I recently saw someone say:     $\sqrt{x}$ is, by definition, positive   Which didn't sit easily with me as we always say $\pm$ for any square root.  Then it got me thinking that if, by definition, the symbol $\sqrt{x}$ meant the positive or the negative root of x, then the $\pm$ symbol would be redundant.  So, have mathematicians defined the root symbol only to include the positive root, thereby necessitating the $\pm$ symbol?  EDIT: This question is different from the one here:  because my question focuses principally on the meaning/convention of the square root symbol rather than on the topic of their being two solutions to $x^2=4$
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Last year in Pre-Algebra we learned about square roots. I was taught then that $\sqrt{64}=8$ and $\sqrt{100}=10$, which I understood and accepted. I was also taught that $\pm\sqrt{64} = 8,-8$ because both of those numbers squared is 64, which I also get. But this year, with a new school and teacher in a different state, our teacher is telling us that: $\sqrt{64}=8,-8$ and $\pm\sqrt{64}$ also is $8,-8$. The way to get the positive root of something is: $+\sqrt{64}=8$ And these seem to contradict each other. I was always taught that a regular square root returned a positive number and only a positive number, but now my teacher is saying a regular square root gives two numbers, and considering the square root of a number $n$ is defined as $y^2=n$ I see where he is coming from. Upon researching this Wikipedia says:  For example, $4$ and $−4$ are square roots of $16$ because $4^{2} = (−4)^{2} = 16$  And Wolfram MathWorld says:  Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of $9$ are $-3$ and $+3$  But on the other side, Wolfram Alpha, when given &quot;The square root of 9&quot; gives only 3.  So, which is right? Is $\sqrt{64}$ considered $8$? or is it $8,-8$?
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. For instance I can define either of the two for square root of $X$: $$\sqrt{X}^{(1)} \equiv \frac{1}{\sqrt{2 i}} \begin{pmatrix} 1 &amp; i \\ i &amp; 1 \end{pmatrix}$$ Or : $$\sqrt{X}^{(2)} \equiv \sqrt{NOT} = \frac{1}{2} \begin{pmatrix} 1+i &amp; 1-i \\ 1-i &amp; 1+i \end{pmatrix} \neq \sqrt{X}^{(1)} $$ We notice that those two matrices do not differ by a global phase: they really act differently on quantum state. Now, it is frequent in the litterature to simply see $\sqrt{X}$ or $\sqrt{Y}$ without further explanations. To be generic I consider $M$ a matrix in which I will consider its square root. As long as we do $\sqrt{M} \sqrt{M}$ whatever the definition taken it doesn't matter. However I if we have calculations involving $\sqrt{M}$ without having it directly squared, the result might depend on the choice we took for the square root. My question Is there a uniform convention in quantum information to define square root of matrices ? Or more specifically square roots of Pauli ? Or each reference must specify which convention they take ?
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He says (direct quote): "In higher mathematics the absolute value of a number, $|x|$, is equal to positive and negative $x$, if $x$ is a positive number."  Then he wrote:  $|x|=\pm x,\quad \text{if}\enspace x&gt;0 $.  I think he misunderstood the definition of absolute value, or did I? From what I understand, absolute value of a number is the distance of a number from zero, so it is always positive. Am I wrong?
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If a real number $a&gt;0$ has an $n$th root, then it is unique: If $x^n=a=y^n$ (with $x$ and $y$ also positive), then $x=y$, because  $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+xy^{n-2}+y^{n-1}),$$  and the factor on the right is positive. This also shows that $x^n&lt;y^n$ if and only if $x&lt;y$.  Now consider the set  $$S=\{x\mid x&gt;0,\;x^n&lt;a\}.$$  This is non-empty: If $a\geq1$, then $(\tfrac12)^n&lt;a$; and if $a&lt;1$, then $a^n&lt;a$.  It's bounded above: If $a\geq1$, then $a^n\geq a$, so $a\not\in S$, and for any $x&gt;a$, it follows that $x^n&gt;a^n\geq a$, so $x\not\in S$. And if $a&lt;1$, then $1^n&gt;a$, so $1\not\in S$, and likewise for any $x&gt;1$.  Therefore, $S$ has a supremum $M&gt;0$. How can we show that $M^n=a$?     suggest that it's easy, but I don't see it.  We could define $\sqrt[n]{a}=\exp(\tfrac1n\ln(a))$, but I don't want to rely on transcendental functions to construct this simple algebraic function. And I don't want to refer to the Intermediate Value Theorem or continuity.    If $M^n&lt;a$, then there is some number between these (for example $\frac{M^n+a}{2}$); if this could be written in the form $x^n$, then $M^n&lt;x^n&lt;a$ would imply $M&lt;x\in S$ and contradict the definition of supremum. But how can we find such $x$? Maybe write $x=M+\varepsilon$ and use the binomial expansion...  Similarly, if $M^n&gt;a$, and something between them could be written in the form $N^n$, then $N$ would be a smaller bound on $S$ than $M$, contradicting the definition of supremum. How can we find such $N$?
