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What I know

If I'm not mistaken:

  • $\pm \sqrt{\frac{x}{y}}=\frac{\pm \sqrt{x}}{\pm \sqrt{y}}$
  • a repeated $\pm$ sign in an equation means make every "$\pm$" sign a plus or make every one of them a minus(you can't take one as a positive and the other as a negative, then $\mp$ sign would be useless[e.i. $\pm 3 \cdot \pm 4 = 3\cdot 4 \,\,\,\,\, \text{ or } \,\, (-3) \cdot (-4)$ which would give one result]).

The Question

I thought if $x=a^2$ and $y=b^2$ then $\pm \sqrt{\frac{x}{y}} = \frac{\pm \sqrt{x}}{\pm \sqrt{y}} = \frac{\pm a}{\pm b}$ and here is where the problem originated... In a fraction like $\frac{16}{4}$ taking it's root would give $\pm \sqrt{\frac{16}{4}} = \frac{\pm \sqrt{16}}{\pm \sqrt{4}}=\frac{\pm 4}{ \pm 2}$ up until this point everything was fine, now to simplify it even more. according to my understanding the solution would be $\frac{+4}{+2} \,\,\text{ or }\,\, \frac{-4}{-2}$ which results in one solution although there must be 2 solutions...

My Explanation

From my understanding, if you altered between the plus and minus signs($\frac{+4}{-2}$ or $\frac{-4}{+2}$) you would give the second solution but this breaks the rule of not altering between multiple $\pm$ signs and would make $\mp$ pointless.

Is there a mistake that I did or is there a rule I haven't heard of, I searched for a while for an answer but it seems no one has asked this question other than me.

So my question is:

How could a square root of a fraction have a negative solution if both cases result in a positive?

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    $\begingroup$ $\sqrt{16}= 4,$ not $\pm 4$ $\endgroup$ – saulspatz Jun 8 '18 at 0:16
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If you're talking about real variables, the convention is that $\sqrt{x}$ is the positive square root if $x > 0$ (and undefined if $x < 0$). Then $\sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}}$ when both sides are defined.

If you're talking about complex variables, $\sqrt{x}$ has two possible values, and one is $-$ the other. Then $\sqrt{\frac{x}{y}} = \pm \frac{\sqrt{x}}{\sqrt{y}}$. Even if you specify the "principal branch" of the square root where the real part is always nonnegative, it could be $-$, e.g. if $x = e^{2\pi i/3}$ and $y = e^{-2\pi i/3}$.

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  • $\begingroup$ to be honest I do not understand how your answer is related $\endgroup$ – Mohannad El-Nesr Jun 8 '18 at 11:52
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The other answers explained you the convention that $\sqrt{x}$ represents the "principal value" of the square root of $x$, i.e. the non-negative solution of $y^2=x$.

I just wanted to tell you about another notation that you are using, that is inconvenient for the computation that you wanted to perform.

If you want to relate the solutions of $x=a^2$ and $y=b^2$ with the solutions of $z=(a/b)^2$ you can do so, but you face another inconvenient notation. The $\pm$, when appearing several times in a formula is usually, as you did assumed to be coordinated. The $+$ and the $-$ are chosen the same in all occurrences of the sign $\pm$. You need to allow them to be non-coordinated such that $\frac{\pm a}{\pm b}$ can give $\frac{+a}{+b}=\frac{-a}{-b}$ as well as $\frac{-a}{+b}=\frac{+a}{-b}$.

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  • $\begingroup$ How could you do so? Isn't that something fixed? I'd like some explanation, when could you make it "non-coordinated" $\endgroup$ – Mohannad El-Nesr Jun 8 '18 at 1:17
  • $\begingroup$ @MohannadEl-Nesr Well, notation is independent of the properties that they represent. The property holds because if $a=a_0$ is one solution of $x=a^2$ and $b=b_0$ is a solution of $y=b^2$, then dividing both equations you get $x/y=a_0^2/b_0^2=(a_0/b_0)^2$. This is independent of which solutions are $a_0$ and $b_0$. Therefore, $z=a_0/b_0$ is a solution of $x/y=z^2$. You get two different solutions for the four choices I listed in the answer above. $\endgroup$ – user568248 Jun 8 '18 at 1:23
  • $\begingroup$ I was talking about how the plus or minus could be un-coordinated $\endgroup$ – Mohannad El-Nesr Jun 8 '18 at 1:26
  • $\begingroup$ @MohannadEl-Nesr I know what you were talking about. What I said is the explanation without using the notation $\pm a$ and $\pm b$ to denote the roots. As you can see above, you can just mention the roots and give an argument without using a notation that most will find confusing. $\endgroup$ – user568248 Jun 8 '18 at 3:07
  • $\begingroup$ Why is it then always referred to the solution of the root with a $\pm$ sign if you could completely ignore it? e.i. $\sqrt{64} = \pm 8$(ignoring notation) not $\sqrt{64} = 8 \,\,\text{ and }\,\, -8$. Excuse me if I'm asking too much, I'm just confused $\endgroup$ – Mohannad El-Nesr Jun 8 '18 at 3:36
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If $x=a^2$ and $y=b^2$ with $b\ne 0$, then

$$\sqrt {\frac {x}{y}}=\sqrt {\frac {a^2}{b^2}}=$$ $$\frac {\sqrt {a^2}}{\sqrt {b^2}}=\frac {|a|}{|b|} $$For example if $x=(-3)^2$ and $y=(-4)^2$ then $$\sqrt {\frac {x}{y}}=\frac {|-3|}{|-4|}=\frac {3}{4} $$

