I want MMA to tell me if two, rather large, expressions are equal. My minimum working example follows. I want MMA to tell me that the following is true:$$x^n = \left(\frac{1}{x}\right)^{-n}$$ I have tried

x^n == (x^(-1))^(-n)


FullSimplify[x^n == (x^(-1))^(-n)]

Neither indicates that the expressions are equal. In fact:

PossibleZeroQ[x^n - (x^(-1))^(-n)]

returns False. Is there any way to get MMA to tell me that the equality holds? Or maybe a better approach/paradigm?

  • 1
    $\begingroup$ It's not generally true, but it is for integers. Try FullSimplify[x^n == (x^(-1))^(-n), n \[Element] Integers]. -- Also for positive bases: FullSimplify[x^n == (x^(-1))^(-n), x > 0] $\endgroup$
    – Michael E2
    Aug 2, 2015 at 2:14
  • $\begingroup$ Exactly what I needed. Thank you. $\endgroup$
    – user39275
    Aug 2, 2015 at 2:20

1 Answer 1


One of the pitfalls new users face, especially if they have analysis (calculus) of only real functions of real variables, is that Mathematica assumes by default that variables are complex and functions are the complex functions. Power functions in particular can seem strange; even those who know it is complex sometimes forget. The cube root is a simple example:

(*  (-1)^(1/3)         -- not -1  *)

(*  0.5 + 0.866025 I   -- it's a complex 3rd root of -1 *)

We can see that the OP's proposed identity does not hold generally:

Block[{x = -1, n = 1/3},
 N@ {x^n, (x^(-1))^(-n)}
(*  {0.5 + 0.866025 I, 0.5 - 0.866025 I}   -- different imaginary parts  *)

There are two cases where it does hold:

FullSimplify[x^n == (x^(-1))^(-n), n ∈ Integers]
FullSimplify[x^n == (x^(-1))^(-n), x > 0]

One can find other cases, too:

Block[{x = I, n = 1/3},
 x^n == (x^(-1))^(-n)]
(*  True  *)



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