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Quadratic root is defined as $\sqrt{ x^2} = |x|$. Easy to remember, but seems to lack logic. And this topic is about you proving me wrong.

1) This definition of a square root is not universal and is restricted to one special case when $x \in R$.

2) And how do we define roots of higher orders then? $\sqrt[4]16 = 2,2i$ (I do not add two more negative answers because of that "absolute value" trick). That means $\sqrt[4]{x^4} = |x|$ is not a function as it returns two values.

3) "Square root has to return only one value, otherwise it is not function. And we want really badly square root to be a function, thus we force it to return an absolute value". Sounds pretty much like the most convincing explanation to me for such a definition. As of my understanding square root (just like roots of other orders) returns more than one value thus is not a function and should not be forced to resemble one

user3600124
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    Complex numbers have no "sign" (positive/negative) as real numbers do, so the above doesn't apply for them, and for negative numbers it works just fine: $;\sqrt{(-2)^2}=2=|-2|;$ , for example – DonAntonio Mar 15 '16 at 20:43
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    By definition, "negative number" is one that is less than zero. Since the complex cannot be ordered, there is no "less than" or "more than" in the complex numbers. What you mean is that we have the notation $;-z;$ for the additive inverse of a complex number. – DonAntonio Mar 15 '16 at 21:02
  • And how do I thumb your remark up now. You are very right – user3600124 Mar 15 '16 at 21:04
  • Thank you, don't worry about that. Most probably you'll have to wait for more reputation points (?) in order to be able to upvote. – DonAntonio Mar 15 '16 at 21:06

2 Answers2

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There are several functions here. As far as I remember:

1)
$\sqrt[2n]x$ for x real is defined for $x \geq 0$ only ($n\geq1$ is an integer).
It's a function returning a single real value which is non-negative.

2)
$\sqrt[2n+1]x$ for x real is defined for all real x ($n \geq 1$ integer).
It's a function returning a single real value (it can be negative).

3) $\sqrt[n]z$ for z complex is defined for all complex z.
It's a multi-valued function which returns $n$ complex numbers.
Note that if you take a real number and use that third definition,
it works OK. One of the roots returned here coincides with the root
/returned by 1) or 2)/, if such a root exists / as per 1) or 2) /.
So this third definition is more general than 1) and 2), it builds upon them.

Of course this idea of root can be further extended and refined,
but at a certain level of knowledge I think this idea of roots is OK
and makes total sense.

peter.petrov
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  • 4 is a complex number (it's imaginary part equals to zero). And when we write $\sqrt4$, what do we refer to? The first definition you have mentioned of a square root or the third one? They return different answers – user3600124 Mar 15 '16 at 20:52
  • When we write $\sqrt4$ we usually mean definition 1) unless we explicitly say otherwise. – peter.petrov Mar 15 '16 at 21:02
  • returns 2 while 3) returns {2,-2}. So ... right, the return different things, but one of the answers returned by 3) is the complex number 2 i.e. the same root returned by 1). The other one returned by 3) is the complex number −2 + 0i (which also happens to be purely real). So 3) is an extension of 1) and 2), something more general. That's what mathematicians want when they extend from reals to complex numbers.
    • – peter.petrov Mar 15 '16 at 21:13