# Talk:Square root

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Wikipedia Version 1.0 Editorial Team / Vital
 To-do list for Square root: Run the first sentence past some non-mathematicians to test for clarity. Other things currently in the lead, such as the origin of the symbol, irrational numbers, could go in later sections. The section on "computation" could start with how to compute a square root on a calculator and then mention trial-and-error methods before going onto the algorithm using e. The references should go into a references section. Sorry if I'm confused, but the page is showing up as:"In mathematics, a square ruth of a". The History section should appear as the first section.

## Trigonometric Definition of a Square Root

Hey I came up with a way to calculate the square root using trig functions:

square root of x = 1/(tan((1/2)arccos((x-1)/(x+1))))

Is this helpful? If so then someone could make a section called "trigonometric definition of the square root" with this equation in there.

Thanks — Preceding unsigned comment added by Knowledge spouse (talkcontribs) 01:18, 10 May 2014 (UTC)

## Request for a graph

Could somebody produce an image of the graph which isn't that humongous and which doesn't use commas as decimal points? AxelBoldt 02:12 Feb 1, 2003 (UTC)

## Taylor series

Can you tell me exactly how to continue the coefficients of the Taylor series? They appear to all have powers of 2 in the denominator.

Hi,

In your Taylor's formula i wrote x^n at the end

62.147.217.21 12:35 May 4, 2003 (UTC)

Why the restriction to normal matrices? I can take any square matrix with a complete set of eigenvectors and just take the square roots of the eigenvalues to get a square root of the matrix. Can't I?

Josh Cherry 01:49, 13 Oct 2003 (UTC)

"Can you tell me exactly how to continue the coefficients of the Taylor series? They appear to all have powers of 2 in the denominator."

I hope my latest edit has addressed this question.

dcljr 23:12, 5 Jul 2004 (UTC)

t762@inforamp.net (J. B. Rainsberger) wrote:

>In article <31C14605.2D3E@laria.u-picardie.fr>,
>   Mustapha Benali  wrote:
>>
>>
>> -1 = (-1)^1 = (-1)^(2/2) = [(-1)^2]^(1/2) = 1^(1/2) = 1
>>
>A great deal has been said about "the square root" of one. I had this
>discussion with a grade five (?) student from Australia on this topic and
>he understood it perfectly, so I present it here:

>"The" square root oftens refers to the principle square root, which is the
>positive quantity whose square is the given quantity. There are *two*
>square roots of any and all real numbers, and they are both complex
>numbers. (Sometimes strictly real.)

>The expression x^(1/2) refers to the principle square root of a number.
>This is verified by the relations:

>    x^(1/2) = y
>    1/2 ln x = ln y

>In order to define this properly for x > 0, we require y > 0. If y < 0,
>then ln y is not strictly a real number, and thus, neither is ln x. If ln x
>is not strictly real, then x is not positive. As a result, if x >= 0,
>x^(1/2) >= 0.

>As for why the above equation is not quite correct, it is commonly known
>that when squaring all members of an equation, extraneous roots are often
>introduced. As an example:

>    sqrt(x) + 2 = x
>    x + 4 sqrt(x) + 4 = x^2
>    4 sqrt(x) = x^2 - x - 4
>    16 x = x^4 - 2 x^3 - 7 x^2 + 8 x + 16
>    0 = x^4 - 2 x^3 - 7 x^2 - 8 x + 16
>    solving for x, x is in {1, 4, -3/2 +/- sqrt(7)/2 i}

>Now I ask you: given that sqrt(x) is the principle (positive) square root
>of x, is x = 1 a solution to this equation? No.

>There you have an example of an extraneous root produced by squaring all
>members of an equation. As a result, if you rearrange to do this:

>-1 = (-1)^1 = (-1)^(2/2) = [(-1)^(1/2)]^2 = i^2 = -1

>then you're cooking.

>Just be careful that you're not adding extraneous roots by squaring both
>sides of an equation. You need to check all roots thus obtained in the
>original equation or expression.


## Simpler proof of the irrationality of the square root of 2

The comment below underscores the fact that "simpler" is a slippery word. This proof seems simpler to the writer, but it depends on knowing the uniqueness of prime decomposition. On the other hand, the usual proof only requires knowing that the square of an even number is even and the square of an odd number is odd. So the "simpler" proof requires much more background.

There is a much simpler proof of the irrationality of the square root of 2 that can be readily grasped in its entirety.

Suppose the square root of 2 is rational, that is, that it can be expressed as a ratio of positive integers. Factor the numerator and denominator of this ratio into primes, and square it. Every prime factor in both numerator and denominator must then appear an even number of times. For this ratio to be equal to 2, all factors in numerator and denominator must cancel, except for a single 2 in the numerator, which is impossible with an even number of factors.

I've seen the proof for this numerous times and done a couple of different ways, a few of which i could even rattle off now without thinking. However, what you have here seems very hard to understand and follow. I would somewhat blame it on the formatting issues, but also on a lack of other information. For example, when you're supposing the square root of 2 is rational, you didn't even list the full definition of rational in your proof. You said the ration of two integers. A better definition would be a/b where a and b are integers and b is not equal to zero. Then, go in to the "without loss of generality, we can assume a and b have no common factors greater than 1". Follow, with what you were saying about squaring both sides and then doing simple math to solve for a squared. But then, remember to point out the previous thereom that gives that a squared must be even. Do more simple pluging in and simple math, but point out the same thereom again when sayin b squared is even. Then, point out that if a and b are even they have a common factor of 2 which condricticts the "without loss of generality...". Therefore, square root of 2 is irrational. Then, finish with Q.E.D. or # or whatever you use to formally end your proofs. In conclusion, i was just saying the proofs wheren't more simple. They were shorter. But they both left out too many important details. —Preceding unsigned comment added by 68.114.97.211 (talk) 03:32, 26 March 2008 (UTC)

Yes, but the uniqueness of prime decomposition is actually quite a difficult theorem and while the ancient Greeks undoubtedly understood it they hardly had the apparatus even to state it (as the English mathematician G. H. Hardy observed). It depends on an induction (itself a relatively recent innovation) on a lemma of Euclid's which states that if an integer is the product of two integers and is divisible by a prime then at least one of the two integers is also divisible by the prime. But that isn't easy to prove, either by the methods used by Euclid or even today (which depends on his famous, but incompetely stated, algorithm to find the greatest common divisor of two numbers). Have a look at the page Euclid's lemma.FightingMac (talk) 01:01, 16 April 2011 (UTC)

The "Square roots using Newton iteration" is very hard for a 12 year old kid to read. I shall try and read it and reduce it to a format which is easier for a 12 year old kid to read.

Ohanian 06:15, 2005 Mar 28 (UTC)

I've added some remarks to the 'Computation' section which notes that the scheme predates Newton-Rhapson by many centuries and offers an informal explanation of it as well as a link to the arithmetic-geometric mean inequality on which it formally depends. FightingMac (talk) 01:01, 16 April 2011 (UTC)

### Square roots using Newton iteration

Basic Newton iteration finds a single root of a function $f(x)$ given a sufficently precise approximation to the root. The nature of which root will be given based on an approximation is dependent on the Newton fractal which we will not discuss here any further. The basic iteration is given by:

$x_{n+1} = x_n - {f(x_n) \over f^\prime(x_n)}$.

There are two widely used functions $f(x)$ used to find the square root of a number, say, "z".

#### First method

The first method finds the square root of "z"

$f(x) = x^2 - z \,$

Note that the $\sqrt{z}$ and $- \, \sqrt{z}$ are roots of the function $f(x)$. ie $f( \sqrt{z} ) = 0$.

The first derivative of $f(x)$ is $f^\prime(x) = 2 x$

Thus iteration for $x_{n+1}$ is derived where:

$x_0 = 1\,\!$        and

 $x_{n+1}\,\!$ $= x_n - {f(x_n) \over f^\prime(x_n)}$ $= x_n - {(x_n^2 - z) \over 2 x_n}$ $= x_n - \frac {x_n}{2} + \frac {z}{2 \,\,\, x_n}$ $= \frac {x_n}{2} + \frac {z}{2 \,\,\, x_n}$.

#### Second method

The second method finds the reciprocal of the square root of "z".

$g(x) = \frac{1}{x^{2}} - z$.

The two roots to $g(x)$ are $\frac{1}{\sqrt{z}}$ and $\frac{-1}{\sqrt{z}}$ .

The derivative of $g(x)$ is $g^\prime(x) = -2 \,\, x^{-3}$.

