1

According to index laws, $(a^b)^c=a^{b\cdot c}=(a^c)^b$

However, if for example we have $a=-1, b=4, c=1/2$, then we get the equation:

$$((-1)^4)^{1/2}=(-1)^2=((-1)^{1/2})^4$$

The first equation is equal to $1$, however, the last one is undefined. How is this possible?

P.S. I'm sorry the formatting of the indexes is not ideal, I'm not sure how to do it properly

Jyrki Lahtonen
  • 133,153
JAS
  • 143
  • Did you think about $i^2=-1$ ? – Claude Leibovici Jun 13 '14 at 09:12
  • 1
    The reason is that $(-1)^{1/2}$ does not exist. – Did Jun 13 '14 at 09:19
  • When fractional exponents are involved the "law" $(a^b)^c=a^{bc}=(a^c)^b$ holds only if $a>0$, and the powers use the principal value. If $a<0$, and the exponents are not integers, then the power has multiple values, and the perils of multi-valued "functions" rear their ugly head. There are plenty of examples like this tagged under fake-proofs. – Jyrki Lahtonen Jun 13 '14 at 09:19
  • Thanks @claude That is helpful in this situation, but what if we consider $b=2$ as oppose to $b=4$. In that case, the first equation is $$((-1)^2)^1/^2)=(1)^1/^2=1$$ and the last is $$((-1)^1/^2)^2=(i)^2=-1$$ have I done something wrong? – JAS Jun 13 '14 at 09:22
  • @Jas: Please check that I fixed your TeX. I guessed that you wanted the exponents to be the ones I wrote. Click "edit" button to see how it is done. You need curly braces {} around the exponent. – Jyrki Lahtonen Jun 13 '14 at 09:23
  • Thanks @JyrkiLahtonen that makes sense – JAS Jun 13 '14 at 09:24
  • For the case $b=2$, working in the complex domain, you have to consider that the square root function has two "branches", so that $1^{1/2}$ is one of $+1$ or $-1$ (similarly, ${(-1)}^{1/2}=\pm i$, but this makes no difference after squaring). –  Jun 13 '14 at 09:30

1 Answers1

3

Attention the exponent law you give is only true under certain assumptions. A few example:

  • If $a \in \mathbb R$ and $b,c \in \mathbb Z$ then $a^{bc} = (a^b)^c$

  • If $a \in \mathbb R$, $a > 0$ and $b,c \in \mathbb{R}$ then $a^{bc} = (a^b)^c$

But $a^{bc} = (a^b)^c$ is not always true for $a,b,c \in \mathbb{R}$... In particular if $a<0$ and $b$ or $c$ is not an integer.

Surb
  • 55,662