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$-1=(-1)^1=(-1)^\frac{2}{2}=((-1)^2)^\frac{1}{2}=\sqrt{(-1)^2}=\sqrt{1}=1$

This proof bugs me for the following reasons:

  • Mathematicians have defined the symbol $\sqrt{}$ (Named the principal square root) to mean 'take only the positive square root of the number under the radical'
  • This makes it so that the $y = \sqrt{x}$ is a function and therefore for every x-value (input) there is only 1 y-value (output).
  • Therefore $\sqrt{x^2}=|x|$ due to the above definition.
  • But then this means $\sqrt{(-1)^2} = |-1| = 1!!!!$

I really don't understand what's wrong with the above proof, the only way this could make sense is to define $\sqrt{a} = \pm k$ where the symbol know gives rise to two solutions to the equation, but then y = $\sqrt{x}$ is not a function and is simply a relation.

If we have $\sqrt{a}=\pm k$ then $\sqrt{(-1)^2}$ can be equal to -1 as well and then $-1=(-1)^1=(-1)^\frac{2}{2}=((-1)^2)^\frac{1}{2}=\sqrt{(-1)^2} = -1 = -1$ and everything works out fine.

Side Note: Why do we even want y = $\sqrt{x}$ to be a function anyway? What's the harm in defining $\sqrt{x} = \pm k$ where y = $\sqrt{x}$ is not a function?

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

Let k = $\sqrt{x^2}$ where k is a constant.

$k^2=(\sqrt{x^2})^2=(((x^2))^\frac{1}{2})^2$

$k^2=x^2$

$k^2-x^2=0$

$(k+x)(k-x)=0$

$k = x$ or $k=-x$

But $\sqrt{a}>0$ for $a>0$ by definition of the principal square root.

$\therefore k \neq-x$

In order to make this identity work properly, an absolute value sign is required.

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

Btw, my question is not the same as why $\sqrt{-1\times-1}\neq(\sqrt{-1})^2$ as I am not using the property $\sqrt{a}\sqrt{b}=\sqrt{ab}$ here and I am not bringing up the concept of imaginary numbers.

Dom Turner
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  • See here and here. – Matthew Cassell Feb 02 '20 at 03:08
  • This has already been discussed in this site. To avoid confusions in this remember one thing, whenever you introduce $-1$ inside a radical, convert that thing to $i^2$ asap (since $i^2$ is defined to be $-1$). So your reasoning becomes; $$-1=\left(-1\right)^{1}=\left(-1\right)^{\frac{2}{2}}=\sqrt{\left(-1\right)^{2}}=\sqrt{i^{4}}=i^{2}=-1$$ – Mourad Feb 02 '20 at 03:15
  • +1 good question, good reasonings, not a "what's wrong with this" with nothing more, and a show of effort to learn. – Simply Beautiful Art Feb 02 '20 at 03:30
  • @Mourad I really don't see how that is helping... – Simply Beautiful Art Feb 02 '20 at 03:30
  • Simply Beautiful Art, I don't mention or bring the imaginary number anywhere in this post, is there a post that explains my exact question without the imaginary number? I just used basic index laws. – Dom Turner Feb 02 '20 at 03:32
  • @SimplyBeautifulArt, The answer to this question has been on this site, since like nine years or so. OP can look into that, and understand that it is undefined to use normal algebra rules in dealing with negative quantities inside radicals. I was referring to how to correctly deal with this by using the imaginary unit (which was made for dealing with such problems) – Mourad Feb 02 '20 at 03:34
  • @Mourad Again, I still do not see how it helps. You're just pushing the problem further down the rabbit hole. How do we conclude $\sqrt{i^4}=i^2$? Exponential rules? That's precisely the problem here. It has nothing to do with the fact that we have negatives inside the radicals here. You could just as easily argue that we have $\sqrt{i^4}=\sqrt{(-i^2)^2}=-i^2$ by that logic. – Simply Beautiful Art Feb 02 '20 at 03:38
  • @SimplyBeautifulArt, Well, $\sqrt{i^4}=\pm i^2$. But here were taking the 'principal value' Take for instance $2=\sqrt{4}=-2$, so $2=-2$? I'm also not very confident in my arguments, seeking to learn. But we can't get conclusive answers by doing this. – Mourad Feb 02 '20 at 03:44
  • @DomTurner Sorry, this and this might answer your question better. – Simply Beautiful Art Feb 02 '20 at 03:46
  • In line $3$, of your proof, it should be $ \sqrt{x^2} = |x|= \begin{cases} x, & \text{if $x$ ≥ $0$} \ -x, & \text{if $n$ < $0$} \end{cases}$ See the wikipedia entry for the same. – Mourad Feb 02 '20 at 04:24

