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Which step in this process allows me to erroneously conclude that $i = 1$

According to the definition of exponentials, $\displaystyle(-1)^{\frac{2}{4}}$ is equivalent to $\sqrt{1}$. However, if we take the $4$ first and the square it. The answer will be $1$. What's the problem?

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Strin
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  • What do you mean by "definition of exponentials" and "sort(1)"? $(-1)^{2/4} = \sqrt{-1} = i$. – Alex Becker Dec 24 '11 at 08:45
  • @Strin You probably noticed that your post was edited (basically to make it more readable). You should check, whether these edits did not change the intended meaning and, if necessary, edit your question again. In particular, did you mean $\sqrt{1}$ or $\sqrt{-1}$? – Martin Sleziak Dec 24 '11 at 10:00

2 Answers2

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I admit, there is a bit of apparent ambiguity here. One might say $(-1)^{(2/4)} = (-1) ^ {(1/2)} = \sqrt{-1} = i$.

But then, according to what we are accustomed to in the real numbers, we might expect to say $(-1) ^ {(2/4)} = (-1^2)^{(1/4)} = 1^{(1/4)} = 1$, in the sense that we are used to saying that the fourth root of $1$ is $1$. Or one might expect to say $(-1)^{(2/4}) = ((-1)^{(1/4)})^2 = (-i)^2 = -1$, since $i$ is a primitive 4th root of unity.

But what we're really doing here is playing with principal values. For example, $\sqrt{4} = 2$. Why isn't it $-2$? So we must not be imprecise. The idea that $b^{p/q} = (b^p )^{1/q}$ is not true in general. And that's our problem. At the end of the day, one should interpret this to mean $\sqrt{-1} = i$, without changing any sorts of order on the exponents.

davidlowryduda
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Notice that $\sqrt{x^2} \ne x$, but instead $|x|$.

yihangho
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