dfrac{1}{\frac{1}{a^4 a^4}}= \dfrac{1}{\frac{1}{a^8}} = a^{8} correct
Exercise 2

The power rules. aa and bb are some numbers, nn and mm are integers. Fill in the boxes.

  1. (SB) Same base rules:

    a−n⋅a−m=a□,a^{-n} \cdot a^{-m} = a^{\square}, \quad

    a−na−m=a□\frac{a^{-n}}{a^{-m}} = a^{\square}

    an⋅a−m=a□,a^{n} \cdot a^{-m} = a^{\square}, \quad

    ana−m=a□\frac{a^{n}}{a^{-m}} = a^{\square}

    a−n⋅am=a□,a^{-n} \cdot a^{m} = a^{\square}, \quad

    a−nam=a□\frac{a^{-n}}{a^{m}} = a^{\square}

  2. (SE) Same exponent rules:

    a−n⋅b−n=(a⋅b)□,a^{-n} \cdot b^{-n} = (a\cdot b)^\square, \quad

    a−nb−n=(ab)□\frac{a^{-n}}{b^{-n}}=\left(\frac{a}{b}\right)^\square

  3. (PP) Power of a power rule: (a−n)−m=a□\left(a^{-n}\right)^{-m} = a^\square

    (an)−m=a□\left(a^{n}\right)^{-m} = a^\square

    (a−n)m=a□\left(a^{-n}\right)^{m} = a^\square

Solution

  1. (SB) Same base rules: a−n⋅a−m=a−(n+m)a^{-n} \cdot a^{-m} = a^{-(n+m)}

    a−na−m=a−n+m\frac{a^{-n}}{a^{-m}} = a^{-n+m}

    an⋅a−m=an−ma^{n} \cdot a^{-m} = a^{n-m}

    ana−m=an+m\frac{a^{n}}{a^{-m}} = a^{n+m}

    a−n⋅am=a−n+ma^{-n} \cdot a^{m} = a^{-n+m}

    a−nam=a−n−m\frac{a^{-n}}{a^{m}} = a^{-n-m}

  2. (SE) Same exponent rules: a−n⋅b−n=(a⋅b)−na^{-n} \cdot b^{-n} = (a\cdot b)^{-n},a−nb−n=(ab)−n\frac{a^{-n}}{b^{-n}}=\left(\frac{a}{b}\right)^{-n}

  3. (PP) Power of a power rule: (a−n)−m=anm\left(a^{-n}\right)^{-m} = a^{nm}

    (an)−m=a−nm\left(a^{n}\right)^{-m} = a^{-nm}

    (a−n)m=a−nm\left(a^{-n}\right)^{m} = a^{-nm}