3 Answers3

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The problem lies in this equality:

$$(-1)^{\frac{2}{2}}=((-1)^{2})^{\frac{1}{2}}$$

which uses the identity $a^{\frac{m}{n}}=(a^m)^{\frac{1}{n}}$. The identity is true if $a$ is positive, but is false in general, and in some cases where $a^m$ is negative, the right hand side is not even defined.

I guess the lesson is that one should pay attention with the notations, their meanings and their conditions-to-apply.

There are ways to make sense of the set of $1/n$th-roots, but I think it is not the central story in this example. However, I suggest you learn more about holomorphic functions, their domains of definition, the analytic continuation of these functions, monodromy obstruction and Riemann surfaces. In these contexts, first we get out of the line $\mathbb{R}$ and come into the whole plane $\mathbb{C}$. Then we discover that the function $\sqrt{z}$ has a singularity at $0$, and each time we analytically prolong it around the point $0$, it changes its size, and the correct way to define it is to change the domain: it's no longer a function $\mathbb{C}\rightarrow\mathbb{C}$, but from a $2$-fold branched covering with a ramification point at $0$, to $\mathbb{C}$.

Another path may be to come back to the definition of $a^\frac{1}{n}$ in classical analysis book (I would recommend Rudin's Principles of mathematical analysis. There are situations where $a^{\frac{m}{n}}=(a^m)^{\frac{1}{n}}$ is true even for $a$ negative.In fact you can find all the triplet of real numbers $(a,m,n)$ such that $a^{\frac{m}{n}}=(a^m)^{\frac{1}{n}}$, and certainly $(-1,2,2)$ is not one of them.