Thus iteration for $x_{n+1}$ is derived where:

$x_0 = 1\,\!$        and

 $x_{n+1} \,\!$ $= x_n - {g(x_n) \over g^\prime(x_n)}$ $= x_n - {x_n^{-2} - z \over -2 x_n^{-3}}$ $= x_n - (-1/2) {x_n^3} (x^{-2} - z)$ $= x_n + (1/2) (x_n - z x_n^3)$ $= \frac{ 3 x_n - z x_n^3 }{2}$ $= 1.5 \, x_n - 0.5 \, z x_n^3 = 0.5 \, x_n \,\, ( 3 - z x_n^2 )$.

#### Example

Find the $\sqrt{7}$ using both methods.

$Z = 7 \,$ Because we are looking for the square root of 7
$f(x)\,$ $g(x)\,$
$x_0\,$ $1\,$ $x_0\,$ $0.5\,$
$x_1\,$ $\frac{1}{2} + \frac{7}{2 \times 1} = 4$ $x_1\,$ $0.5 \times 0.5 \,\, ( 3 - 7 (0.5)^2 ) = 0.312$ $\frac{1}{x_1}= 3.2\,$
$x_2\,$ $\frac{4}{2} + \frac{7}{2 \times 4} = 2.875$ $x_2\,$ $0.5 \times 0.312 \,\, ( 3 - 7 (0.312)^2 ) = 0.362$ $\frac{1}{x_2}= 2.762\,$
$x_3\,$ $\frac{2.875}{2} + \frac{7}{2 \times 2.875} = 2.654$ $x_3\,$ $0.5 \times 0.362 \,\, ( 3 - 7 (0.362)^2 ) = 0.376$ $\frac{1}{x_3}= 2.652\,$
$x_4\,$ $\frac{2.654}{2} + \frac{7}{2 \times 2.654} = 2.645$ $x_4\,$ $0.5 \times 0.376 \,\, ( 3 - 7 (0.376)^2 ) = 0.378$ $\frac{1}{x_4}= 2.645\,$
$\sqrt{7} \approx 2.645$ $\sqrt{7} \approx 2.645$

#### Comparison

The iteration for $f(x)$ involves a division which is more time consuming than a multiplication in computer integer arithmetic. The iteration for $g(x)$ involves no division and is thus recommended for large integers z.

This iteration using "g" involves only a squaring and two multiplications, as opposed to a division in the case of "f". In practical implementations of large integer square roots, the iteration involving "g" is faster for large integers "z" since division is at best $O(M(n))$, a constant times the time function of multiplication. The constant term is almost always 3 or more, meaning that a single division can almost never be faster than 3 multiplications.

I was after the history of where the square root character is derived like how the page on eight shows where the character was derived from. Question by unknown user.

I believe it was created from radix symbol. Do a search on radix on google. Ohanian 00:57, 2005 May 3 (UTC)

The square root notation originated in Germany, 16th or 17th century or so. They denoted a radical with an "r" in front of the number, like r2 for square root of 2. They then took to putting a line over the top of the number they were talking about. Mathematicians-in-a-hurry eventually took to writing the r and the line without raising their pens from the paper, and the radical sign was born.

Contrary to popular belief, the use of the radical to represent roots can actually be traced to a corruption of the Greek letter "nu" (similar to a Latin v) - which was placed next to the number in question to create a "new" number, the root.

The use of an adjoining bar across the top is analogous to the use of a bar by statisticians to represent the "mean" of a collection of numbers; it indicates the part of the expression we "mean" when creating the new number. Ohanian 01:02, 2005 May 3 (UTC)

## Calculating Square Roots - Mentally

Hi, I find the explanation of calculating square roots using Pell's equation hard to follow. Would it be possible to explain what is happening in each step more clearly? -- MattW, 05 June 2005

## continued fraction rewrite

### Continued fraction methods

Quadratic irrationals, that is numbers involving square roots in the form (a + √b)/c, have periodic continued fractions. This makes them easy to calculate recursively given the period. For example, to calculate √2, we make use of the fact that √2 − 1 = [0; 2, 2, 2, 2, 2, ...], and use the recurrence relation

an + 1 = 1/(2 + an) with a0 = 0

to obtain √2 − 1 to some specific precision specified through n levels of recurrence, and add 1 to the result to obtain √2.

#### Steps for finding continued fractions

Find $\sqrt{r}$ using continued fractions

$\sqrt{r} = N_0 + \frac{1}{N_1 + \frac{1}{N_2 + \frac{1}{N_3+\,\cdots}}}$

Let $\sqrt{r}=\sqrt{N_0^2 + d}= N_0 + \frac{1}{X_0} \qquad where \quad X_0 > 1$

$X_0 = N_1 + \frac{1}{X_1}$
$X_1 = N_2 + \frac{1}{X_2}$
$X_2 = N_3 + \frac{1}{X_3}$

Step 0. We shall assume that r is not a perfect square. In other words:

$\sqrt{r} = \sqrt{N_0^2 + d} \qquad \mbox{ and } \quad N_0 \in \mathbb{Z} \quad \mbox{ and } 0 < d < 1$

Step 1. Find $N_0$ using some other method. The best method is $N_0 = intsqrt(r)$ using some other algorithmn to determine the integer square root.

$N_0 = intsqrt(r)$

Step 2. Find the lower bound (L) and uppper bound (U) for $\sqrt{r}$ where both (L) and (U) are integers.

$L \qquad < \qquad \sqrt{r} \qquad < \qquad U$

Hence

$L \qquad = \qquad N_0$
$U \qquad = \qquad N_0 + 1$

Step 3. Write $X_0$ in terms of $\sqrt{r}$.

$X_0 = \frac{1}{\sqrt{r}-N_0} \qquad from \qquad \sqrt{r} = N_0 + \frac{1}{X_0}$
$X_0 = \frac{1}{\sqrt{r}-N_0} \times \frac{\sqrt{r}+N_0}{\sqrt{r}+N_0} = \frac{\sqrt{r}+N_0}{r-N_0^2} = \frac{\sqrt{r}+N_0}{d}$
$\downarrow$
$\mbox{ substitute } \quad \sqrt{r} \quad \mbox{ with } \quad L \quad \mbox{ or } \quad U$
$\downarrow$
$X_{0 LowerBound} = \frac{L+N_0}{d} = \frac{2 \, N_0}{d} \qquad \mbox{ Calc the numeric value }$

and

$X_{0 UpperBound} = \frac{U+N_0}{d} = \frac{2 \, N_0 + 1}{d} \qquad \mbox{ Calc the numeric value }$

Step 4. Substitute $X_0$ with $N_1 + \frac{1}{X_1}$

$X_{0 LowerBound} \quad < \quad X_0 \quad < \quad X_{0 UpperBound}$

after substitution

$X_{0 LowerBound} \quad < \quad N_1 + \frac{1}{X_1} \quad < \quad X_{0 UpperBound}$

Since $\frac{1}{X_1}$ is less than one, we can determine $N_1$ from the numeric values of $X_{0 LowerBound}$ and $X_{0 UpperBound}$ because $N_1 \quad \in \quad \mathbb{Z}$

Step 5. Once we know the value of $N_1$, we can rework the equation for $X_1$

$X_0 = N_1 + \frac{1}{X_1} \quad \longrightarrow \quad X_1 = \frac{1}{X_0 - N_1}$
$X_1 = \frac{1}{ \frac{ \sqrt{r} + N_0 }{ d } - N_1} = \frac{1}{ \frac{ \sqrt{r} + N_0 + d\,N_1}{ d } } = \frac {d}{\sqrt{r} + N_0 + d\,N_1}$
$X_1 = \frac {d}{\sqrt{r} + (N_0 + d\,N_1)} \times \frac {\sqrt{r} - (N_0 + d\,N_1)}{\sqrt{r} - (N_0 + d\,N_1)} = \frac {d\,(\sqrt{r} - (N_0 + d\,N_1))}{r - (N_0 + d\,N_1)^2}$
$X_1 = \frac {\sqrt{r} - (N_0 + d\,N_1)}{ \frac {r - (N_0 + d\,N_1)^2} {d}} = \frac {\sqrt{r} + M_1 } {D_1}$
$\mbox { where } \quad M_1 = - ( N_0 + d\,N_1 )$

and

$\mbox { where } \quad D_1 = \frac { r - ( N_0 + d\,N_1 )^2 } { d }$

Step 6. Write $X_1$ in terms of $\sqrt{r}$.

