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Is it wrong to say $$ \sqrt{x} \times \sqrt{x} =\sqrt{x^2}= \pm x$$ I am quite sure that $\sqrt{(x)^2} = \pm(x)$

But, does $\sqrt{x } \times \sqrt{x} =- (x)$ doesn't holds in $\mathbb{R}$ but if we assume $\mathbb{C}$ it holds right?

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Yes ! Your last step is invalid. –  Theorem Jun 15 '12 at 7:52
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Note that you could only get the result $-x$ if you chose one positive and one negative sign for $\sqrt{x}$ on the left-hand side, and this would mean that the square root function had not been properly defined. This is true in both $\mathbb{R}$ and $\mathbb{C}$. On the other hand, there are contexts where $x=-x$, particularly where you are working in characteristic 2 (familiar as mod 2 arithmetic). –  Mark Bennet Jun 15 '12 at 8:19
    
There should also be no comma before $\forall$, assuming standard English grammatical style. –  Alexander Gruber Jun 1 '13 at 3:02
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6 Answers

up vote 9 down vote accepted

Recall that a function returns a unique value. For this reason in the real numbers we have to choose positive or negative values for every number.

However it is vastly more useful to use positive numbers because then we can define composition in a meaningful way, e.g.: $$\sqrt{\sqrt{x}}=\sqrt[4]{x}$$

Had we chosen $\sqrt x<0$ then $\sqrt{\sqrt x}$ would not be a real number, whereas $\sqrt[4]x$ is still a real number (positive or negative).

These are good enough reasons to always require that when taking a root of even order we always choose the positive value, and therefore $\sqrt{x^2}=|x|$.

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Yes it is indeed wrong, because

$\sqrt{x} \times \sqrt{x}=(\sqrt{x})^2\neq\sqrt{(x)^2}$ (for $x < 0$, because here the lhs is not defined, but the rhs is)

We know that $(\sqrt{x})^2\geq0$, in particular for $x\geq0$ we have $(\sqrt{x})^2=x$, otherwise the term is not defined.

$\sqrt{(x)^2} = \pm(x)$ is also false, because a positive number always has a positive root. You probably mean $\sqrt{(\pm(x))^2} = |x|$

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Aha got it!-Thanks :) –  Quixotic Jun 15 '12 at 7:57
    
Good to hear that. –  Listing Jun 15 '12 at 7:58
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@Listing: at highschool we learnt that when dealing with real numbers he have that $\sqrt(x^2) = |x|$. –  Chris's sis Jun 15 '12 at 8:07
    
$\sqrt(x^2) = |x|$ does holds the problem here is that I interpreted $\sqrt{x} \times \sqrt{x}$ as $\sqrt{x^2} $ –  Quixotic Jun 15 '12 at 8:17
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The last line should be $\sqrt{(\pm(x))^2} = |x|$. –  Arturo Magidin Jun 15 '12 at 16:07
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The squareroot function takes non-negative values on non-negative arguments. For $x\ge 0$, $(\sqrt{x})^2=\sqrt{x^2}=|x|$.

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I think the source of your question is the confusion between "the square root" and "a square root" of a number.

It is the usual convention that the square root function on positive reals returns a positive real number (and it's not really all that arbitrary, it is in many ways very convenient). And that is what is usually meant when "the square root" is mentioned.

However, if $a\geq 0$, then $\sqrt a$ and $-\sqrt a$ are both square roots of $a$ in the sense that they are solutions of the equation $x^2-a=0$. Thus it is not wrong to say that the product of two square roots (but not THE square roots) of $a$ is $\pm a$ ($-a$ if they are distinct roots, $a$ otherwise -- this is also true for any complex $a$).

On the other hand, it is wrong to say that the square of a square root of $a$ is $\pm a$, since it is $a$, whichever square root you use (in fact, it is the very definition of a square root), which also applies to the product of two numbers, both of which are THE square root of $a$, so $\sqrt a \cdot \sqrt a=a$.

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This is IMHO the best answer to this question. –  ShreevatsaR Jul 14 '12 at 15:06
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It is wrong, because $\sqrt{x}$ is only defined for $x\geq0$ (if $x$ is a real number), so it's pointless to say $\sqrt{x}·\sqrt{x}=\pm x$.

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What about $\sqrt{0}$? –  Théophile Jun 17 '12 at 1:50
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@Théophile ook... $x\geq0$ –  Barranka Jun 17 '12 at 3:39
    
Notice that $\sqrt{x^2}=|x|$, so it can be said that $\sqrt{x^2}=\pm x$ –  Barranka Jun 20 '12 at 14:35
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No, there is an important distinction to be made between these: $|x|$ is a single, positive value, while $\pm x$ refers to two values, one positive and one negative. –  Théophile Jun 21 '12 at 3:01
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Short answer: Yes, it is wrong.

Long answer: It depends on what you mean by your equation. As @Asaf Karagila points out, a function maps an input to one single output (though it may not necessarily be unique for each input). Therefore, to identify the function

\begin{equation} \begin{array}{rCc} f: \mathbb{R} & \longrightarrow & \mathbb{R}\\ x & \longmapsto & \sqrt{x} \cdot \sqrt{x} \end{array} \end{equation}

as $f(x) = \pm x$ is wrong, since $\pm x$ is not a single value. In this sense, it is wrong to say that $\sqrt{x} \cdot \sqrt{x} = \pm x$.

However, you can also interpret the equation $$ \sqrt{x} \cdot \sqrt{x} = \pm x $$ as identifying all $x$ that solve at least one of the following equations: $$ \sqrt{x} \cdot \sqrt{x} = x $$ $$ \sqrt{x} \cdot \sqrt{x} = -x. $$

For instance, the quadratic formula for solutions of $x^2 + ax + b = 0$ is often stated as

$$ x = -\frac{a}{2} \pm \sqrt{\left(\frac{a}{2}\right)^2 - b},$$

which is to be interpreted as "any $x$ that solves the above equation when $+$ or $-$ is substituted for $\pm$ also solves the equation $x^2 + ax + b = 0$".

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