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The rules of powers are in highschool books often briefly stated in the following way:

  1. $\displaystyle a^n \cdot a^m = a^{n+m}$
  2. $\displaystyle \frac{a^n}{a^m} = a^{n-m}$
  3. $\displaystyle \left (a\cdot b\right )^n = a^n \cdot b^n $
  4. $\displaystyle \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$
  5. $\displaystyle \left(a^n\right )^m = a^{n\cdot m}$

I sometimes try to explain to my highschool students that those rules are not always true. For example, $0^{-2} \cdot 0^{2} = 0^0$ or I give other interesting false deductions such as: $$\left(-1\right)^3=(-1)^{6\cdot \frac{1}{2}}=\left((-1)^{6}\right)^{\frac{1}{2}}=\sqrt{1}=1 $$

However I could not find an exact reference to where those rules are true.

Steward's Review of Algebra states that those rules are true if $a$ and $b$ are positive (real) numbers, and $n$ and $m$ are rational numbers. This is of course very conservative. Those rules are also true if $a\ne$, $b\ne 0$ and $n,m$ integers. Besides that I think many of those rules are also true if $n,m$ are real numbers.

So my question is, when are the above rules correct?

Kasper
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    The issue with $0$ is that $0^0$ is not "clear"; in some context it is adopted the convention $0^0=1$ but your "equivalence" support the definition of "indeterminate", due to the fact that $0^{-2}= \dfrac {1}{0^2}= \dfrac 1 0$. – Mauro ALLEGRANZA Mar 16 '18 at 09:30
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    My advice is to avoid writing $x^y$ unless either $x>0$ or $y$ is a non-negative integer. If you really must do it for other values of $x$ and $y$ please be very specific about what you mean by it. – Angina Seng Mar 16 '18 at 09:57
  • Anything involving $0^{-2}$ is surely a bad example, because $0^2$ is zero and you can't divide by zero. – Kevin Mar 16 '18 at 16:59
  • Things get much easier when all base are $>0$... It is not surprising that these rules are not always true for negative base, if you know the proof of them.(See Baby Rudin Chapter 1, for a consrtuction of $a^b$. – Tony Ma Mar 17 '18 at 01:01

2 Answers2

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Provided $a,b>0$, all the rules are true for real $a,b,m,n$.

If $a=0$ or $b=0$, no negative power may appear.

For $a<0$ or $b<0$, irrational exponents are excluded. Rational ones are possible provided the denominator of the simplified fraction is odd. This can cause rule 5 to fail ($(-1)^1\ne((-1)^{1/2})^2$).

  • Rational ones are possible provided the denominator of the simplified fraction is odd --- Need to include the proviso that for rational exponents the evaluation process begins by writing the exponent as a ratio of integers in lowest terms. Otherwise, the output is not uniquely defined for rational numbers, since $(-1)^{\frac{2}{6}}$ and $(-1)^{\frac{1}{3}}$ are different despite the fact that $\frac{2}{6}$ and $\frac{1}{3}$ are the same rational number. (Continued) – Dave L. Renfro Mar 16 '18 at 09:37
  • In the case of $(-1)^{\frac{2}{6}}$ we could be taking an even root of a negative, or we could be left not knowing how to proceed (your instructions don't tell us what to do if the fraction is NOT reduced). – Dave L. Renfro Mar 16 '18 at 09:38
  • @DaveL.Renfro: the rules are not written with exponents in the form $p/q$ with $p,q$ integers, but in the form $m$ rational. So your distinction between $2/6$ and $1/3$ is irrelevant. In any case, I wrote the simplified fraction which is the condition to apply the parity rule. –  Mar 16 '18 at 09:41
  • @DaveL.Renfro: once again, the exponents are not given in reduced or not reduced form. They are given as rationals, not as fractions. $2/6$ (if it made sense to "input" this) is an acceptable exponent, as the denominator of its simplified fraction representation is $3$, and $(-1)^{"2/6"}=-1$ because $2/6=1/3$. –  Mar 16 '18 at 09:49
  • @YvesDaoust Could we summarise this as, that the rules are true if the expressions in the rules are defined? If $a=0, a^n$ is not defined for negative powers. If $a<0$, $a^n$ is not defined for $n$ irrational. And only if $n$ is odd and $m$ and integer the expression $a^{\frac{m}{n}}$ is defined. – Kasper Mar 16 '18 at 13:55
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    @Kasper Nothing is undefined about ((-1)^6)^0.5 or (-1)^(60.5). It's just that a^(bc) = (a^b)*c is not in general true. – Joren Mar 16 '18 at 14:14
  • In light of your second paragraph, then what is $0^0$? – Allawonder Nov 04 '19 at 04:58
  • @DaveL.Renfro Apart from non-reduced rationals, we still need to specify how to choose one of the possible roots, especially when the denominator is not odd. – Allawonder Nov 04 '19 at 05:00
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Whenever the base is positive and the exponent is real, or the base is zero and the exponent is positive, or the base is negative and the exponent is an integer.

Rócherz
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