$X_1 = \frac {\sqrt{r} + M_1 } {D_1}$
$\downarrow$
$\mbox{ substitute } \quad \sqrt{r} \quad \mbox{ with } \quad L \quad \mbox{ or } \quad U$
$\downarrow$
$X_{1 LowerBound} = \frac{L + M_1}{D_1} \qquad \mbox{ Calc the numeric value }$

and

$X_{1 UpperBound} = \frac{U + M_1}{D_1} \qquad \mbox{ Calc the numeric value }$

Step 7. Substitute $X_1$ with $N_2 + \frac{1}{X_2}$

$X_{1 LowerBound} \quad < \quad X_1 \quad < \quad X_{1 UpperBound}$

after substitution

$X_{1 LowerBound} \quad < \quad N_2 + \frac{1}{X_2} \quad < \quad X_{1 UpperBound}$

Since $\frac{1}{X_2}$ is less than one, we can determine $N_2$ from the numeric values of $X_{1 LowerBound}$ and $X_{1 UpperBound}$ because $N_2 \quad \in \quad \mathbb{Z}$

Step 8. Once we know the value of $N_2$, we can rework the equation for $X_2$

$X_1 = N_2 + \frac{1}{X_2} \quad \longrightarrow \quad X_2 = \frac{1}{X_1 - N_2}$
$X_2 = \frac{1}{ \frac { \sqrt{r} + M_1 } { D_1 } - N_2 } = \frac{1}{ \frac { \sqrt{r} + M_1 - D_1 \, N_2 } { D_1 } } = \frac { D_1 } { \sqrt{r} + M_1 - D_1 \, N_2 }$
$X_2 = \frac { D_1 } { \sqrt{r} + ( M_1 - D_1 \, N_2 ) } \times \frac { \sqrt{r} - ( M_1 - D_1 \, N_2 ) } { \sqrt{r} - ( M_1 - D_1 \, N_2 ) } = \frac { D_1 ( \sqrt{r} - ( M_1 - D_1 \, N_2 ) ) } { r - ( M_1 - D_1 \, N_2 )^2 }$
$X_2 = \frac { \sqrt{r} - ( M_1 - D_1 \, N_2 ) } { \frac { r - ( M_1 - D_1 \, N_2 )^2 } { D_1 } } = \frac { \sqrt{r} + M_2 } { D_2 }$

$\mbox { where } \quad M_2 = - ( M_1 - D_1 \, N_2 )$

and

$\mbox { where } \quad D_2 = \frac { r - ( M_1 - D_1 \, N_2 )^2 } { D_1 }$

Step 9. Repeat step 6 , 7 and 8 .

Not I fancy how the rest of us do it ... ;-) FightingMac (talk) 15:50, 16 April 2011 (UTC)

## Rough estimate?

I'm confused by the "rough estimate" section:

"Take the integer part of the number n. Z = int(n) Count the number of digits in Z. Let D be the number of digits. Calculate the value of 3D. The rough estimate is half the value obtained in step 3. E = (3D) / 2 "

But the examples given, the estimate is way out, e.g. the "estimate" for √723.47 is 13.5, when the actual value is 26.89. Why half the value obtained in Step 3, when the unhalved value (27) is much closer? Same for all the other examples, the unhalved value is much nearer the actual √n

Also the estimates obtained in this section do not match the estimates given as a comparison in the next sectiond etailing the more accurate estimate. There has to be an error here somewhere? 143.252.80.124 09:53, 15 Jun 2005 (UTC)

A rough estimate is just that, a rough estimate. It merely provides a starting value for the seed of other algorithmns to start converging on the actual value of the square root of a number. You are mistaking the rough estimate for a "near value". The rough estimate can be nowhere near the actual value, it just have to get the magnitude of the value correct. Choosing the correct seed value will greatly help reduce the number of iterations for the other algorithmns. Ohanian 07:58, 2005 Jun 16 (UTC)

Also, the question of which is closer - with or without halving - depends on the mantissa of the number in question. It can be shown that for numbers with few digits, halving gives a better result in most cases. However, for larger numbers (6+ digits) the unhalved value quickly becomes more dominant. Considering this and the additional operation (in writing as well as in calculation) required for the halved version, I think I'll edit it to the non-halved version. --Meni Rosenfeld 14:20, 22 December 2005 (UTC)

## Computing Square Roots w/ calculators

I'm not a mathematician, and it would be awfully nice if somebody could insert something here describing this process in language an educated layman can understand - maybe some sort of step-by-step thing? Little facts like this are interesting, after all. ZacharyS 22:38, 31 August 2005 (UTC)

In about para 5, there is the following sentence. Square roots of positive integers are often irrational numbers, i.e., numbers not expressible as a quotient of two integers. Whilst I accept that quotient may have a specific mathematical meaning, I believe that its use here is too technical for this article and that ratio would be more meaningful to a greater % of readers. The reason is that most non-mathematicians consider a quotient to be an integer, created by a division calculation along with a remainder. I propose to change this back to ratio unless good reason not to is posted here within about a week. -- SGBailey 21:44:22, 2005-09-02 (UTC)

## Cleanup?

After skiming this article, I found it to be very comprehensive, but way too messy, long, and clumsy. I suggest cleaning it up by spliting the detailed mathematics to seperate page and maybe only state the result(or roughly state the algorithm) directly in this page. --Lemontea 02:41, 11 September 2005 (UTC)

I agree that this article needs some trimming and tightening up. Some stuff could be indeed moved to a computational article, and some other cut out or shortened. Anybody willing to do the work? Oleg Alexandrov 15:50, 14 September 2005 (UTC)
I've moved the computation section to Methods of computing square roots and have left a single paragraph and a link to the that page. I have attempted to clean up the markup on it a bit. Eric119 01:28, 18 September 2005 (UTC)

## Riemann surface

could someone add a discussion of the square root function on the appropriate Riemann surface? Then I think it would be true that sqrt is a multiplicative homomorphism. --MarSch 15:00, 1 November 2005 (UTC)

## Square roots of squares

I've recently had an argument with someone who claims that $\sqrt{x^2} = x$, even for negative x. Absurd as this may sound, she even said she consulted with mathematicians who supported her claim. This convinced me that a little clarification of the matter is in order, and I have added a note relating to this issue. As this is my first wikipedia edit, I wish to receive some feedback about the relevance and format of this note.

--Meni Rosenfeld 17:57, 22 December 2005 (UTC)

Thank you for your note. Good point. I removed then the absolute value property from a bit above in the text, while putting back your text; so now that property is still stated only once (as I wanted) but in the way which addresses your concerns (as you wanted. :) Oleg Alexandrov (talk) 18:41, 22 December 2005 (UTC)

--Nathan VanFelix 06:15, 03 April 2006 (UTC)

The statement $\sqrt{x^2} = x$, is false; the =x must be changed to =±x because of negatives.

Of course it's false, and this is exactly what I tried to clarify in the edit mentioned here. And $\sqrt{x^2} = \pm x$ is not the best way to write it either - The accurate statement is $\sqrt{x^2} = |x|$ (for reals). -- Meni Rosenfeld (talk) 18:36, 4 April 2006 (UTC)

More accurately, $|\sqrt{x^2}| = x$, I think. D. F. Schmidt (talk) 01:46, 17 August 2009 (UTC)
No, that's wrong. -- Meni Rosenfeld (talk) 19:28, 5 March 2011 (UTC)
Meni Rosenfeld is correct. $\sqrt{x^2}$ is correctly pronounced as "the principal square root of x2" — that is, in the case of reals the positive one, which is |x|. Duoduoduo (talk) 19:48, 5 March 2011 (UTC)

## The exponential identity

The main article gives a paragraph about the exponential identity for computing square roots. Is this really necessary when there's an entire comprehensive article for methods of computation (With this one appearing at the top)?

I have half a mind to delete it. Are there any objections?

