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Here's my train of thought:

I. $4^2=4^{(4/2)}$

II. $4^{(4/2)}=(4^4)^{(1/2)}$

III. $(4^4)^{(1/2)}=256^{(1/2)}$

IV. $256^{(1/2)}=\sqrt{256}$

square root of 256 has 2 solutions: 16 and -16

Can you explain me where I made the mistake and why am I not allowed to do whatever I did?

I thought that maybe (II) is wrong and only $4^{(4/2)}=(4^{(1/2)})^4$ is true, but if $(x^a)^b=x^{(a\cdot b)}$ and numbers can switch places during multiplication and this does not change the outcome then: $x^{(a\cdot b)}=x^{(b\cdot a)}$, and $(x^b)^a=x^{(b\cdot a)}$

Also, please don't insult me in the replies. I feel stupid enough already for not getting this.

Nazmul Hasan Shipon
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fipao
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    Well, you have essentially shown that $a^2=b^2$ does not imply $a=b$. – lisyarus May 18 '21 at 09:30
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    You are correct that $x^2=256$ means either $x=16$ or $x=-16$. So this means that just because $x^2=y^2$, that doesn't necessarily mean $x=y$. It could also be the case that $x=-y$. – Douglas Molin May 18 '21 at 09:31
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    You just calculate $\sqrt{a}$ for some positive $a$, in your case $a=256$. The two solutions appear if you want to SOLVE $x^2 = 256$ but $\sqrt{256}$ is exactly one value, namely $16$. – Jfischer May 18 '21 at 09:33
  • 1)Nobody can insult you here. We have site rules, we have moderators. 2) Not understanding something doesn't mean being stupid. You just have to try again. – lone student May 18 '21 at 09:48
  • @simon I am actually beginning to learn about complex number and was trying to rise complex numbers to non integer power, for example i^pi and I tryied to do it by doing better and better aproqimations of pi and then braking it down in 2 parts, first integer powers and then roots but, then i realized than x^10=i has 10 solutions and I didn't know witch one is the "real one", This is when this question came up. Does this mean that my approach is wrong? – fipao May 18 '21 at 10:03
  • Also @lone student Thank you but this is my third or so post on the internet and I am a little insecure – fipao May 18 '21 at 10:05
  • For what it's worth, I upvoted your question, because it was excellently presented, with nice work shown. Other responses have indicated your analytical error. Having an analytical error in a mathSE question does not (in my opinion) in any way detract from the quality of the question's posting. – user2661923 May 18 '21 at 11:21
  • @fipao You are correct that $x^{10} = i$ has 10 solutions. There is no "real one" -- they are all equally valid (and none are on the real axis). This is one of many reasons why this idea of "principal square root" is problematic: it does not extend to the complex numbers. As you explore this area of math, I urge you to frame these problems in terms of finding roots of polynomials, rather than applying "nth root functions". – JounceCracklePop May 19 '21 at 01:43
  • Here is one discussion worth reading and here is another discussion worth looking at – Steven Alexis Gregory May 19 '21 at 04:33

2 Answers2

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Your I to IV are all correct. Here is what is incorrect: "square root of $256$ has two solutions." That is not true.

It is true that the equation $x^2 = 256$ has two solutions. But by convention, the square root (written as $\sqrt{256}$ or $\sqrt{256}$ or $256^{1/2}$) is defined to be the non negative solution. It would have been possible to define it to be the negative one, but nobody does that. But it can never be both at the same time. A function like "sqrt" always needs to produce a single result.

amWhy
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Simon
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    This called sometimes principal square root. – lone student May 18 '21 at 09:50
  • You're saying that the mistake in OP's proof that $4^2=-16$ is that they should not have considered the negative square root of 256 as a possible answer, simply because that's how square root is defined. Yet a trivial change to the question makes the negative root the correct one to consider, not the positive one. Aren't you leading OP astray here? – JounceCracklePop May 19 '21 at 15:18
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    @JounceCracklePop Yeah, this might be misleading. In that case the equation (II) is the critical one. Essentially it states $x = \sqrt{x^2}$ which is only true if $x$ is positive AND sqrt is definied as the positive root. For general $x$ you can only say that either $x=\sqrt{x^2}$ or $x=-\sqrt{x^2}$. – Simon May 19 '21 at 15:54
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OVERALL THEORY

If $x$ is a positive real number and $m$ and $n$ are integers, then

$$x^{\frac mn} = (x^m)^{\frac 1n} = (x^\frac 1n)^m = \sqrt[n]{x^m} = (\sqrt[n]x)^m$$

When $x$ is negative, all heck breaks loose. We can say that

$$\text{$x^m$ is positive when $m$ is even and is negative when $m$ is odd.}$$

So there is no sense in which $4^2 = -16$.

It is OK to work with negative numbers and integer powers. For example, $$((-2)^3)^4 = (-2)^{12} = 4096$$

But, weird things can happen with a negative $x$ and fractional exponents. For example $$ \begin{array}{c} (-8)^\frac 13 = \sqrt[3]{-8}= -2 \\ (-8)^\frac 26 = \sqrt[6]{((-8)^2)}= \sqrt[6]{64} = 2 \\ \end{array} $$

Since $\frac 13 = \frac 26$, we see that we have a problem.

There is no way to fix this problem. There is no way around it. Raising a negative real number to a fractional power is not well-defined.

PRINCIPAL SQUARE ROOTS

The roots of an expression, $f(x)-n$, is the set of all $x$ for which $f(x)=0$

The solutions to the equation $x^2 - n = 0$ are called the square roots of $n$.

  • Negative numbers have no (real-valued) square roots.
  • $0$ has one square root.
  • Positive numbers have two square roots.

The principal square root of a number is defined to be the non negative root of that number. There is a really cool equation that encapsulates this.

$$\sqrt{x^2} = |x|$$

Hence, for example,

$$\sqrt{(-3)^2} = \sqrt{3^2} = |3| = |-3| = 3$$

So, if $x^2 = y$, then the principal value of $\sqrt y$ is $|x|$.

ORIGIN OF THE WORD ROOT

Steven Alexis Gregory
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