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Can the square root of a real number be negative?

Dealing with the questions of functions in eleventh class my maths teacher says that square root of a real number is always positive. How is it possible?

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Every positive number $a$ has two solutions to the equation $x^2 = a$. So people generally write $\sqrt a$ to denote the positive solution. So it is always positive - but by convention or definition, not by any mathematical reasoning. – Stephen Montgomery-Smith May 26 '14 at 1:02
@StephenMontgomery-Smith - I disagree. The square root of a number has to be positive, or you end up with $a = \sqrt{b} = -a \implies 1= \sqrt{1} = -1 \implies 1=-1$. That would create a mathematical contradiction, not just a problem with convention. – Ephraim May 26 '14 at 5:24
@tpb261 - no, because we are saying that $x$ has two possible solutions ($2$, and $-2$), however, they are not both the solution "at the same time". Since any number squared by definition must be positive, $2^2 = 4 = (-2)^2 \implies 2^2 = (-2)^2 \implies 4 = 4$ which is correct. $x$ can be positive or negative, however, if we define a number squared as positive, than it's square root must be positive in order to avoid the $2=-2$ contradiction. $\sqrt{a^2} \neq a$, instead, $\sqrt{a^2} = \vert a \vert$ The idea is that a square root does not actually reverse the number squared. – Ephraim May 26 '14 at 6:59
@Ephraim "they are not both the solution "at the same time" Agreed. But, both are acceptable as solutions - that is all I wanted to say to the OP. – tpb261 May 26 '14 at 7:33
@Ephraim "a=√b=−a⟹1=√1=−1⟹1=−1" It's pointless to begin with — until you decide which of two a or -a is a square root you cannot use √b in equation. – Cthulhu May 26 '14 at 11:26

Given a positive real number $a$, there are two solutions to the equation $x^2 = a$, one is positive, and the other is negative. We denote the positive root (which we often call the square root) by $\sqrt{a}$. The negative solution of $x^2 = a$ is $-\sqrt{a}$ (we know that if $x$ satisfies $x^2 = a$, then $(-x)^2 = x^2 = a$, therefore, because $\sqrt{a}$ is a solution, so is $-\sqrt{a}$). So, for $a > 0$, $\sqrt{a} > 0$, but there are two solutions to the equation $x^2 = a$, one positive ($\sqrt{a}$) and one negative ($-\sqrt{a}$). For $a = 0$, the two solutions coincide with $\sqrt{0} = 0$.

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Would the person who downvoted this care to explain why? – Michael Albanese May 26 '14 at 1:02
My guess is that (s)he would not. Good answer. – André Nicolas May 26 '14 at 1:04
@AndreNicolas: It was worth a try. Thanks for your comments. – Michael Albanese May 26 '14 at 1:24
@MichaelAlbanese In most of the time, it is not worth a try, because people do -1, close the tab and never come back. Alas. – glglgl May 26 '14 at 10:51
I did not downvote, but do sense there's a potentially important omission. When I was a school student (way back in the days of New Math), the term "square root of $a$" was used to mean "a real number $b$ such that $b^{2} = a$", as in $\sqrt{4} = \pm2$. (I'm not saying this is good notation...!) Only in university was "$\sqrt{a}$" systematically used to denote the non-negative square root of a real number $a \geq 0$. It's possible the OP's teacher and/or textbook are either unspecific or in implicit conflict on the matter. – Andrew D. Hwang May 27 '14 at 12:07

It is just a notational matter. By convention, for positive $x$ (real clearly), $\sqrt{x}$ denotes the positive square root of the real number $x$. Likewise we agree by way of notational convention that $-\sqrt{x}$ is the negative square root of $x$. Of course, every positive real number, $x$, has two square roots, $\sqrt{x}$ and $-\sqrt{x}$, positive and negative real numbers respectively.

I worry sometimes about what gets taught by way mathematics in secondary school these days.

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Technically this statement is wrong. He could say, "The square root of a positive number is positive (by definition)". E.g. for 0 you get $\sqrt{0}=0$ which is neither positive nor negative. And for negative numbers you even get complex solutions which are neither positive nor negative nor 0.

The definite article and the singular in "the square root" is also important to imply the conventional definition of $\sqrt{}$. But more correctly he should say "the principal square root", because mathematically the expression "the square root" doesn't make sense, since there are two different roots in general.