--Meni Rosenfeld 14:36, 23 December 2005 (UTC)

I thinks it's good to have a short piece of information on computation in the article with a link to the more comprehensive one. The paragraph gives the most important (IMO) information about computing square roots; those who want more can consult the longer form. Eric119 18:18, 23 December 2005 (UTC)
Agree with Eric. Oleg Alexandrov (talk) 21:52, 23 December 2005 (UTC)

## List in math tags

Why should the list of square roots be in latex? The images are very wide on a small screen, and one can't copy/paste the numbers (especially if one is reading a saved copy). (If the reason is the radical signs, and that's considered important, a possibility is making the "sqrt(2)" part in tex and the rest in plain text.) Frencheigh 14:14, 8 March 2006 (UTC)

About why latex, I think you'll agree that the text version looked awful. About separating the radical part from the digits, I agree and have changed accordingly. -- Meni Rosenfeld (talk) 15:44, 8 March 2006 (UTC)

## Question about sign of root

Why is the square root defined to be only the positive root? It would make much more sense to me if a square root has two roots. In this way the definition can be simplified, and anything involving a square root can forgo the plus/minus notation. The only thing that would need to be noted is a reminder that the square root is an operator that in most cases has multiple solutions. Why is the extra complexity added? Fresheneesz 02:08, 19 March 2006 (UTC)

Actually, that would make things much more complicated. In fact, you wouldn't be even able to write $\sqrt{4}$ because every mathematical expression is expected to have a well-defined value. Would you say $\sqrt{4}=2$ or $\sqrt{4}=-2$? It can't be both. How about $\sqrt{4}=\pm2$? But $\pm2$ is not a number, so you'll need to make additional definitions which complicate everything by far. The words "the square root of", the √ sign, and most of this article, refer to the square root function, and these, by definition, have a unique value. That's the only way for any of these to be useful. Of course, you can always say that -2 is a square root of 4 - but $\sqrt{4}$ will always mean +2. -- Meni Rosenfeld (talk) 14:31, 19 March 2006 (UTC)
"you wouldn't be even able to write $\sqrt{4}$"
Why not? Should I, then, not be able to write {1,2,45} union {3} ? There is no unique value to that expression, yet it is 100% mathematically sound. The solutions to a square root can be thought of as a set of two elements. I don't see the complication in doing that.
And whats so inherintely useful about defining the square root as a function? Like, an equation can have multiple solutions - infintely many in fact. I don't see the complication in limiting the square root to being a function. Fresheneesz 19:24, 23 March 2006 (UTC)
Okay, you've chosen $\sqrt{x}$ to mean "the set of all square roots of x". Of course, there's no problem with that, but I fail to see how is this more useful than the common definition. Functions are useful. They are used everywhere in mathematics. You can use them to write expressions. With your suggestion, you can no longer write that the distance between points (x1,y1) and (x2,y2) is $\sqrt{(x1-x2)^2+(y1-y2)^2}$, since distance is positive - or any other similar formula. You wouldn't be able to talk about functions like $\sqrt{x^2+1}$ and calculate their derivatives, etc. You wouldn't be able to write that the length of the curve of the function f(x) is
$\int_a^b \sqrt{1+f'(x)^2}$
In short, most of mathematics breaks down, and we have gained absolutely nothing - we could always just write {x|x^2=2} or $\pm\sqrt{2}$ in those cases when we are interested in both roots. -- Meni Rosenfeld (talk) 19:45, 23 March 2006 (UTC)
Well, first of all, distance is not neccessarily positive: in almost every mathematical case it is much more useful to allow distances to have a sign, so that one direction is positive and the other is negative. For example, the two points (x1,y1) and (x2,y2) that you mentioned could have different orientations that the distance's could be associated with. The sole reason why there are two roots per square, is that in squareing a number, information is lost. In the case of the distance formula you wrote, two peices of information are lost, the sign of the x component and the sign of the y component. The distance in that case is a vector, but when you get absolute distance, the direction is lost. In any case, if my suggestion was carried out, a simple absolute value sign would do the trick to get absolute distance (sorta makes sense right?). Simply adding in aboslute value signs can solve any of those problems you gave me, but allowing a root to have two solutions allows for simpler writing in cases where you want both roots to be there - like almost all cases i'm familiar with. The quadratic formula wouldn't need that +/- sign, for example. My main point is that it would be easier to write an abolute value sign in the cases where its needed, rather than write {x|x^2=2} every time *thats* needed. The most important thing is learning about roots though. People are taught that the square root of x is a number whos square is x. But then the "real" definition turns this around and says its only the positive ones of those. Doesn't it make more sense if one of the solutions to $sqrt{x^2} = y\!$ is y=x, regardless of x's sign? 68.6.112.70 04:57, 24 March 2006 (UTC)
My main problem is that a definition like yours will have to deal with objects that are not numbers. Something like √4 has to be some definite object, say the set {-2, 2}. So you can define arithmetic operations as element wise operations, for example 1 + √4 = 1 + {-2, 2} = {1 + -2, 1 + 2} = {-1, 3}. Likewise, you'll get:
$\sqrt{4}^2 = \{-2, 2\}^2 = \{(-2)^2, 2^2\} = \{4, 4\} = \{4\}$
But I would want something like √4^2 to be the number 4, not the set {4}. This mixture of numbers and sets doesn't make me very comfortable. Of course, with suitable definitions you can make it work - but I personally don't believe this is worthwhile. The fact that most books (that I know of) use the √4 = +2 idea, suggests that there are many who share my opinion.
In any case, your pedagogical comment reminds me of the fact that people are "taught" (incorrectly) that a prime number is one that is only divisble by 1 and itself, and are then surprised why 1 isn't prime - as it seems to fit this definition. Teachers should teach the correct definition, and if they don't, it's their problem. Likewise, teachers should emphasize the difference between "a" square root and "the" square root. Besides, people should be taught to understand that a mathematical expression has a definite interpretation. This interpretation can be a set or any other of your favorite objects, but when people learn about square roots they usually don't know anything about sets, do they? -- Meni Rosenfeld (talk) 10:08, 24 March 2006 (UTC)
I see what you're saying, but you don't have to think about sets to do it. For example, kids use the set of natural numbers all the time, but the don't know about sets. I think it would be quite reasonable to teach that it has two solutions - it would be a good intro to that, since many equations end up having multiple solutions. What about x^(1/2)? Is that considered a square root? The whole issue of there being "two square roots" but only one solution to the "square root function" seems arbitrary. It just seems like a function isn't appropriate for the square root. Fresheneesz 19:00, 24 March 2006 (UTC)
I'm fairly opposed, but I guess at this point it amounts to personal preference - Which we will definitely not be able to convince each other. About "arbitary", that's not necessarily bad - Imagine a horse standing equidistantly between 2 stacks of hay. If it arbitarily chooses to approach one of them, it will have a meal. If it tries to figure out which stack would be more logical to approach, or refuse to make any choice because each choice would be arbitary, it will starve to death. In my opinion, refusing to adopt √x = the positive square root, just because it's arbitary, would make us starve. -- Meni Rosenfeld (talk) 19:24, 24 March 2006 (UTC)
Well, about the horse thing, its more like the horse is standing between two stacks of hay, and instead decides to push them together so it can eat em both. Personally, I use the radix in just such a way. I've been pretty fine with it. But I suppose the main reason i'm discussing it here is that it would be something good to add to this page - the "reason" positive root was chosen. Since to many people it does seem counterintuitive (and for those that have been taught incorrectly). Maybe a note on the usefulness of a single-output function would be good (i'm pretty fuzzy on that myself). Btw, is x^(1/2) considered the same way that √x is, do you know? Fresheneesz 19:40, 25 March 2006 (UTC)
About x^(1/2), I suppose so, but I thınk ıt depends on the context. In real numbers, I thınk ıt almost always means the posıtıve root. For complexes. ın general a^b = exp (b ln (a)), where ln ıs a gıven branch of the natural logarıthm - whıch unless mentıoned otherwıse, ıs usually taken to mean the prıncıpal branch - gıvıng, agaın, the posıtıve square root for a real number. -- Meni Rosenfeld (talk) 15:09, 26 March 2006 (UTC)
I agree on non-integer exponents: they are only defined for positive radicants (else you get once again errors like (-1)^(1/2) = (-1)^(2/4) = ((-1)^2)^(1/4) = 1^(1/4) = {1,i,-1,i} but their square is not equal to -1 for all of these.
I don't agree on
$\sqrt{4}^2 = \{-2, 2\}^2 = \{(-2)^2, 2^2\} = \{4, 4\} = \{4\}$
because for me, A² = A A = { xy ; x ∈ A, y ∈ A }, and then, { -2, 2 }² = { 4, -4 }.
Concerning the subject, the symbol of the square root is also used for sets, more precisely for the radical of ideals (then not only the result, but also the argument are sets). Also, one can develop a calculus for "ambiguous" functions as set-valued functions, but of course the result is then no more a real number. This is also used for a "fuzzy" calculus where numbers are replaced by intervals, e.g. [1.9,2.1]+[3.4,3.6]=[5.3,5.7], which allows to deal with "uncertain quantities" (measurements with errors...), but in any case one must careful redefine all operations and relations ((in)equalities...) known for the real numbers, and investigate about the properties they have, before using them. — MFH:Talk 19:31, 2 June 2006 (UTC)

## Complex square root

Is the formula for complex square root correct?

