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I have this problem:

$$\sqrt{\sqrt{16}}$$ would it be positive 2 or it would be $\pm$2?

In this class we do not deal with complex numbers, so all the square roots are positive, thus for the class the answer is 2.

However, I am asking this question: in real mathematics, doesn't one consider the negative root as well?

rschwieb
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yiyi
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5 Answers5

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The symbol $\sqrt{x}$ always denotes the non-negative square root of $x$, assuming that $x\ge 0$. Thus,

$$\sqrt{\sqrt{16}}=\sqrt4=2\;.$$

Brian M. Scott
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  • what about $\sqrt{x^2} = \vert x \vert$ – yiyi Aug 27 '13 at 09:15
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    @yiyi: It’s true: that simply says that the $\sqrt$ sign always chooses the non-negative square root. – Brian M. Scott Aug 27 '13 at 09:25
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    @yiyi it is true that $ - 2^4 = 16 $ but if we would allow this as answer $ \sqrt x $ would not be a function anymore because a function in a mapping and a mapping is always to ONE answer – Willemien Aug 27 '13 at 11:01
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    @Willemien: Actually, $-2^4=-16$; you mean that $(-2)^4=16$. – Brian M. Scott Aug 27 '13 at 19:20
  • -1: The square root does not "always" denote the non-negative square root. On the contrary, if not defined elsewise by the context of the equation (f.e. lengths in geometry) you should always assume that both the negative and positive value are valid. Your answer confused me enough to look this up in Wikipedia. –  Aug 28 '13 at 06:30
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    @Chris: You are mistaken. The symbol $\sqrt{}$ does always denote the non-negative square root. When both roots are desired and that symbol is used, it is used in conjunction with $\pm$ or $\mp$. I suspect that you are confusing the statement $x^2=4$, which is satisfied by both $x=2$ and $x=-2$ and does require taking both into account, with the statement $x=\sqrt4$, which is not equivalent and which unambiguously says that $x=2$. – Brian M. Scott Aug 28 '13 at 19:09
  • @Brian: I've never heard of that and also Wikipedia states otherwise, do you have any evidence? –  Aug 28 '13 at 19:51
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    @Chris: Several decades of doing and teaching mathematics as a professional mathematician. Please give me the Wikipedia citation in question. – Brian M. Scott Aug 28 '13 at 19:55
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for the solution to $x^4=16$ the answers are $x\in \{2,-2,2i,-2i\}$

this is because you need to combine the answers of $x^2=4$ and $x^2=-4$

ratchet freak
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6

As Brian said by $\sqrt{x}$ we usually (I wouldn't say always) mean the non-negative square root of x. But because this is for class needs and you'll usually meet this in solving equation you should always consider both negative and positive root, because they can lead to two distinct solution.

Why we usually mean only the non-negative square is because in some fields (e.g. geometry) there aren't negative values. So if you take square root of given area of a square, you'll get the side of the square, but the negative value is left out, because square with negative sides doesn't exist.

So it depends on where you use it. In geometry we use only the positive root, but in algebra we must consider both roots.

Stefan4024
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  • There is too such a thing as negative values in geometry. The issue is that they don't represent what you might think they represent - if a line segment has negative length, for instance, that means that our 'segment' is actually every point on the line they define except those between the two points. Likewise, if something has a negative area, that means it contains the entire plane/space/etc except the area we would normally consider to be in the figure. – AJMansfield Aug 27 '13 at 23:02
  • I must admit I didn't know this before. But as you say we can't have length of segment with negative value, because the length of the segment is give by the distance formula, which only gives positive values, because it takes absolute value of the square root. – Stefan4024 Aug 27 '13 at 23:40
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Equivalently, we know that:

$$\sqrt{\sqrt{16}}=\sqrt[4]{16}=\sqrt[4]{(\pm 2)^4}=|\pm2|=2$$

Mikasa
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  • $\quad +1\quad$ – amWhy Aug 27 '13 at 11:59
  • Wait, is $$(-2)^4$$ not also 16? – 11684 Aug 27 '13 at 12:27
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    Wow, that was quick! But how can -2 equal 2? (Sorry, I'm not a maths expert.) Ah, I overlooked the vertical bars and saw that they mean 'absolute value' after a quick search on wikipedia. Now I understand your answer. – 11684 Aug 27 '13 at 12:30
  • @11684: That's Ok. Which $-2$ do you mean? $(-2)^4=16$ but $\sqrt{16}=\sqrt{(-2)^4}=|-2|=2$. – Mikasa Aug 27 '13 at 12:32
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    AFAIK ±2 means +2 (=2) OR -2. So considering that, and overlooking the vertical bars you get $2 = 2 ∨ -2 = 2$. (We use a V like symbol for or in case of multiple possibilities - found the right one.) – 11684 Aug 27 '13 at 12:34
  • @11684: Yes. It means $2\vee (-2)$ – Mikasa Aug 27 '13 at 12:36
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Since square root of any positive number is positive. Let $\displaystyle x = \sqrt{\sqrt{16}}$.

Now squaring both side we will find $x^2 = \sqrt {16}$.

Since we know that after squaring any equation the number of roots increase now again squaring the above equation we will find that $$x^4 = 16$$ $$x^4 - 16 = 0$$ Now factorizing this equation we will get $$(x^2 + 4)(x^2 - 4) = 0$$ $$(x^2 + 4)(x + 2)(x - 2) = 0$$ $$(x + 2i)(x - 2i)(x + 2)(x - 2) =0$$

This implies that $x = 2, -2, 2i, -2i$.

Since $x$ is real and positive hence $\displaystyle x = \sqrt{\sqrt{16}} = 2$.

Ludolila
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user157973
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