anonymous67
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    Why is $a^\frac{m}{n}=(a^m)^\frac{1}{n}$ only true for a > 0? Is there a way to demonstrate this? I understand why $a^m$ can't be negative if you are dealing with the real number system, but I don't see why $a$ can't be negative. – Dom Turner Feb 02 '20 at 03:20
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    @DomTurner It is wrong for the exact reason you posted this question –  Feb 02 '20 at 03:28
  • The identity can be true elsewhere (but not in the case of the question). I choose to talk about the situation where $a>0$ because it is a standard situation considered in maths books. I don't think this matter is important enough to be axiomatized, but if you go deep enough, yes, all the facts in number theory are derived from axioms in number theory. – anonymous67 Feb 02 '20 at 03:31
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    Actually, after thinking about it, I can see why in some cases $a$ can't be negative, is it because if we have a = -5 and if m = 1 and n = 2 then we have $(\sqrt(-5))^\frac{1}{2}$ which is undefined? Therefore it can only be guaranteed to be defined and true if a>0 – Dom Turner Feb 02 '20 at 03:44
  • @DomTurner It is possible to consider such things, but it really doesn't remove the point that even then you will not have exponential rules as you would normally expect from the positive reals sometimes (depending on precisely how one "chooses" what $\sqrt\cdot$ and whatnot means). – Simply Beautiful Art Feb 02 '20 at 03:50
  • Is there a proof showing that $a^\frac{m}{n} = (a^m)^\frac{1}{n}$ only when a>0? – Dom Turner Feb 02 '20 at 03:54
  • @Dom Turner, I believe it can be considered close to be an axiom as BQT stated. – Mourad Feb 02 '20 at 03:57
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    Theorem: there cannot be such a proof. Proof: $(-1)^{\frac{1}{3}}=((-1)^1)^{\frac{1}{3}}$. In fact $a^{\frac{1}{n}}$ can be easily defined for $a$ negative if $n$ is a odd integer. – anonymous67 Feb 02 '20 at 03:58
  • Or what about an intuitive understanding by letting n ∈ $ℝ$, Then $n^2≠a$, where $a<0$. Hence $\sqrt{a}$ is undefined or not possible in the real plane. – Mourad Feb 02 '20 at 04:00
  • So I guess it's not a matter of proof? But just a matter of definitions and notations? Because it depends on what $m$ and $n$ are right? I mean, you can't proof that $2^2 = 2 \times 2$ we just defined squaring a number to mean that. – Dom Turner Feb 02 '20 at 04:01
  • Is this a similar case here? – Dom Turner Feb 02 '20 at 04:01
  • @DomTurner that's exactly the case. it's indeed a matter of definitions and notations. but it doesn't mean that it's not important. – anonymous67 Feb 02 '20 at 04:02
  • as I stated in my answer, these little bugs in the notation can give you some hints about interesting ideas staying behind, but for the matter at hand (which concerns only two numbers $-1$ and $1$), they are just little bugs. – anonymous67 Feb 02 '20 at 04:04
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    I don't think it's not important, I'm just trying to specifically pinpoint what went wrong with the above proof. If anything Definitions and notations are the most important things in maths because they are the basic building blocks that make theorems and other major results. If you have definitions, axioms, and notations that make no sense and have no consistency, then the rest of the mathematics you build upon that is meaningless, as they do not accurately represent sound and agreeable concepts. – Dom Turner Feb 02 '20 at 04:06
  • To answer to your question, the exact thing that went wrong is the third equality. – anonymous67 Feb 02 '20 at 04:07
  • Ok, thanks, that's all I needed to know. – Dom Turner Feb 02 '20 at 04:08
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The fallacy is due to the two-valued 'function', namely square root. Denote by $y$, the square of a real number $x$.

That is $x^2= y$.

We also know $ (-x)^2= y$.

From this concluding $x= -x$ is the wrong logic. Only conclusion that can be drawn is $(-x)^2= x^2$, which is not really a new statement.

P Vanchinathan
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  • Yes, actually a simple argument like this is the only one required for OP's question. Again, doing the OPs problem, we have $$-1=(-1)^1=(-1)^\frac{2}{2}=((-1)^2)^\frac{1}{2}=\sqrt{(-1)^2}=\sqrt{1}=\color{red}{\pm}1$$ – Mourad Feb 02 '20 at 03:52
  • That would make sense, but I thought $\sqrt{1} = 1$ as we are using the $\sqrt{}$ symbol. – Dom Turner Feb 02 '20 at 03:57
  • Actually $\sqrt{x}=\pm y$. Since $(\pm y)^2=x$. But for example $\sqrt[^3]{a}=+b$, because $(-b)^3=-a$. – Mourad Feb 02 '20 at 04:06
  • Actually, now that I think about it, I think there is a problem with defining $\sqrt{x} = \pm y$, because if you have $\sqrt{x} = \pm y$ then $\sqrt{x} = y$ or $\sqrt{x} = -y$ but then $\sqrt{x} = y = -y$ A negative number can never equal a positive number. The definition of that form of the square root doesn't hold up with the other notation of mathematics. – Dom Turner Feb 02 '20 at 04:14
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    @Dom Turner, you're forgetting that $y$ $-y$, are actually matched together by assigning an equality with $\sqrt{x}$, so we can only say that $y^2=(-y)^2$ – Mourad Feb 02 '20 at 04:26
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If I may be extreme, let me say once more:

In Mathematics, Nothing is True, unless there is a proof that it’s true.

The formula $(a^b)^c=a^{bc}$ when $a$ is a positive real number, and $b,c$ are real numbers, is true because there’s a proof. It can easily fail when $a\not>0$, and there is no proof when $a$ is not positive real.

Lubin
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