From the article:

$\sqrt{x+iy} = \sqrt{\frac{\left|x+iy\right| + x}{2}} \pm i \sqrt{\frac{\left|x+iy\right| - x}{2}}$

I think it should be:

$\sqrt{x+iy} = \pm\left (\sqrt{\frac{\left|x+iy\right| + x}{2}} + i\ \sgn y \sqrt{\frac{\left|x+iy\right| - x}{2}}\right )$

Can anyone verify this?

Pengwy 02:15, 1 May 2006 (UTC)

You're right that the first formula here is wrong; And the formula you give is almost right: Since sgn(0) is usually defined as 0, this formula fails for y = 0. However, if you read the line after the formula in the article, you'll see that the article only gives one root, with a formula identical to the one you give. So this isn't a question of correctness, more of clarity. Besides, the formula should give only one root - the same way $\sqrt{4} = +2$ and not $\sqrt{4} = \pm2$. But I agree the formula should be self-contained and not depend on external explanations. Any ideas how to write a formula which holds for all cases (including y = 0)? -- Meni Rosenfeld (talk) 08:12, 1 May 2006 (UTC)
An alternative to the formula suggested above is (if the square root is interpreted as a multivalued function)
$\sqrt{x+iy} = \pm\frac{|x+iy|+x+iy}{\sqrt{2|x+iy|+2x}}$
which degenerates to the indeterminate form 0/0 when x+iy is on the negative real axis. (I'm not too optimistic that there is a simple formula for the square root that also includes the negative real axis.) — Revond 18:09, 25 November 2006 (UTC)
It's intresting problem. Can you write more about formula for the square root that also includes the negative real axis ? Maybe new article or wikibooks ? --Adam majewski (talk) 13:09, 22 November 2008 (UTC)

## Too technical?

Is it just me or is this article way too technical? Say for someone who left school years ago and now needs to mug up on some basic maths? I suggest adding some basic examples at the beginning, including ones that show two negative numbers multiplying to a positive (will ring bells for some readers). Most people, if they understand that bit, will be satisfied and leave it at that. Roots of negative numbers would be dealt with next.Itsmejudith 22:54, 18 July 2006 (UTC)

It seems fine to me, but I may be sort of biased. Let's break it down.
In mathematics, the principal square root of a non-negative real number $x$ is denoted $\sqrt x$ and represents the non-negative real number whose square (the result of multiplying the number by itself) is $x.$
That sentence seems fine to satisfy mathematicians and non-mathematicians alike. The meaning of non-negative is obvious, real number is wikilinked if it is not understood, and it defines square.
For example, $\sqrt 9 = 3$ since $3^2 = 3\times3 = 9.$
Then it gives an example which helps solve any confusion.
Are there any parts in particular you are thinking about? — 14:54, 19 July 2006 (UTC)
Thanks for your reply. The first sentence is obviously crucial. But I suspect that if it were written to be comprehensible to non-mathematicians then it would not satisfy the mathematicians. Still, I will run it past some non-mathematicians to see what sense they can make of it. I don't know what second sentence is meant to add, even though I know what both square roots and quadratic equations are. There is one example, which is good. It might be even better to include a second example, like 10 is the square root of 100, and to write these both out in words as well as figures. The article on square should be linked at the beginning. (Square is a much simpler article.) Other things currently in the lead, such as the origin of the symbol, irrational numbers, could go in later sections. The section on "computation" could start with how to compute a square root on a calculator and then mention trial-and-error methods before going onto the algorithm using e. The references should go into a references section. That's a lot of criticisms, sorry, but that's just how it all seems to me. Itsmejudith 15:36, 19 July 2006 (UTC)
Don't be sorry, the criticisms are great! I'm putting them in a todo list at the top of this page. — 15:49, 19 July 2006 (UTC)
I edited the beginning and calculation section now. What do you think? I didn't really remove anything, I just clarified it a bit. — 16:11, 19 July 2006 (UTC)
A great improvement! Now I've instigated the to-do list, I'll try and help out with bits of it in the next few days.Itsmejudith 17:13, 19 July 2006 (UTC)

## Babylonian method

In step 2 of the algorithm, the variable r is used without being defined. Perhaps the words "To calculate the square root of a nonnegative number r" (or some such) could be inserted in the introductory phrase (before the colon).--Emoll 12:50, 20 July 2006 (UTC)

Ah, yes, I've fixed that now. — 13:02, 20 July 2006 (UTC)

## Non-mathematician test

So far I've had one response. And he liked the article and could understand nearly all of it. He suggested that the proof of root 2 being irrational could go in the article because it is easy to follow and neat. I like that idea too. It is in "The Number Devil", a maths book for children, by Hans-Magnus Enzensberger. Itsmejudith 20:15, 24 July 2006 (UTC)

I'd be glad to put it in (or you can), but where? And which proof? Can you outline the one you had in mind so I can figure out which one you're talking about? Thanks. — 21:17, 24 July 2006 (UTC)
P.S. I'm glad that the other non-mathematician liked the article and understood basically all of it. :-) — 21:17, 24 July 2006 (UTC)
I'll find the one in Enzensberger's book and post it here on the talk page first. I remember it being very straightforward but don't remember what it was! I hope it can be taken from the book and cited to Enzensberger without copyvio. Itsmejudith 05:39, 25 July 2006 (UTC)
It is straightforward. I don't think there's a need to cite it from anywhere. The outline is: Assume √2 = p / q in lowest terms. Then p2 = 2q2. So p2 is even, and therefore so is p: p = 2m. So 4m2 = 2q2, or 2m2 = q2. Therefore q is even, and 2 is a common factor of p and q - a contradiction. -- Meni Rosenfeld (talk) 08:03, 25 July 2006 (UTC)
Yeah, that's a basic one, but it works. One of you can add it or I'll add it when I have time. — 11:45, 25 July 2006 (UTC)
Actually, that's already in the square root of 2 article. I wikilinked to it in the intro. — 15:08, 10 August 2006 (UTC)

## Introductory Sentence(s)

For clarity, particularly for non-mathematicians, I suggest starting by defining a square root in general before mentioning the principal square root. For example, it could say something like "In mathematics, a number y is a square root of another number x if y multiplied by itself, or squared, equals x. When x is a non-negative real number, a square root that is non-negative is said to be the principal square root of x and is designated as ..."--LoboSooner 04:11, 19 August 2006 (UTC)

I agree. I've made some modifications to the introduction. -- Meni Rosenfeld (talk) 11:41, 19 August 2006 (UTC)

## Position of the graph of the function

I was wondering if the graph of the function could be brought further up, to just under the first mention of the function. On the grounds that a graph is for many people the best way to understand a function. I haven't enough experience of handling the mathematical mark-up to attempt the move myself.Itsmejudith 22:42, 22 December 2006 (UTC)

## Surds

Is the single sentence in the second paragraph of the article actually making a point about surds rather than about square roots as such? If not, then isn't the statement a tautology? I'm bearing in mind that "x squared = 4" is a quadratic equation and of course the answer is "plus or minus 2". And while "plus or minus 2" may be "a square root" (or two square roots!), that is only in the sense that every number is a square root of something. What I suspect the sentence is really getting at is that often the result of a quadratic equation is an irrational square root, in which case the choice arises either to leave it in the answer as a surd or to approximate it as a decimal. In fact the word "surd" doesn't currently appear in this article. I wondered if any other editors had a view on the explanatory purpose of the sentence.Itsmejudith 22:55, 22 December 2006 (UTC)

What does "surd" mean? It is certainly a very obscure word (if a word at all) which is one reason not to use it. JRSpriggs 07:18, 23 December 2006 (UTC)
A number is written in surd form if it is written with a radical symbol, as for example $\sqrt{4}$ or $13+\sqrt[7]{2}$. The term surd is usually reserved for a number that must be written in surd form, for example $\sqrt{2}+\sqrt{3}$. Such a number is necessarily irrational. The article nth root uses the term surd to refer to any unresolved radical whether or not it can be simplified to avoid use of the radical sign. Michael Slone (talk) 14:32, 23 December 2006 (UTC)
To Michael: Thanks for the explanation. To judith: I think the article is clear enough. Solving a quadratic equation numerically often requires extracting a square root as in Methods of computing square roots. JRSpriggs 11:28, 24 December 2006 (UTC)

