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Ok, here is what I think. Please correct me if I am wrong. $$\sqrt{9} \neq 3$$ and also $$\sqrt{9} \neq -3$$

Now let's assume, that above statements are false, then we have $-3 = \sqrt{9} = 3$ and since $3 \neq -3$ the assumption must be wrong. Ok, square root must be equal to 3 and -3 at the same time. As of my understanding a set of numbers is not a number itself and that leads to a conclusion, that a square root of a number is not a number. Right?

Both answers are of the same importance - neither of them is superior. Then here is the question - when we plot $f(x) = \sqrt{x}$, why do we always plot positive answers? Can someone give me a proof, that plotting negative answers is not allowed?

Antonio Vargas
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user3600124
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  • No, $\sqrt{9}$ is defined to be the positive square root. If you are allowed to define multivalued functions, then you have to lose a few things like transitivity of you write $f(9)=\pm 3$. But that's why we dislike multi-valued functions. – Thomas Andrews Feb 02 '16 at 01:18
  • If you want to plot $y^2=x$, that's an entirely different thing, but it is not a function. – Thomas Andrews Feb 02 '16 at 01:19
  • When you get to complex analysis, you'll see that square root "branches." There a complicated geometry going on. – Thomas Andrews Feb 02 '16 at 01:21
  • And, by the way, why do you assume both of those statements are false? What contradiction is raised if you assume one of them is false? – Thomas Andrews Feb 02 '16 at 01:26
  • You say that neither $3$ nor $-3$ is "superior" to the other (as a square root of $9$). But $3$ is greater than $-3$.... – Barry Cipra Feb 02 '16 at 01:27
  • Thomas, square root has two answers and non of them is superior. That means they must be both false or true. Otherwise on of the answer is for some reason superior... is it? – user3600124 Feb 02 '16 at 01:33
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    There are two square roots, but only one $\sqrt{9}$. @user3600124 The symbol does not mean "all the square roots." – Thomas Andrews Feb 02 '16 at 02:31

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You are partially correct:

Now let's assume, that above statements are false, then we have $−3=\sqrt{9}=3$ and since $3≠−3$ the assumtion must be wrong.

Yes. It is all correct. The key is that the prhase "above statements are false" means $\sqrt{9}\neq3$ AND $\sqrt{9}\neq-3$ are both false. Then, since this statement is false, the true is $\sqrt{9}=3$ OR $\sqrt{9}=-3$, which indeed is true cause the first is true ($\sqrt{9}$ means, by definition, the positive root of 9)

sinbadh
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The equation $x^2 = 9$ does indeed have two solutions, and therefore it is not incorrect to say that "$9$ has two square roots", namely $x=3$ and $x=-3$. But the notation $\sqrt{9}$, by definition, refers to the positive square root of $9$. There is only one of those, and it is equal to $3$, not to $-3$.

mweiss
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  • Ok, let's solve that equation $x^2 = 9$. Let's put both sides of the equation under the square root sign. We get $\sqrt{x^2} = \sqrt{9}$, $x = \sqrt{9}$. And now we get only one answer 3, because "by definition" square root has only one positive answer, right? And the answer is wrong, it's incomplete. What do we have to do with the function to get the second answer -3 then? What steps do we take? – user3600124 Feb 02 '16 at 01:46
  • You have made a common mistake: You "simplified" $\sqrt{x^2}$ as $x$. But these two expressions are not equal, for precisely the reason you identify. The correct way to simplify $\sqrt{x^2}$ is as $|x|$. So we end up with $|x|=3$, which has two solutions. – mweiss Feb 02 '16 at 02:05
  • @user3600124 $\sqrt{x^2}=|x|$ – Akiva Weinberger Feb 02 '16 at 04:07
  • @AkivaWeinberger Isn't that what I said? – mweiss Feb 02 '16 at 15:17
  • @mweiss So it is. I was tired. – Akiva Weinberger Feb 02 '16 at 15:19
  • I can see, that you are right over here, but still do not understand, where this absolute value thing comes from. Here is what I think $\sqrt{x^2} = (x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x^1 = x$ – user3600124 Feb 03 '16 at 10:58
  • Yes, we know that is what you think. But it is incorrect: $(a^b)^c = a^{bc}$ is only true for positive values of $a$. – mweiss Feb 03 '16 at 17:41
  • If you square a negative number and then take the square root of the result, you don't get back the original number you started with -- you get its absolute value. That's because the squaring operation loses information; if you know the result of squaring a number is $25$, there is no way to tell whether you started with $5$ or $-5$. But the square root of $25$ is $5$, which in either case is the absolute value of the number you began with. – mweiss Feb 03 '16 at 17:44
  • Related: http://matheducators.stackexchange.com/questions/10488/student-converted-sqrtx2-and-ended-up-with-just-x-instead-of-x – mweiss Feb 03 '16 at 17:44
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It's not a mistake to plot the negative branch of the graph $\sqrt{x}$. Rather it arises from the convention that the sign $\sqrt{x}$ means the positive root of x, whilst $-\sqrt{x}$ means the negative root of x. If we wish our graph to represent a function, with the important property that it be single-valued (one value of $x$ gives a unique value of $y$), then we have to restrict the graph to one or the other of the branches.

AmpleMimic
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  • Are you sure, that $\sqrt{9} = 3$ by definition? That is a square root of something always gives us one positive answer. – user3600124 Feb 02 '16 at 01:39
  • As long as $x$ is a non-negative real number, then yes, $\sqrt{x}$ by definition is the positive square root of the number. If $x$ is negative, nor not a real number, then there are still two square roots, but there is no unambiguous way to choose which one will be meant by $\sqrt{x}$, and things get complicated. – mweiss Feb 02 '16 at 02:03