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As far as I know, $x^{1/2} = \sqrt{x}$. On the other hand, $\sqrt{i^4} = \sqrt{1} = 1$ and $(i^4)^{1/2} = i ^ {4/2} = i^2 = -1$. Does this mean that $\sqrt{x} \neq x^{1/2}$?

Soroush khoubyarian
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  • Are you sure @Rajiv Kaipa? – Juniven Acapulco Apr 13 '17 at 12:09
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    @RajivKaipa That's unusual, the common definition of $\sqrt{a}$ is the (unique) positive number $b$ such that $b^2=a$. – StackTD Apr 13 '17 at 12:10
  • @ΘΣΦGenSan $(-1)\cdot (-1) = 1 \cdot 1 $, thus these are the solutions to $\sqrt 1$ – Fabian Schn. Apr 13 '17 at 12:10
  • Yes @ Fabian Schn,, there is no problem with $(-1)(-1)=(1)(1)$, but it does not imply that $\sqrt{1}=\pm 1$. – Juniven Acapulco Apr 13 '17 at 12:13
  • @FabianSchn. No, if you define $\sqrt{1}=\pm 1$, then the assignment $\sqrt{ }:\mathbb{R}^+\rightarrow \mathbb{R}$ is not a single-valued function. Certainly that's not what you want. – Mathematician 42 Apr 13 '17 at 12:13
  • Also, does that fact that normal rules of exponents does not apply to complex number help? Because, if they do, we have $i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} = 1$ –  Apr 13 '17 at 12:16
  • Now i am confused. $\sqrt x$ asks for the values that multiplied by itself equals x, right? Thus, we have two solutions, a positive and negative... – Fabian Schn. Apr 13 '17 at 12:16
  • @FabianSchn. If you use that definition you cannot define the squareroot function. It is true that solving $x^2=a$ for some positive $a$ yields two solutions, this does not imply however that $\sqrt{a}$ takes two values. – Mathematician 42 Apr 13 '17 at 12:18
  • @FabianSchn One thing more, the symbol $\sqrt{x}$ means the positive square root of $x$. If you want the negative square root, then you write $-\sqrt{x}$. – Juniven Acapulco Apr 13 '17 at 12:24
  • sorry, Im so slept today, lol. I didnt read carefully your question. The problem is that in general for $a\in\Bbb C$ and $p/q\in\Bbb Q$ we have that $(a^p)^{1/q}\neq (a^{1/q})^p=: a^{p/q}$. So the problem is not with the notation $x^{1/q}=\sqrt[q]{x}$ if not with the swapping of order in the exponentiation. – Masacroso Apr 13 '17 at 13:00

3 Answers3

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You use the exponent law $\left( a^b \right)^c = a^{bc}$ in the following step:

$\color{red}{(i^4)^{1/2} = i ^ {4/2}} = i^2 = -1$

but that rule, while true for real numbers and a positive base $a$, does not hold in general. You can get in trouble with negative bases and more generally with complex numbers.

For more details and some examples, you can check Failure of power and logarithm identities.

StackTD
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That is a question of definitions. Already in the reals, you have $\sqrt{(-1)^2}$, that could be either one or minus one. In the reals, we solve this problem by defining that the square root of $x > 0$ should always be the unique positive number $y >0$, solving $y^2 = x$. However, we don't have an ordering -- i.e. a concept of positive or negative -- in the complex field. Thus here we run into problems.

It is still true that $\sqrt{x}$ and $x^{1/2}$ mean the same, however, you have to define what a square root is. As the complex field is algebraically closed, the equation $y^2 = x$ will always have two different solutions for $x \neq 0$ and thus you need a way to distinguish them to talk about a well-defined root. There are ways to do it, if you are interested in it you might want to study complex analysis.

Either way, you should remember that the rules for computing with these functions are not always true, i.e. the rule $(a^x)^y = a^{xy}$ only holds when both $x$ and $y$ are integers.

Dirk
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  • Thank you I appreciate it. – Soroush khoubyarian Apr 13 '17 at 12:16
  • In many contexts, but not necessarily all, $\sqrt{2}$ will be assumed to be the positive square root. This is especially likely with computer languages and calculators. As Bemte says, there is no similar simple rule when you are working with complex numbers. His last point is important, here is another common mistake: if you know that $x^2 = y^2$ then you cannot deduce that $x = y$. – badjohn Apr 13 '17 at 12:21
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The problem is the application of the usual power laws to complex numbers.

The principal complex power is defined as $$ z^\alpha=e^{\alpha \,\text{Log}z} $$ with Log(z) the principal logarithm (note the capitals in Log and Arg).

The principal square root of $i^4$ is then $$ (i^4)^{1/2}=e^{1/2 (\text{Log}\,i^4)} =e^{1/2(\log|i^4|+i\text{Arg}(i^4))} =e^{1/2(\log|1|+i0))} =e^0=1 $$ Both square roots of $i^4$, $\pm1$, follow from the multi-valued nature of the non-principal log function (note the lower case spelling in $\log$ and $\arg$)

$$ (i^4)^{1/2}=e^{1/2 \left( \text{log}\,i^4\right)} =e^{1/2(\log|i^4|+i\,\text{arg}(i^4))} =e^{1/2(\log|1|+i2k\pi))} =e^{ik\pi}=\pm1 $$ where $k$ is an integer.

PM.
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