2
$\begingroup$

Quick easy question! I was always taught at school that $(x^m)^n=x^{mn}$. However, I've only just noticed when computing something (two years into an undergrad course..) that for example $(x^2)^{\frac{1}{2}}=|x|$ rather than $x$. I've never really thought about it before. A lot of the time when simplifying I would have blindly done things like $\sqrt{x^4+x^2}=x\sqrt{x^2+1}$ which is clearly wrong now I've looked at it.

Is there a general rule I have not been taught before?

$\endgroup$
4
  • 4
    $\begingroup$ The rule $(x^m)^n=x^{mn}$ is true if $m$ and $n$ are integers. $\endgroup$
    – user85798
    Dec 28, 2013 at 21:48
  • 1
    $\begingroup$ I think that pretty much answers my question. $\endgroup$
    – Mike Miller
    Dec 28, 2013 at 21:50
  • 2
    $\begingroup$ Alternatively, $(x^m)^n = x^{mn}$ is also true when $m$ and $n$ are arbitrary reals but $x > 0$. $\endgroup$
    – Erick Wong
    Dec 28, 2013 at 21:55
  • $\begingroup$ Taking non-integer powers of non-positive numbers is often a non-safe thing to do. Caveat exponentiator. $\endgroup$
    – ncmathsadist
    Dec 28, 2013 at 22:47

1 Answer 1

Reset to default
4
$\begingroup$

$(x^2)^{1/2} = |x|$ is correct for real $x$. But incorrect (in general) for complex $x$.

For example, if $x=i$, the equation would be $(-1)^{1/2}=1$ which is wrong.

For complex $x$, write $\overline{x}$ for the complex conjugate of $x$, then we do have $\left(x \;\overline{x}\right)^{1/2} = |x|$ for all complex $x$.

$\endgroup$
1
  • $\begingroup$ It was more a question about what I (thought!) I was taught rather than the example, but a good answer nonetheless. In the complex example, would $\sqrt{z^4+z^2}=z\sqrt{z^2+1}$ be correct then? $\endgroup$
    – Mike Miller
    Dec 28, 2013 at 22:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .