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I've always been told that if

$\ x^2 = 4,$

$ =>x = \pm2$

But recently, Prof. mentioned that if $x = \sqrt{4}$,

Then $x = +2(only)$

I am very skeptical about this because they both mean the same thing and still yield different results! So how is the above statement justified?

Ishan Banerjee
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Graviton
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    $\sqrt{x}$ is a function. By definition a function can only take one value for a particular $x$. We have defined $\sqrt{x}$ to be non-negative. This means that if $x^2=a$, then $x=\pm\sqrt{a}$. – robjohn May 03 '13 at 16:19
  • This has been asked many times here. Let's see if I can find one...related – The Chaz 2.0 May 03 '13 at 17:11
  • I reference two comments from two other posts to help: 1. See Clive Newstead's comment under the Question at http://math.stackexchange.com/questions/343894/if-pm-x2-x-then-sqrtx-x $\qquad$ 2. See the first sentence of Arturo Magidin's answer at http://math.stackexchange.com/a/41882/53259. –  Aug 30 '13 at 14:42

5 Answers5

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It is correct that if $x^2 = 4$ then $x$ is either $2$ or $-2$.

But by definition $\sqrt y$ (for a positive real number $y$) is the $positive$ real number $z$ such that $z^2 = y$.

Tobias Kildetoft
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  • What about $4^{\frac{1}{2}}$? Is it equal to $±2$ or is it too defined as just $2$? – User Aug 09 '18 at 16:23
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Why doesn't one define $\sqrt{4}$ to be the two-pronged number $\pm2$? There is a good reason: what value would you assign to

$$\sqrt{9}+\sqrt{4}$$

among $5$, $1$, $-1$ and $-5$? And $\sqrt{4}-\sqrt{4}$ would be $0$, $4$ or $-4$?

We need a function returning a single value, so we can algebraically manipulate it. By convention, it has been decided to use the non-negative solution of the corresponding equation, because so algebraic manipulations are simpler.

No law could disallow somebody to define the square root of $x$ (for $x\ge0$) by taking the non-positive value; or even the positive value when $x$ is irrational and the non-negative value when $x$ is rational; or any other variation thereof. But, of course, computations involving square roots would be rather complicated.

egreg
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If we consider the equation $$ x^2 = 4 $$ then indeed you have two solutions to the equation: $x = \pm 2$. When we talk about the squareroot we have a function. And the key thing about a function is: one input gives one output. We have not defined $\sqrt{y}$ as any solution to $x^2 = y$.

So then it is a matter of choice (definition) that we have said that (for $y\geq 0$) $\sqrt{y}$ is the non-negative solution to the equation $x^2 = y$. So $\sqrt{4} = 2$.

Thomas
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Fix $x\in\Bbb R_{\geq 0}$. The symbol $\sqrt x$ returns the (unique) non-negative number $y$ such that $y^2=x$. The symbol $-\sqrt x$ returns the (unique) non-positive number $y'$ such that $y'^2=x$. It's a convention, and mainly because we want $x\mapsto \sqrt x$ to be single valued.

Of course in the above we can say positive instead of non-negative, and negative instead of non-positive by ruling out the $\sqrt 0 =0$ case.

Pedro
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Good question.

I wondered about that myself while we talked in mathematics about the parabola and why the teacher only allowed the upper half of its inverted function.

While the answers here repeat the conventional wisdom that we need to define a "function", this is not necessarily true. In the complex plane we can assign a "holomorphic function" to the complex square-root by Riemann surfaces. Because it is not so easy to define the exact difference between a "true" function and a "function" which is locally injective, mathematicians invented the stopgap of "multi-valued functions".

I think the real reason to permit only the positive solution is because it allows the operation:

$$\sqrt{a*b} = \sqrt{a} * \sqrt{b} $$

As you can easily prove yourself, this is only possible if you define the result of the real square root function as strictly positive or zero.

This operation is not allowed if we enter the complex plane.

CommonerG
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Thorsten S.
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