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I'm feeling confused. If I square 1 and -1, the answers should be equal:

$1^2 = (-1)^2$

Then I take both sides to the power of $\frac12$:

$\left(1^2\right)^\frac12 = \left((-1)^2\right)^\frac12$

This next step seems to make sense according to the simple arithmetic rule about multiplying exponents:

$1^\left(2\cdot\frac12\right) = (-1)^\left(2\cdot\frac12\right)$

And then comes the weirdness:

$1^\left(\frac22\right) = (-1)^\left(\frac22\right)$

$1^1 = (-1)^1$

$1 = -1$

Obviously I did something wrong... Every step seems perfectly reasonable except going from step 2 to step 3. That seems reasonable too, that's what I was taught about exponents, but that's the only step which I could conceive has special constraints I violated. Is $\left((-1)^2\right)^\frac12 = (-1)^\left(2\cdot\frac12\right)$?

at01
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    http://math.stackexchange.com/q/664383/1242 – Hans Lundmark Feb 25 '14 at 09:10
  • @HansLundmark, so $\left(x^a\right)^b = x^\left(a\cdot b\right)$ only when x is positive? Why is that? I guess I intuitively felt that, but still don't understand why. If a number is multiplied by itself m number of times and that whole thing is multiplied by itself n number of times, then the original number must've been multiplied by itself mn times... – at01 Feb 25 '14 at 10:03
  • That's true for integer exponents, but how do you multiply something with itself a half number of times? – Hans Lundmark Feb 25 '14 at 11:17

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Note that $\sqrt{1} = \pm 1$. By squaring, you end up with extraneous solutions.

Another example is trying to solve $x=2$. By squaring both sides, we get $x^2 = 4$, which has solutions $x=2$ and $x-2$, hence $-2=2$, which is nonsense.

naslundx
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    the radical is defined as the positive square root, so it can't be -1 just by definition. In your example, it's not a problem for x to be both 2 and -2, that doesn't mean 2 = -2. – at01 Feb 25 '14 at 10:14
  • It is a problem for $x=-2$, since this is not a solution to the original equation $x=2$. – naslundx Feb 25 '14 at 10:16
  • I see what you're saying. However, squaring x does naturally change the equation somewhat. In the example I gave, we're just squaring 1 and -1. – at01 Feb 25 '14 at 10:44
  • The point is that $x^2 = y^2$ does not imply $x = y$. (In this case, $x=1$ and $y=-1$.) – naslundx Feb 25 '14 at 10:48