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Problem: Is there an irreducible $f\in \mathbb{Z}[x]$, whose image in every $(\mathbb{Z}/p\mathbb{Z})[x]$ has a root for $p$ prime? If there is, what is the minimal degree possible?   I can only prove that $x^2-c$ is impossible, by quadratic reciprocity and Chinese remainder theorem. Even the case of $a x^2 - c$ is unknown to me.  Meanwhile, if $f$ is not required  to be irreducible, but only have no root in $\mathbb{Z}$, then $(x^2-a)(x^2-b)(x^2-ab)$ is a solution for non-sqaure $a,b,ab$, since if both $a,b$ are not squares in $\mathbb{Z}/p\mathbb{Z}$, then $ab$ is a square. Yet I still don't know if this is of minimal degree.  Also, it is natural to pose generalizations   When $p$ is not necessarily prime (equivalently for all prime powers). When $p$ is odd prime. When $p$ represents sufficiently large primes (equivalently all but a finite number left).
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Determine number of real roots of $f(x)=x^{5}+x^{3}-2x+1$.   I used Descartes's rule of sign to determine the number of positive and negative roots; there is only 1 negative (hence minimum 1 real root) and two positive roots. These 2 positive roots can be anything- a complex pair, or two real roots- but what about the other 2 roots? Is there any flaw in my understanding of rule of signs? How can I find the exact number of real roots?  Thanks in advance.
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Possible Duplicate:          The square root of a number squared is equal to the absolute value of that number. Why is $\sqrt{x^2} = |x|$? Why not just $x$? Please give me a reason and also help me prove it.
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I would like an efficient algorithm for square root of a positive integer.  Is there a reference that compares various square root algorithms?  Thanks.
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This may seem like a very basic question, but:  What exactly is the difference between a root of a polynomial, and a zero?  Of course I realise that they are technically exactly the same thing, but there seem to be subtle rules as to when to use each term, and a couple of times in the past I have been told I am using "root" where I should be using "zero".  Is it generally accepted that one should use "root" in an algebraic context,  and "zero" in a analytic context? If not, when should one use one or the other...and does it really matter?!
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Q1 - Why is $\sqrt{-x}*\sqrt{-x}=-x?$  Q2 - I was thinking it would be:  $\sqrt{-x}*\sqrt{-x}=\sqrt{-x*-x}=\sqrt{x^2}$ but apparently not (why not?)  Q3 - What are the formal algebra rules to use? Can I calculate this without using i such as in: $\sqrt{-x}*\sqrt{-x}=i\sqrt{x}*i\sqrt{x}=-\sqrt{x^2}=-x$.
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If $x^2-bx+a = 0$ and $x^2-ax+b = 0$ both have distinct positive integers roots, then what is $(a,b)$?   My Try: $$\displaystyle x^2-ax+b = 0\Rightarrow x = \frac{a\pm \sqrt{a^2-4b}}{2}$$  So here $a^2-4b$ is a perfect square.  Similarly  $$\displaystyle x^2-bx+a = 0\Rightarrow x = \frac{b\pm \sqrt{b^2-4a}}{2}$$  So here $b^2-4a$ is a perfect square.  But I did not understand how can I solve after that.
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Using the siunitx package, if I specify an error with the \pm symbol then it automatically converts it to a form using parenthesis. I would like it to not do this and keep the \pm symbol, but still space the elements of the quantity and symbol in a good way.  e.g. \SI{10 \pm 2}{\kelvin} would produce 10(1) K  I would like it to produce: 10\pm1 K or something similar with spacing taken care of.
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Inspired by  on the math StackExchange, suppose you want to typeset the square root of the number 144...4, that is, 1 followed by n 4's. If you write $\sqrt{1\underbrace{44\ldots4}_{n\ \text{times}}}$, as in the current version of the question, you get a huge radical sign that reaches down to match the "n times" text below the number. Is there an elegant way to typeset this so that the radical sign looks like it would with $\sqrt{144\ldots4}$, but also keeping the underset text in the right place?
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As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.  But then I remember that $\sqrt(x)$ is the same as $x^{1/2}$ and thus, $\sqrt{x^2}$ is the same as $x^{1}$.   So, as far as I get it, $\sqrt{x^2} = x^{2/2}$.  But then, I can cancel $\frac{2}{2}$ to $1$. So: $\sqrt{x^2} = x^1 = x$.  Where does the absolute value in this derivation of the calculation come from?  I get it why its there: Because $(-x)^2 = x^2$, so the radicand is always positive. But from $\sqrt{x^n} = x^{n/2}$ I dont understand how it gets there.
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Can someone provide a proof that every quadratic residue, when working in $\mathbb Z_p$, where $p$ is a prime, has exactly two roots? Indeed, there cannot be only one root as for any $a^2$, we know $a$ and $-a$ are both roots. So then we need only show that there cannot be more than two- does this follow from some argument for the number of roots a quadratic can have modulo a prime? I'm not well-acquainted with the theorems pertaining to prime moduli.
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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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In class, we were simply given that this limit is undefined since along the paths $y=\pm x$, the function is undefined.  Am I right to think that this should be the case for any function, where the denominator is $x^2-y^2$, regardless of what the numerator is?  Just wanted to see if this is a quick way to identify limits of this form.  Thanks for the discussion and help!