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  • $\begingroup$ what about the other solution? "$-\frac{3}{4}$" $\endgroup$ – Mohannad El-Nesr Jun 8 '18 at 1:13
  • $\begingroup$ There is no other solution. The question was $\sqrt{\frac {(-3)^2}{(-4)^2}}$ and, if $x>0$ then $\sqrt{x}$ ALWAYS ALWAYS ALWAYS means the positive square root. So $\frac 34$ is the ONLY solution. $\endgroup$ – fleablood Jun 8 '18 at 1:18
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deep breaths and calming thoughts

Let's not worry about the symbols $\sqrt{}$ and $\pm$ but lets address the core issue.

If you have a ration $\frac xy$ (assume $x > 0;y>0$) what are its square roots; and are the set of square roots of $\frac xy$ and the ratios of the square roots of $x$ and $y$ the same thing.

The answer is yes. If $x$ has two square roots $a$ and $-a$ (assume $a > 0$) and $y$ has two square roots $b$ and $-b$ (assume $b > 0$) then:

The two square roots of $\frac xy$ are $\frac ab$ and $-\frac ab$ as $(\frac ab)^2 = (-\frac ab)^2 = \frac {a^2}{b^2} = \frac xy$.

Furthermore $\frac {-a}{b} = -\frac ab$; $\frac {a}{-b} = -\frac {a}b; \frac {-a}{-b} = \frac ab$ and the fourth ratio is $\frac ab$.

So the square roots of $\frac xy$ and the ratios of the square roots of $x$ and the square roots of $y$ are the same two numbers.

...

So now let's bring in the $\sqrt{}$ symbol.

If $w > 0$ then $\sqrt{w}$ is the non-negative (positive because $w > 0$) number $v$ so that $v^2 = w$, that is to say the positive square root. There is also another negative square root that happens to be $-v$ and we right that as $-\sqrt w$.

Note: $\sqrt{w}$ is always and always will be the NON-negative square root, whereas there will (if $w > 0$) be two square roots; one negative the other positive.

So $\sqrt{xy} =\frac {\sqrt x}{\sqrt y}$. And using your example $\sqrt{\frac {16}{4}} = \frac {\sqrt{16}}{\sqrt{4}} = \frac 42$. That's all there is too it becase square roots are not negative.

But what of the other square roots?

$\frac {16}{4}$ has two square roots: $\sqrt{16}{4} = \frac 42$ and $-\sqrt{16}{4} = -\frac 42$.

$16$ has two square roots: $\sqrt 16 =4$ and $-\sqrt {16}=-4$ and $4$ has two square roots: $\sqrt{4} =2$ and $-\sqrt 4 = -2$. So there are four possible combination of ratios.

$\frac {\sqrt{16}}{\sqrt {4}} = \frac 42$

$\frac {-\sqrt{16}}{\sqrt {4}} = -\frac 42$

$\frac {\sqrt{16}}{-\sqrt {4}} = -\frac 42$

and $\frac {-\sqrt{16}}{-\sqrt {4}} = \frac 42$

Those are the two values.

And that's that.

....

Okay what about the symbol $\pm$.

So we can condense the above by:

$\frac {\pm 4}{\pm 2}$ and that can be any of the four $\frac {4}{2},\frac {-4}{2},\frac {4}{-2},\frac {-4}{-2}$.

But as $\frac {4}{-2} = \frac {-4}{2}$ and $\frac {-4}{-2} = \frac {4}{2}$ we can "fix" the numerator to be positive and simply write $\frac {\pm 4}{2}$ (although it'd probably by better to write it as $\pm \frac 42$.

....

So what about this "breaking" rule?

Well, ther's really no such thing. If the numerator can be one or the other and the denominator can be one or the other there is now reason you can break them.

What matters is if the terms are dependant upon each other. So as $\pm(a-b)$ This is eithere $a-b$ or $-a + b$. So the first term is $\pm a$ and the second term is $\mp b$. But they depend on each other. So we can't "break" them".

But that would not be the case if , say, we were asked $x^2 = 9$ and $y^2 = 4$ what is $x + y$. Well, $x = \pm 3$ and $y = \pm 2$ so $x+y$ maybe any of the four values $\pm x \pm y$ ($3+2; 3-2; -3+2; -3-2$). And we certainly can break them as they are completely independent of each other.

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  • $\begingroup$ Why did you turn this personal? I am asking a question and you're here to answer, that's it. If you're going to start mentioning old problems and personal arguments and start insulting then you could keep your answer to yourself $\endgroup$ – Mohannad El-Nesr Jun 8 '18 at 1:21
  • $\begingroup$ I'm mentioning the old problems because they are relevant to this problem. And if you had listened to what eveyone had said instead of arguing back, you would never have had this question in the first place. The first seven times you say $\sqrt{16}=\pm 4$ we will politely explain to you why you are mistaken but the eighth time and after yous say something mindbogling arrogant likes "that's just a convention and it doesn't mean $\sqrt{16}\ne -4$" then it's time to lose our patience and start beating our heads against the wall until it sticks. $\endgroup$ – fleablood Jun 8 '18 at 1:31

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