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I have a equation $(-1)^{2m}$=$1$ so for this $m$ has to be integer but if i write it as $((-1)^2)^m$=1 , $m$ can be rational as well. so which solution is right and why? thanks for your help

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  • $\begingroup$ Usually $m$ denotes an integer. Are you looking for rational solutions? $\endgroup$ – Angelo Rendina Sep 24 '14 at 13:51
  • $\begingroup$ well if i take 2 inside all real values of m are accepted but thats not the case when 2 is not taken inside , which feels a bit contradictory to me $\endgroup$ – avz2611 Sep 24 '14 at 13:53
  • $\begingroup$ It's the old mess due to radicals. If you bring 2 inside you are changing equation. $\endgroup$ – Angelo Rendina Sep 24 '14 at 13:55
  • $\begingroup$ Intuitively, $(-1)$ raised to any even power is $1$, so we have all integers as a solution to $m$. $\endgroup$ – Daccache Sep 24 '14 at 13:55
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    $\begingroup$ It is. The problem is exactly that you are squaring the $-1$. $\endgroup$ – Angelo Rendina Sep 24 '14 at 14:04
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As a start consider $m = \frac12$ - then $$(-1)^{2m} = (-1)^1 = -1\ne 1$$

In general, we do not allow rational powers of negative numbers for this very reason - if $x$ is negative and $a,b \in \mathbb Q$ then in general $$x^{ab} \ne (x^a)^b$$

Some textbooks allow rational powers of negative numbers if the denominator is odd - in this case you are correct, and any $m \in \mathbb Q$ with odd denominator will be a solution.


The occurs because we define $x^{\frac12}$ to be the positive values $y$ satisfying $y^2 =x$. But $-y$ will also satisfy this equation. For example, $$\sqrt{x^2} = x$$ is only true if $x$ is non-negative.

In addition, it is not immediately clear how to define something like $(-1)^{\frac14}$. It could be any of $\pm \dfrac1{\sqrt2}\left(1\pm i\right)$ and there is no obvious way to choose which of these we should assign to be $(-1)^\frac14$.

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  • $\begingroup$ oh well i did not know that , but why does this happen ? $\endgroup$ – avz2611 Sep 24 '14 at 13:56
  • $\begingroup$ In short because $(-1)^\frac12$ is not a real number. For a similar problem, you would expect $\sqrt {x^2}$ to be $x$, but this is not true if $x$ is negative $\endgroup$ – Mathmo123 Sep 24 '14 at 14:01
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Obviously, $2m$ must be an exponent that you accept as valid for a negative base.

Irrational numbers are ruled out.

The case of integers depends on the parity. Let use denote as $e$ an even integer and $o$ an odd one: $(-1)^e=1$ and $(-1)^o=-1$.

The case of integer inverses is clear as well: $(-1)^{1/e}$ is not defined and $(-1)^{1/o}=-1$.

The case of rational needs more care as $(-1)^{p/q}$ could be interpreted both as $((-1)^p)^{1/q}$ or $((-1)^{1/q})^p$.

Anyway,

  • $((-1)^o)^{1/o'}=(-1)^{1/o'}=-1$ and $((-1)^{1/o'})^o=(-1)^o=-1$, so $(-1)^{o/o'}=-1$,

  • $((-1)^e)^{1/o}=1^{1/o}=1$ and $((-1)^{1/o})^e=(-1)^e=1$, so $(-1)^{e/o}=1$,

  • $((-1)^o)^{1/e}=(-1)^{1/e}$ and $((-1)^{1/e})^o$ are both undefined and so is $(-1)^{o/e}$.

In this sense, raising a negative to a rational power is a well defined operation.

Now it makes sense to accept $$2m=\frac eo,$$ or $$\color{blue}{m=\frac io},$$ where $i$ is any integer.

As has been said by others, we may not use $((-1)^a)^b)=(-1)^{ab}$, so that $(-1)^{2m}=1$ and $((-1)^2)^m=1$ are two different equations, with different solutions.

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