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I believe your confusion is coming from assuming that if $a^2 = b$ then $\sqrt{b} =a$, but this is actually not the case. The correct form of this would be $\sqrt{b} = \vert a \vert$. Because of this, $a^2 = b = (-a)^2$ however $\sqrt{b} = \vert a \vert \neq -\vert a \vert$. This can be proven by contradiction:

If we were to say that $a=\sqrt{b}=-a$, that would therefore imply that $a = -a$, and for example $1= \sqrt{1} = -1 \implies 1=-1$ which we know to be false. This contradiction does not show up when saying $a^2 = b = (-a)^2 \implies a^2 = (-a)^2$, because, if we were to use the same example, we would get $1^2 = 1 = (-1)^2 \implies 1^2 = (-1)^2 \implies 1 = 1$, which is true (since by definition, any number squared must be positive). This is why, if you were to evaluate $a^2 = b$, you would get two possible solution for $a$ (one positive, and one negative). However, if you were to evaluate the equation $a = \sqrt{b}$, $a$ can only have one solution at any given time, and for convention, a square root was defined to always be positive. This is an important distinction because it allows us to look at the equation $4=a^2$, and find that $a=2\oplus a=-2$, thus avoiding the contradiction $a=2 \wedge a=-2 \implies 2=-2$ by saying $2^2=4=2^2 \implies 2 = 2 \oplus (-2)^2=4=(-2)^2 \implies -2=-2$. This idea might seem to get lost when graphing equations such as a circle. The equation $x^2 + y^2 = 1$ appears to have 2 $y$ values for every $x$, and 2 $x$ values for every $y$. This can be better understood by instead looking at the parametric equation for a circle: $x=cos(t); y=sin(t)$. For any given value of $t$, there is only one corresponding value of $x$, and only one corresponding value of $y$. If you are given a value for $x$, and told to solve for $t$, the most you can do is find possibilities of $t$, since the $(x,y)$ points on the graph repeat themselves every $2\pi*t$. The same idea is true for square roots. When you square a number, it always creates a positive number, therefore it is impossible to reverse definitively. The most we can do is say that there are two possibilities of what the original number was. For convention, it has been established that for an equation $a^2 = b$, where $\sqrt{b}=c$, we say that $c=\vert a \vert$. It would work just as well to defined a square root by $c=-\vert a \vert$, but I guess the mathematicians that decided on it liked working with positive numbers more.

The point in all of this was to simply establish that taking the square root of a squared number, does not reverse its exponent, because it cannot be reversed definitively. As user86418 put it:

If a and b are real numbers, then the conditions $a^2=b$ and $a=\sqrt{b}$ are not logically equivalent; the second implies the first, but not conversely.

Therefor, for the purposes of convention, a square root has been defined to be the absolute value of the original number that was squared. This why, if you plug the functions $y^2 = x$ and $y = \sqrt{x}$ into a graphing calculator or Wolfram Alpha, you will find that you get two very different looking graphs. Notice how the the graph of $y=\sqrt{x}$ never goes below the $x$-axis. Had a square root been defined as always negative, the graph of $y=\sqrt{x}$ would simply be flipped about the $x$-axis.

y^2 = xy = sqrt{x}

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can the person who downvoted please explain? – Ephraim May 26 '14 at 6:49
I didn't downvote, but don't understand your usages of time. It sounds as if you're asserting that a functional equation with multiple solutions, such as $\sin x = 0$, "can only have one solution at any given time" (since otherwise mathematics would contain a contradiction, e.g. $\pi = 0$). Does this paraphrase really express your view accurately? – Andrew D. Hwang May 26 '14 at 10:33
@user86418 - not exactly. $\sin(x) = 0$ is true for an infinite number of $x$ values, however, that does not mean that $x$ is equal to each of those values simultaneously. What I mean by this is that when you say that the equation is true for an infinite number of $x$ values, it means that $x$ is equal to any one of them, but not all of them, otherwise you would end up with $\pi = 0$. – Ephraim May 26 '14 at 15:35
eg. for the equation $sin(x) = 0$, we can say that $x=0$ or $x=\pi$ or $x=2\pi$ or $x=$... However, what we cannot say is that $x=0=x=\pi=x=2\pi=...$ because that would yield $0=\pi=2\pi=...$ which creates a contradiction. – Ephraim May 26 '14 at 15:54
@user86418 - I have edited my answer to try to better explain what I mean by "can't have two values at the same time" – Ephraim May 26 '14 at 23:14

THE square root is defined to be the positive root. In reality there are always 2 roots, one positive and one negative. Think about the sqrt of 4, what this is asking is "what two numbers that are the same multiply to 4?" The answer would be 2 or -2 because 2x2=4 and (-2)x(-2)=4.

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"Always 2 roots, one positive and one negative"... Except $0$, which has one root with multiplicity $2$ that is neither positive nor negative. – vrugtehagel Feb 18 at 8:59

Recall the vertical-line test: if you graph an equation and if there is a vertical line that intersects the graph at more than one point, the equation does not correspond to a function.

The vertical-line test is, although not particularly rigorous nor very extensible, a good demonstration of how to show that a function is well-defined. Although yes, the square of any positive real is equal to the square of the corresponding negative, the square root function cannot be well-defined if it generates both a positive and negative value.

As such, $f(x)=\sqrt{x}$ is defined to be the positive root so that it is a function. When it is desired that both the positive and negative roots are referred to, one has the option of using either $\pm\sqrt{x}$ or $\surd x$.

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The vertical line test determines whether or not a "function" is well-defined, not injective. – Dustan Levenstein May 26 '14 at 1:13

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