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Let $x^2=y$, then $x=\pm\sqrt y$. But why can't it be $\pm x=\sqrt y$? I started thinking about this when i encounter this answer to a question but I didn't really understand it. In the answer, it says that there are two possible definitions for the notation $√$:

  1. For any positive real number $a$, $\sqrt a$ is defined as the square roots of $a$
  2. For any positive real number $a$, $\sqrt a$ is defined as the positive square root of $a$

By the first definition, $\sqrt {16}= \pm 4$. It also says that if I use the first definition,then I will encounter some problems in the future and that’s why we use the second definition. In the answer it says that solving for $x$ in $x^2 - \pi =0$ for $x>0$, by doing $x=\sqrt \pi$, would be incorrect. I still can’t understand why $x=\sqrt \pi$ would be incorrect if I use the first definition.

For me, $\pm x = \sqrt y$ and $x=\pm \sqrt y$ looks like they mean the exact same thing, but I suppose they don’t. Why is that?

Mohammad muazzam ali
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  • I think the source of confusion here is not the definition of $\sqrt{\cdot}$, but what you mean by $\pm$. $$ x = \pm \sqrt{y} \Leftrightarrow x = \sqrt{y} \vee x = -\sqrt{y}$$

    but

    $$ \sqrt{y} = \pm x \Leftrightarrow \sqrt{y} = x \vee \sqrt{y} = -x $$

    So, the second expression only gives a valid alternative (depending on the sign of $x$) while the firs always gives two alternatives.

     – PierreCarre Nov 03 '21 at 09:17
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    $(1)$ the square root is the non-negative solution by convention $(2)$ $\pm x=\sqrt{y}$ is the same as $x=\pm \sqrt{y}$ , but it is stange to have $\pm x$ on the left side if we want to have the value of $x$. $(3)$ If $x>0$ , then we only have one solution which is the square root without a sign. – Peter Nov 03 '21 at 09:17
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    @Peter I think your (2) is the true answer here: It's correct but strange. Why not make an answer? – Arthur Nov 03 '21 at 09:49
  • Why so much downvotes on the question (-3) ? It is in mathjax, OP explains thoroughly his source of confusion and ask for clarifications, this is fine for me. – zwim Nov 03 '21 at 10:34
  • @zwim I agree. I see no reason for a downvote , except it is considered as a duplicate, but this does not seem to be the case either. However, $2$ downvotes is not "much". – Peter Nov 03 '21 at 10:37
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    @PierreCarre so from what I understand, $\pm x = \sqrt y$ is pretty much the same thing as $x = \pm \sqrt y$? – Mohammad muazzam ali Nov 03 '21 at 14:11
  • Related: $|x|=\pm x;$ versus $;\pm x=|x|$ – ryang Jan 28 '22 at 19:35
  • @PierreCarre Your second equivalence means exactly the same as your first. If that final equality feels wrong to you, it's because you are forgeting that $x$ can be negative, in which case it is that right disjunct that is true, rather than the left disjunct. – ryang Feb 16 '23 at 16:02

4 Answers4

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Neither is incorrect. $x =±\sqrt{y}$ is the same as $\sqrt{y}=±x$.

However, you'd usually want to have the variable on one side (usually left) and its solution(s) on the other side (usually right).

Thus, one would normally interpret $x=±\sqrt{y}$ as $x$ is the variable and $±\sqrt{y}$ are its solutions while $\sqrt{y} = ±x$ would be interpreted as $\sqrt{y}$ is the variable and its solution is $±x$. Both are of course, equivalent ways of saying the same thing.

Ibrahim
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  • so, it’s just a convention? saying $\pm x = \sqrt y$, for $x ≥ 0$, wouldn’t be wrong? – Mohammad muazzam ali Nov 03 '21 at 14:09
  • @Mohammadmuazzamali Why $x≥0$? Did you mean $y≥0$ instead? Because $x$ can be any real number and only $y$ needs to be non-negative (for $\sqrt{y}$ to be defined) in the equation $x^2 =y \Leftrightarrow x = ± \sqrt{y} \Leftrightarrow ±x = \sqrt{y}$. In that case, yes. It's a convention to want to have the unknown on one side and its solutions on the other. – Ibrahim Nov 03 '21 at 23:57
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Let $y\geq0$. If $\sqrt{y}$ is defined as the set of solutions to the equation $y=x^2$ and $x$ is one solution then I would say that $\sqrt{y}=\pm x$ is perfectly sensible notation (since $x$ and $-x$ are the only solutions). More commonly, however, $\sqrt{y}$ is defined as the unique non-negative solution to $y=x^2$. In this case, if $x^2=y$ then $x=\sqrt{y}$ or $x=-\sqrt{y}$. In my opinion one should avoid writing either of the expressions "$x=\pm\sqrt{y}$" and "$\sqrt{y}=\pm x$". My point: $x$ is some number, but what is $\pm\sqrt{y}$, and similarly, $\sqrt{y}$ is some number but what is $\pm x$?

jakobdt
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$\sqrt {x}$ is the "principal root." That means that $\sqrt x \ge 0.$ (Or, at least it is until you learn about complex numbers.)

$x^2 = 4$ is multivalued and has two solutions. That is $x = \pm \sqrt 4 = \pm 2$

When you write $\pm x = \sqrt y$ this requires that it be possible that $\sqrt y$ be negative, which it can't.

$x = \pm \sqrt y$ says $\sqrt y$ is always positive, but $x$ can be negative.

user317176
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    "When you write $±x = \sqrt{y}$ this requires that it be possible that $\sqrt{y}$ be negative, which it can't." I don't think so. Writing $\sqrt{y} = ±x$ means $\sqrt{y}$ is "$x$ or $-x$", not "$x$ and $-x$" so it doesn't require anything of the sort. – Ibrahim Nov 03 '21 at 09:40
  • @Ibrahim When we say $x = \pm 2$ then that means that both $2,$ and $-2$ are acceptable values for $x.$ – user317176 Nov 03 '21 at 09:44
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    Not quite, the sign $±$ means "plus or minus" where or is inclusive. So $x=±2$ simply means $x=2$ or $x=-2$ or both. – Ibrahim Nov 03 '21 at 09:52
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There has been a lot of nonsense posted here. The facts are these:

The notation

$$a=\pm b$$ is simply a universally accepted shorthand for $$a=b\text{ or }a=-b$$

The notation $$\pm a=b$$ has no universally recognised meaning, and should never be used.

TonyK
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  • Can $\pm a = b$ be understand as $a = b$ or $-a = b$? that also seems pretty reasonable. – Mohammad muazzam ali Nov 04 '21 at 07:05
  • @Mohammadmuazzamali: There is no need for it, and it is not normal, or expected. Also, if we allow $\pm a=b$, then we should also allow $c=\pm a=b$; but what would that mean? $c=b$, or $c=\pm b$? Better not to use it $-$ I can't see why you would want to. – TonyK Nov 04 '21 at 11:17