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I am now teaching mathematics and in the math book for the students, it says that a 4th root is always positive. I have never thought about this rule before but actually, say for example:

$\sqrt[4]{16} = 2 $ because $ 2^2=4$.

But also

$(-2)^4 = 16$

But the rule for even number roots are clear. It says that even number roots are always positive, without giving any explanation on why. Could someone here answer the reason for this rule?

Mr. P
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  • It is a convention so that we get a function, giving only one value. One can, without any problem, choose any other branch (except for the obvious issue of inconsistencies with math texts). – ultralegend5385 Sep 12 '21 at 14:54
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    Basically that is the definition. For $\sqrt{\cdot}$ to be a function it needs to assign just one value $\sqrt{x}$ for each $x$, so we take de positive one. It is important to know that the equation $x^2=1$, for instance, has two real solutions, but it does not mean that $\sqrt{1}$ has two possible values. – Mateus Rocha Sep 12 '21 at 14:54
  • For any even number $n > 0$ and $a > 0$, the equation $x^n = a$ has two real solutions which are negatives of one another. We denote the positive solution by $\sqrt[n]{a}$ (and hence the negative solution is $-\sqrt[n]{a}$). For $n = 2$, this has been discussed before, see here for example. – Michael Albanese Sep 12 '21 at 14:55
  • Anyway, you obtain the other 4th roots multiplying any 4th root of a number by the comples 4th roots of unity. – Bernard Sep 12 '21 at 14:59
  • It's not so much that eventh roots are positive (they aren't: $16$ has four fourth roots, not all positive), but that for nonnegative $x$, the radical sign has a clear and unambiguous definition to output just nonnegative values. Please see Points 2 & 3 here. – ryang Sep 12 '21 at 15:39
  • Did you mean "because $2^4 = 16$"? – Phil Frost Sep 12 '21 at 23:32

2 Answers2

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As others said in the comments, it is from the definition that $\sqrt{{}\cdot{}}$ must assign one value of $\sqrt{x}$ for each $x$. Think of the function $x = y^2$, if you were to graph it would look like a sideways parabola, however, in practice this function actually takes the value of $y = \sqrt{x}$ where $y \geq 0$.

Arthur
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Igor
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$$x^4=16\implies 16e^{2ik\pi} \implies x_k=2 e^{(2ik\pi)/4}, k=0,1,2,3$$ So all the roots are $2, 2i, -2, -2i$, only the first one is real and positive.

Z Ahmed
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