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I am a math major and often I doubt simple and fundamental definitions, consequently creating confusions about them. One day I was working on a linear approximation problem (the derivation led to the square root of $47$, something like that) which I got the answer to be $6.85...$, but I wondered since $6.85$ squared gives me $47$, why wouldn't I take the negative of $6.85...$ as well. After this I started doubting the whole square root definition and how it is derived from the axioms.

My question is, why doesn't the square root of a number equal to a negative value? I have to let everyone understand, I know the absolute value definition of a square root, it's domain, range, inverse etc., but what I'm asking for is more of a fundamental understanding of the consequences of a negative result from a square root. Does anyone have an idea how can I clear this up?

P.s. I have asked professors but everyone refers to definitions, for which they are correct, but I'm trying to go a little further from the definitions. The best way I could phrase the question was through the consequences aspect of it (i.e. I cannot take the square root of a negative number if I want the result to be in the domain of real numbers, in this way what would be the consequence if I said that the square root of $49$ is plus and minus $7$?).

Thank you for your time!

TShiong
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Tommy
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    Not quite sure what the question is here. It is certainly the case that $\pm 1$, say, are both solutions to $x^2=1$. However, when we write, e.g., $\sqrt 1$ then, by definition we only take the positive value. Thus $\sqrt 1=1$. It's just the definition of that function. – lulu Dec 12 '22 at 21:07
  • I don't understand your question. You ask why, and people have already answered you by citing definitions. You say that you don't want to hear about definitions... but without definitions how could we even communicate? – JMoravitz Dec 12 '22 at 21:07
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    It is a matter of convention (agreement) and history. Negative numbers arose late, and the notion of square root already existed. We pick the nonnegative one because it makes some formulas easier. For instance, with the nonnegative value we have that for $a,b\geq 0$, $\sqrt{ab} = \sqrt{a}\sqrt{b}$. If you picked the negative value, the formula would be $\sqrt{ab} = -\sqrt{a}\sqrt{b}$. Nothing wrong with it, just more cumbersome. – Arturo Magidin Dec 12 '22 at 21:07
  • "I want to know why $4\times 3$ is equal to $12$ and not equal to $69420$.... I know that the definition says that it should equal 12 but why not let it equal 69420 instead?" Because then we wouldn't be talking about the same thing... Do you see why your question sounds to me and others just as nonsensical? – JMoravitz Dec 12 '22 at 21:08
  • Does this answer your question? Square root and principal square root confusion. If not, there is a long list of other choices before it's worth keeping this open separately. – Nij Dec 12 '22 at 21:10
  • @JMoravitz I guess question is that, how come we decided that square root to be single-valued function rather than multi-valued one? – Salihcyilmaz Dec 12 '22 at 21:14
  • Because singlevalued functions are very useful for many scenarios. That isn't to say that we might not also refer to what you are talking about, the "multivalued one"... just we give it a different name... just like how we might refer to one color as sky-blue rather than navy-blue... They are both talked about, they are both used... but if you are living inland rather than by the ocean you'll be using the one more often than the other... Your complaint is to me like complaining about why we named the sky "the sky" versus something else... If the ocean was instead called "the sky" so what? – JMoravitz Dec 12 '22 at 21:17
  • @ArturoMagidin In short, we need only $f(x)=\sqrt x$ be a single real valued function, since we need inverse of $f(x)=x^2$. For instance, if we define $f(x)=\sqrt {x^2}=-|x|$, then "everything" still works, right? Thus, $f(x)=-\sqrt x$ would be inverse of $f:\mathbb R^+\to\mathbb R^+, f(x)=x^2$. Is it true? – lone student Dec 12 '22 at 21:18
  • @lonestudent It's very convenient that $\sqrt {ab}=\sqrt a \times \sqrt b$. That's not true if you always take the negative root. – lulu Dec 12 '22 at 21:21
  • Yes, by naming the ocean "the sky" and the sky "the ocean" instead... we would have "sky-blue" would refer to the darker shade rather than the lighter shade of blue... Does this stop a painter from painting pictures? Of course not. Is it worth musing on for more than a second? Hardly. Historically we called the sky "the sky" and so that is what we call it now. We can still talk about the ocean instead just fine... its still there. Historically we let $\sqrt{~}$ refer to the positive root. We can still talk about the negative root instead just fine... its still there. – JMoravitz Dec 12 '22 at 21:21
  • @lulu If you define $\sqrt {x^2}=-|x|$ then, $\sqrt {ab}=-\sqrt a×\sqrt b$. Everything still works. (But we don't need this) – lone student Dec 12 '22 at 21:27
  • @JMoravitz please calm down my fellow mathematician, I have not committed a crime by asking a question which is in fact being debated by others as well. I believe that what I was aiming for is to understand if it is a convention or a fundamentally defined concept. I am not going against definitions or speaking of their need in math, rather if you would read the question before breaking the keyboard you would understand that I was asking for the answer that the person below me and above you gave. – Tommy Dec 12 '22 at 21:28
  • @ArturoMagidin thank you for the answer, that was what I was looking for apparently! – Tommy Dec 12 '22 at 21:31
  • "being debated" There is no debate... Do you debate what the "sky" refers to? – JMoravitz Dec 12 '22 at 21:35
  • @JMoravitz people like you are ruining this great page of knowledge with their ignorance, arrogance and absolutely ridiculous analogies of square root definitions with the sky while trying to belittle someone for asking a question. Nonetheless, have the day you deserve my friend! – Tommy Dec 12 '22 at 21:45

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The squaring function $s(x) = x^2$ maps $\mathbb{R} \to [0, \infty)$, so given any $y \geq 0$, it's natural ask about its preimages. This is the set of all real numbers $x$ such that $s(x) = y$. But if you want this inverse process to yield a specific real number $x$ for every $y \geq 0$ that you feed it, then you have to make a choice. It's a convention, nothing more, to choose the positive square root and assign it to the radical notation: $x = \sqrt{y}$.

A nice way to make this more explicit, is to take the squaring function, and first restrict its domain as much as possible without losing any of the range. This defines a new function, let's say $s_+: [0, \infty) \to [0, \infty)$ that agrees with the original function, i.e. $s_+(x) = x^2 = s(x)$ for all $x \geq 0$ but is one-to-one. This means that for any $y \geq 0$, there is precisely one $x \geq 0$ such that $s_+(x) = y$.

This issue comes up all the time, by the way, when we want to "undo" a function that maps many inputs to a single output.

For example, the sine function: $\sin: \mathbb{R} \to [-1, 1]$, because of periodicity and other symmetry, there are infinitely many $x$ such that $\sin(x) = y$ for a given $y \in [-1, 1]$. So, we restrict the domain to the interval of angles $\bigl[-\tfrac\pi2, \tfrac\pi2 \bigr]$. Now there is precisely one angle $x$ whose sine is $y$.

And this is just for functions whose domain and codomain are subsets of the real numbers. We get even more interesting examples if we use complex numbers or "numbers" from other algebraic fields or rings that are found throughout mathematics.

Sammy Black
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  • I believe that my true question lied on whether it is a convention or not, thanks for the explanation, cleared it up! – Tommy Dec 12 '22 at 21:29