## Two little questions

These relate to the start of the "properties of the square root function" section. 1) Does the second property, that the function always returns a unique value, add anything that is not already implicit in the term "function"? 2) Does the use of set notation to define the set of non-negative real numbers add anything? I am very confused by the union with the set {0}, since (in my very limited understanding) if 0 is included then the set does not map onto itself. I can just about understand that the function maps the set of positive real numbers onto itself. Itsmejudith 14:44, 13 May 2007 (UTC)

(1) It is necessary to emphasize that the principal square root is unique because people have heard that each number has two square roots. (2) The principal square root of zero is zero. The principal square root of a positive real number is a positive real number. Furthermore, the inverse of this is the square function which takes zero to zero and positives to positives. So the principal square root function is indeed onto the non-negative reals when the domain is restricted to the non-negative reals. JRSpriggs 07:42, 14 May 2007 (UTC)
(1) Clarifying: Every number (except 0) does have two square roots, but only one of them is the principal square root which is denoted $\sqrt{x}$. Anyway, I disagree that there should be a separate property which states the result is unique. It should either be removed or merged into the property that it is a function.
(2) Perhaps, instead of $\mathbb{R}^+ \cup \{0\}$ we should write $[0, \infty)$? -- Meni Rosenfeld (talk) 16:26, 14 May 2007 (UTC)
Thanks for bearing with my ignorance. My motivation of course is that if I can't understand what the article means then those even less numerate than me won't be able to either. Itsmejudith 21:49, 14 May 2007 (UTC)

I would actually like to know what the term in category theory(?) for describing how the square root function $f(x)$ has multiple $y$ values for each $x$. Seems like a bit of an omission. --65.183.152.86 (talk) 05:17, 20 August 2009 (UTC)

## Electric current?

Mets501: I suggest you read again Nonagonal Spider's addition. It is legitimate. I have not restored it, since I have no opinion on whether it actually improves the clause. -- Meni Rosenfeld (talk) 23:12, 26 May 2007 (UTC)

## Decimals?

Can Square Roots be decimals? Wilsonbiggs 15:16, 1 July 2007 (UTC)

I'm not sure I understand what you mean. The square root of a positive real number is a real number, so it always has a decimal expansion. -- Meni Rosenfeld (talk) 16:30, 1 July 2007 (UTC)
The user may have meant: can a square root be a number less than one. The answer to that is yes, for example the square root of 0.25 is 0.5. As the article points out, square roots are often irrational numbers - they can be written as decimals but the digits to the right of the decimal point go on forever without any repeating pattern. Itsmejudith 17:30, 1 July 2007 (UTC)

## History of Nomenclature "Square Root"

The square root operation is disjoin from the notion of a root, and the "square" of a variable has taken on a meaning altogether different from the geometrical square. Perhaps someone has knowledge of the history of these words as they appeared in history? This would be a lovely thing to add to the history section. Jameson 16:54, 26 July 2007 (UTC)

## History of square root

I read many articles about historical origin of square root and I did find something interesting. The owner of the article says that an ancient mathematician wrote this: "The number is like a plant, it grows from a root" and he put an example "Number 16 has grown from a root 4". I think this one of the best explanation I found. — Preceding unsigned comment added by Qusayqusay (talkcontribs) 16:47, 2 February 2013 (UTC)

Added information on discovery of non-existence of square roots of negative numbers in the history section- "Mahāvīra, a 9th century Indian mathematician, was the first to tell that square roots of negative numbers don't exist." isoham (talk) 12:52, 16 September 2013 (UTC)

Added a different reference. Apart from the one cited there exist more like this and this isoham (talk) 13:43, 19 September 2013 (UTC)

## Incorrect requirement for exponential identity

The exponential identity section incorrectly states that $\sqrt{wz}$ is equal to $\sqrt{w}$$\sqrt{z}$ only for the special case of $w>0, z>0$

and leads to an incorrect conclusion to the equation $-1=i*i=$$\sqrt{-1}$$*\sqrt{-1}=\sqrt{-1*-1}=\sqrt{1}=1$

In fact, the problem is elsewhere in the proof. $\sqrt(-1)$ has a principle square root of $i$, but its roots are $+/-i$. Therefore, both $\sqrt{wz}$ and $\sqrt{w}$$\sqrt{z}$ have the same set of answers, $+/-\sqrt{wz}$. The error in the proof above is the assumption that $1=\sqrt{1}$, when in fact $+/-1 = \sqrt{1}$.

Dr gnow 17:11, 21 August 2007 (UTC)

In this article (and I believe in most mathematical texts) √ represents one particular branch of the square root function, one for which, among other things, $\sqrt{1}=1$ and $\sqrt{-1}=i$. The expression $\sqrt{1} = \pm1$ is not really meaningful. -- Meni Rosenfeld (talk) 12:39, 23 August 2007 (UTC)
Bummer, Meni! The text of the article seems to have been changed to support the idea that +/- 1 is a valid solution to this particular text. Don't worry, I'll make sure your mistake is appropriately documented on your discussion page.

Dr gnow 05:29, 3 September 2007 (UTC)

## "Objects other than numbers"

Can an example or two be added to clarify this statement for those of us who are non-mathematicians. I think "object" probably doesn't mean "thermos flask" or "bookcase" but perhaps "matrix" or "vector". Would it be correct to say "mathematical objects"? Itsmejudith (talk) 12:26, 7 July 2008 (UTC)

good point - done.— MFH:Talk 13:06, 7 July 2008 (UTC)
great, thanks. Itsmejudith (talk) 13:24, 7 July 2008 (UTC)

## "This link to new high-order methods for computing square roots should be added in the external links"

http://mipagina.cantv.net/arithmetic/rmdef.htm Such webpage contains a section with links to books, papers, workshops, math-websites, etc. relating such methods. If any doubts I can bring you detailed info on that. These are true high-order arithmetical methods so it is a futil attempt to try to ignore them. Many thanks. Arithmonic (talk) 19:14, 6 November 2008 (UTC)

## Computing sqrt of complex number in c

Hi, I have to compute sqrt of complex number. Here is code in c ( not mine). I do not understand what it does. It seems to use different formula the in this article. Is it an error or some advanced trick ?

complex cx_sqrt(z)
complex z;
{
complex w;
double fabs(), sqrt();
/* Worry about numerical stability */
if (z.x == 0.0 && z.y == 0.0) return(z);
else
if (z.x > fabs(z.y))
{
w.x = sqrt((z.x+sqrt(z.x*z.x+z.y*z.y))/2);
w.y = z.y/(2*w.x);
}
else
{
w.y = sqrt((-z.x+sqrt(z.x*z.x+z.y*z.y))/2);
w.x = z.y/(2*w.y);
}
return(w);
}


--Adam majewski (talk) 13:14, 22 November 2008 (UTC)

Why don't you just use csqrt? Dmcq (talk) 13:52, 22 November 2008 (UTC)
By the way it is a proper complex square root but its branch cut is different from normally defined, I'm not sure where it has wandered to or even if it is connected - an exercise for the reader :) The return y should be negative if the input y is. It looks like they probably stuck in a bit to cope with some problem with accuracy and changed the meaning. This is why people should normally use a library function rather than roll their own. Dmcq (talk) 14:29, 22 November 2008 (UTC)
Thx for answer. So for csqrt and formula in article

$z = \sqrt{x+iy} = \sqrt{\frac{r + x}{2}} + i \frac{y}{\sqrt{2 (r + x)}}$

• branch cut is along the negative real axis
• principal value is :
• in the range of the right half-plane (including the imaginary axis)
• −π < arg(z) ≤ π

( maybe I'm wrong, but it is hard to find it in article ) and for cx_sqrt :

• branch cut is (?)
• principal value is the same as above ?

