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I recently saw someone say:

$\sqrt{x}$ is, by definition, positive

Which didn't sit easily with me as we always say $\pm$ for any square root.

Then it got me thinking that if, by definition, the symbol $\sqrt{x}$ meant the positive or the negative root of x, then the $\pm$ symbol would be redundant.

So, have mathematicians defined the root symbol only to include the positive root, thereby necessitating the $\pm$ symbol?

EDIT: This question is different from the one here: Square root confusion? because my question focuses principally on the meaning/convention of the square root symbol rather than on the topic of their being two solutions to $x^2=4$

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    $\begingroup$ yes, typically $\sqrt{}$ denotes the principal root $\endgroup$
    – costrom
    Commented Dec 10, 2015 at 18:05
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    $\begingroup$ Sometimes it is $0$. $\endgroup$ Commented Dec 10, 2015 at 18:07
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    $\begingroup$ If $x$ is negative we got imaginairy numbers. Example: $\sqrt{-4}=\sqrt{4}i=2i$ $\endgroup$ Commented Dec 10, 2015 at 18:08
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    $\begingroup$ For $x\ge 0$, the quantity $\sqrt{x}$ is a square root, and is non negative by convention/definition. The value $-\sqrt{x}$ is also a square root. The quantity $\sqrt{x}$ is not meant to signify both roots. $\endgroup$
    – copper.hat
    Commented Dec 10, 2015 at 18:10
  • $\begingroup$ It enetirely depends on the context and the definition. The usual definition used by mathematicians - that $\sqrt{x}$ is a non-negative number always (and only defined when $x\geq 0$) is in part due to a preference for "functions" - things that take in one input and return one output. It also keeps you from making errors if there is only one possible value for something. If you say $x=\sqrt{4}$, and that means $x=\pm 2$, then you have something that looks definitive, but is not. (Basically, $\pm$ is a red flag to take care in computations, and it is nice to have fewer red flags.) $\endgroup$ Commented Dec 10, 2015 at 18:12

2 Answers 2

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I will explain the difference by example.

The equation $x^2 = 4$ has solutions $x = \pm 2$. However, if I were to just say $\sqrt{4}$, that would be $2$ and not $-2$.

Also, $\sqrt{x}$ is nonnegative by definition.

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It's correct that if look at, say, $4$, then both $2$ and $-2$ are square roots of $4$, and in general, for any positive $x$, that $x$ has two square roots, where each is the negative of the other. But in the case of $4$, because of that second thing I mentioned, if we know $2$ is a square root of $4$, then we know immediately that $-2$ is as well.

We use $\sqrt{x}$ in two ways. Sometimes, in the sense that you mentioned, we want to use $\sqrt{x}$ to refer to the family of all numbers, let's call them $t$, such that $t^{2} = x$. That is, we use it to refer to a collection of (two) numbers satisfying a particular condition.

However, we also note that if we have one root, then we can easily determine the other by taking the negative. So in a sense, we only actually need one of them to "communicate" the solution set. So we often choose a convenient "representative", and often pick the positive one. So here, we use $\sqrt{x}$ as a function that spits out the "seed" of the solution set for $t^{2} = x$.

If the use of "by definition" bugs you, then replace it with "by convention".

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