Am I right ? --Adam majewski (talk) 15:23, 22 November 2008 (UTC)

What you say about the article is right. And as for cx_sqrt the value is the negative of the principal value when y < 0 and x <= -y. In fact its branch cut wanders off along the line x>=0 y=-x. It is easy to fix but I don't see why you should try, just call csqrt. Dmcq (talk) 15:40, 22 November 2008 (UTC)

## The section "As expansions in other numeral systems"

Copied here from my talk page. Paul August 19:10, 26 May 2009 (UTC)

I don't understand why you've twice reverted the article square root to make it contain "Even in non-standard positional numeral systems, the only notable exception is golden ratio base." What do you think that means, and why is it preferable to attaching the dependent clause "Even in non-standard positional numeral systems" to the clause it actually refers to, which is in the previous sentence? 128.100.5.116 (talk) 15:13, 26 May 2009 (UTC)

Hello 128.100.5.116. Sorry I had to revert your edits, and thanks for discussing this rather than just continuing to insist on your edit, that's the best way to handle editorial disagreements. Let me try to explain why I think your edits are problematic.
The square roots of the perfect squares (1, 4, 9, 16, etc.) are integers. In all other cases, the square roots are irrational numbers, and therefore their representations in any standard positional notation system are non-repeating. Even in non-standard positional numeral systems, the only notable exception is golden ratio base.
You changed the second sentence above so that the frist two sentences read:
The square roots of the perfect squares (1, 4, 9, 16, etc.) are integers. In all other cases, the square roots are irrational numbers, and therefore their representations in any standard positional notation system are non-repeating, even in non-standard positional numeral systems.
This reads as asserting that in all numbering systems, both standard and non-standard, every irrational number has a non-repeating representation, which is simply false. The article in fact provides a counterexample, the Golden ratio base.
Even in non-standard positional numeral systems, the only notable exception is golden ratio base.
I concede this sentence is a bit problematic. What it is trying to say is that, not only is it the case that irrational numbers have non-repeating representations in standard positional notation systems, they also have non-repeating representations in most non-standard systems as well, the Golden ratio base, being the only notable exception.
Does this make sense? In any case, I'm happy to discuss this more, but the best place for such discussions is on the talk page of the article in question here, that way other interested editors of that article can participate as well. If you have more to say on the matter, we can copy the above discussion to the talk page and continue it there.
Regards,
Paul August 16:32, 26 May 2009 (UTC)
Well, first of all, I don't see this paragraph as talking about any irrational numbers other than the square roots of non-square integers, so I don't know how you get from either wording of it to a statement that all irrational numbers have nonterminating representations in all bases. The more limited statement that all irrational square roots of integers have nonterminating representations is also as you say "problematic" (indeed, false) - the golden ratio base isn't the only exception, you can easily construct, say, a number system with base root-two in which root-two will terminate. I don't know why golden-ratio is a more notable nonstandard exceptional base than root-two; I'm pretty sure I've read about both in the recreational math literature, though I don't have references handy. Anyway, my only real interest was in the grammar. I think the wording I proposed is no less factually correct than the wording you reverted it to, and has the advantage of being closer to English. But I'm not willing to spend time writing new material to correct the factual problems and make the article really cover the material perfectly; I meant to make a simple punctuation fix and was quite surprised anyone would bother to disagree. I'll leave it alone at this point. 128.100.5.116 (talk) 17:30, 26 May 2009 (UTC)

End of copied text.

Well, first of all, I don't see this paragraph as talking about any irrational numbers other than the square roots of non-square integers, so I don't know how you get from either wording of it to a statement that all irrational numbers have nonterminating representations in all bases.

I get there by interpreting the "therefore" in that sentence as saying that 1) The square roots of non-perfect squares are irrational, 2) all irrationals have non-repeating representations in standard positional notation systems, 3) therefore the square roots of non-perfect squares have non-repeating representations in standard positional notation systems.

In any case, you are correct that any irrational base works as an exception, and I now think that it would be best simply to remove the "Even in ..." sentence altogether, so I will.

Paul August 19:10, 26 May 2009 (UTC)

## Square root of zero

It probably seems trivial to most people that sqrt(0) = 0. But earlier today, the way I was thinking of it was x = x^2, which makes this impossible. (As a joke, I posed the question to my brother-in-law who is not exactly a math student. He replied, "I think the square root of 0 equals chuck norris.") I mean, 0 = 0^2, or so it seems. After all, 0x = 0, even if x itself is 0. But if you take x = x^2 and divide each side by x, you get 1 = x, not 0,1 = x. So that leads me to believe that either there is more than one zero or algebra is not only insufficient but defunct in the arbitrary case.

Could someone possibly shed some light on this point? D. F. Schmidt (talk) 01:29, 17 August 2009 (UTC)

The division doesn't work for x = 0. See Division by zero#Fallacies based on division by zero Dmcq (talk) 06:53, 17 August 2009 (UTC)
Division by zero does work for Chuck Norris. DVdm (talk) 08:38, 17 August 2009 (UTC)
Dmcq, can you clarify how that list of fallacies applies to the topic of sqrt(0)? I looked at the article, but somehow it doesn't ring any bells with respect to this particular topic. If 0 = x = x^2, then 0 = 0 = 0^2, and 0/0 = 0^2/0. But I don't think that's even what I'm talking about. D. F. Schmidt (talk) 03:29, 18 August 2009 (UTC)
As it say "It is possible to disguise a special case of division by zero in an algebraic argument". Just sticking in x doesn't mean that dividing by x isn't dividing by zero. The correct conclusion from x = x^2 is that x is either 0 or 1. Dmcq (talk) 08:40, 18 August 2009 (UTC)
True enough, I suppose. But it remains true that if x = 1, then x = 1. If you multiply each side by 0, then sure you get 0 = 0. But that doesn't mean x is anything but 1.
And what of the problem of logarithms? ln(x)/2 = ln(sqrt(x)), right? I suppose it is true that ln(0)/2 = ln(sqrt(0)), but neither is a number, right? or is it a complex number? In the same way that infinity/infinity is undefined, it seems that zero is similarly ill defined. Imagine taking a mole of a metal, say iron, and removing one atom (or even thousands) at a time. Eventually you would in fact wind up with none left. But in the meantime, realistically, the delta mass would be negligible, basically 0. 10^3 is 1000, hardly 0, but that's still nothing compared to 6.23*10^23 (atoms per mole).
To me, mathematical zero basically looks like that to me. It's not precisely nothing, but comparatively. Just as it's common to say "5*0 = 0", I figure the expression zero is used in higher math than algebra, zero is equivalent to 10^3/10^23. And ln(10^3/10^23) = -20*ln(10) = approx. ln(0). But sqrt(10^-20) = 10^-10 != 10^-20.
Is my rationale rational? D. F. Schmidt (talk) 02:35, 19 August 2009 (UTC)

Is all this leading in the direction of a proposal of some modification or addition to the article? If not, please note that the talk page has a rather specific and well documented purpose. Thanks - DVdm (talk) 07:26, 19 August 2009 (UTC)

The queries seem to be about Zero and possibly Infinitesimal. Queries in general on maths should go to the maths reference desk Dmcq (talk) 08:29, 19 August 2009 (UTC)
Fair question. It's simply that I had the problem in mind that I had mentioned before, and hoped that Wikipedia could shed some light on the subject. I had hoped the subject of sqrt(0) was prominent enough to be treated in the article. (I don't remember seeing a treatment of it under 0 (number). I will try asking at the reference desk. Thanks. D. F. Schmidt (talk) 14:30, 19 August 2009 (UTC)
The first line of the article says: "... a square root of a number x is a number r such that r^2 = x...". So 0 is a square root of 0 since 0^2 = 0. That's all there is to it.
The article's definition does not say that x^2 = x implies x = 0. We know that x^2 = x implies that x = 0 or x = 1, which means by definition that 1 is a square root of 1, and 0 is a square root of 0. Nothing more, nothing less.
I don't think you will find more at the reference desk. Cheers - DVdm (talk) 15:41, 19 August 2009 (UTC)

Ok. As I was posing the question to the math reference desk, I remembered algebra 2 from too many years ago. Well duh, basically you wind up with $x (x - 1) = 0$. Divide by x you get x = 1. Divide by x - 1 and you get x = 0. lol. That sort of thing is the variety of answer I had been looking for. An algebra question solved by an algebra solution. I had of course been thinking of only one half of the process.

I don't suppose this warrants putting it in the text of the article. I'm not sure how such a trivial problem and trivial solution would be introduced, but it might be worth noting since 0 is in too many cases a singular case, and since it borders the negative numbers whose square roots are nonreal. D. F. Schmidt (talk) 05:00, 20 August 2009 (UTC)

## Page seems complex

Any person that can read the opening sentence ""In mathematics, a square root of a number x is a number r such that r2 = x, or, in other words, a number r whose square (the result of multiplying the number by itself) is x."" already knows what a square root is. Any person who doesn't, cannot comprehend this sentence. I was curious about this because a young man I know told me he read this page and was lost.

Are we helping or telling everyone how smart we are. —Preceding unsigned comment added by Mark4131 (talkcontribs) 01:49, 29 August 2009 (UTC)

Do you have any concrete suggestions, or even specific criticism of what is wrong? Abuse isn't particularly helpful. Shreevatsa (talk) 04:16, 29 August 2009 (UTC)
In my argument above, I explained my problem and solution for the square root of zero. Yours might be a similar problem, Mark4131. $x^2=x$, which yields $x^2-x=0$, which is factored down to $x(x-1)=0$ which means that x = 0, 1. So the square root of 0 is 0, and the square root of 1 is 1, from this path of math. (Philosophically, I dissent; $\sqrt 0$ and therefore $0/\sqrt 0$, and even 0 should be undefined, I reckon--after all, it is ill defined in so many ways already.)
So basically, my thought is that the definition so given (if it is agreed upon by proper math experts) should remain. Otherwise, it should be redefined (again, if it is so agreed upon) as $r:r=x/r$.
I do agree that it does sound a tad pretentious to give that definition. But so many mathematical and scientific definitions are worded similarly. And as Shreevatsa points out, is there an alternative possibility? I can't think of one myself. It's a matter of how the math community defines the square root. D. F. Schmidt (talk) 18:56, 29 August 2009 (UTC)

I find the 2nd and 3rd paragraph contradicting each other. Every non-negative real number x has a unique non-negative square root,... Every positive number x has two square roots.... So is it one or two?--Gciriani (talk) 21:22, 20 January 2010 (UTC)

They are both correct. One of the roots of a positive number is negative. Dmcq (talk) 21:28, 20 January 2010 (UTC)
I find the two statements combined at least ambiguous. I'm not disputing the fact that one root is negative, but the confusing explanation: one statement says "one", the other says "two". The explanation ought to be clear and not confusing.--Gciriani (talk) 17:09, 21 January 2010 (UTC)
Actually it is not true that one statement says "one", the other says "two". One statement says "one non-negative root", the other says "two square roots". DVdm (talk) 18:29, 21 January 2010 (UTC)

The second statement is incorrect and very misleading. Every positive number has exactly one (not two) square root by the definition of roots in your cited book of Algebra, http://books.google.com/books?id=Z9z7iliyFD0C&pg=PA120#v=onepage&q&f=false. This legitimate, mathematical source defines the root where the "nth root of a nonnegative number a is a nonnegative number x such that x^n = a." So the singular implies only one root and the nonnegative tells us that we cannot consider negative numbers as roots. Even though it is true that a nonnegative number could be a *solution* to the equation x^n = a, it is not in face a root. This is further emphasized by the question, "What happens if there are many numbers x having this property?" The answer, this cannot happen, tells us that every nonnegative number has one and only one root. I have an M.S. in mathematics, and I teach statistics at the university level. — Preceding unsigned comment added by Ellensmyth (talkcontribs) 23:22, 12 June 2011 (UTC) To comment on this topic, though, I would (and have been) strongly argue that the square root of 4 is just 2. If the square root of 4 is both 2 and -2, then every calculator in the world is wrong, as well as every mathematical program like Mathematica, Maple, and Wolfram Alpha. Whenever we do sqrt(4), where sqrt() is that radical symbol, we read that is the square root of 4. And the answer to that is always 2. My argument above and link to the mathematical definition of root also tell why the square root of a nonnegative number (e.g., 4) has only one, nonnegative root, which is 2. — Preceding unsigned comment added by Ellensmyth (talkcontribs) 23:29, 12 June 2011 (UTC)

The expression "square root" is indeed in some contexts usually interpreted as meaning "positive square root". However, that does not preclude the use of the expression in other contexts without such restriction of meaning. JamesBWatson (talk) 15:28, 13 June 2011 (UTC)

Does the idea that the square root of 4 gives us -2 and 2 not imply a mathematical paradox? That is, is it not a proof for a=-a? Tristan Tondino (talk) 23:08, 5 February 2010 (UTC)

No. They are square roots but that doesn't mean they are the same. Just because 2×2 = −2×−2 doesn't mean 2 is the same as −2. There is no 'the square root' when either is allowed, or sometimes 'the square root' only means the positive one. Dmcq (talk) 01:37, 6 February 2010 (UTC)
Would this count as a "proof?" Where a=2; a(a)=-a(-a); since a=2 we can cancel one "a" leaving a=-(-a) or a=a. Is this formally acceptable, despite being fairly trivial?Tristan Tondino (talk) 16:47, 6 February 2010 (UTC)

## Bivalence

Is the square of a positive Integer considered to be bivalent in some number contexts? Any thoughts anyone?Tristan Tondino (talk) 17:23, 9 February 2010 (UTC)

I wonder what makes a person so "priviliged" that his/her external-link is allowed to appear at wikipedia. I mean, if Wikipedia does not like EXTERNAL LINKS then WIKIPEDIA SHOULD NOT mantain the external-links section. I think it is an obvious public-outrage for many people to realize that only two or three guys are "priviliged" by wikipedia, i do not see any justice on that. Wikipedia should either mantain the EXTERNAL LINKS in another related page or definitely eliminate all of them. What makes some few guys so "privileged"? What makes some guy to be the judge for selecting only two or three external links? Mirificium (talk) 23:07, 20 March 2010 (UTC)

## for kids

can someone make a page with a kid can understand —Preceding unsigned comment added by 98.162.205.238 (talk) 21:31, 7 April 2010 (UTC)

There is pointer for them in the left column. Look for the "Simple English" entry. DVdm (talk) 21:53, 7 April 2010 (UTC)

## Consistency in naming the argument?

In the lede and first section "Properties", the letter x is used for the number whose square root is being discussed (as in √x  ), but half-way in the section "Computation" the x gets to stand instead for the square root to be computed, whereas the argument is now called r – a letter I'd expect to stand for root – (as in x = √r  ). In my opinion this is confusing, but an attempt to introduce uniform names was reverted ("not a good idea"). What do others think? Reverted at the same time were several – in my opinion – valid and valuable additions and clarifications, as well as many little corrections,[1] as editors haven't always been careful to add necessary limitations to the ranges of the argument (like "positive" or "non-negative"); I'd like to urge editors to always consider and indicate the range of validity or applicability.  --Lambiam 13:57, 28 April 2010 (UTC)

Hi Lambiam, glad you brought it up here. I agree that in some places we can indeed use "a" in stead of "x", but I would prefer that we keep using the letter x where functions (and Taylor series) are involved (and/or mentioned nearby). This is most specially the case in the computation section on which I was concentrating. You have indeed a point with the "r" for root and that crossed my mind too. And indeed you made some valid and valuable additions and clarifications indeed. I propose I make that computation section in complete agreement with the source and replace the "r" with "a", and make the remainder (of that section) consistent. How about? DVdm (talk) 14:12, 28 April 2010 (UTC)
I went ahead with this - it was less work than I had anticipated. DVdm (talk) 14:31, 28 April 2010 (UTC)
Ok, you went ahead as well. Good job. I hope I didn't cause too much trouble with this :-) Cheers - DVdm (talk) 16:29, 28 April 2010 (UTC)

## Formula

Even before the +/- were taken out, the formula wasn't correct for negative reals. Now, it may work if the sgn function is redefined so that sgn 0 = 1, rather than the usual 0. I'm not quite sure how to specify that. — Arthur Rubin (talk) 09:29, 9 November 2010 (UTC)

Actually, the second +/- was OK but of course only in combination with the (sourced) text that explained what it meant. If we don't have a source for this new version, I think we must go back to the original properly sourced version - see equation 3.7.27 at [2] DVdm (talk) 09:43, 9 November 2010 (UTC)
Meanwhile I have restored the correct and sourced version, but I changed the order of the clarifications a bit. Tricky :-)

DVdm (talk) 10:08, 9 November 2010 (UTC)

## Impressive Textbook Chapter, But Useless for Beginners

This is an impressive article. But it also never answers my questions. I came here to read this article because I wanted to learn the significance of the square root -- why do I see it in so places when I'm reading scholarly and not-so-scholarly articles and books? Why is the square root important? What type of applications is it used in?

You know, practical information.

I don't expect the entire article to be in the "Dummies" format, but an introductory paragraph with two or three sentences that lay out out some of the questions -- and the answer to those questions -- would be wonderful. The information above the table of contents is waaaay too technical for mere mortals to understand. It should, in simple language, introduce the reader to the topic. You can save the master's level facts for further down in the article.

I would hope you read this the way it was intended, as helpful, constructive criticism. And I also hope you will apply my suggestions to all fundamental mathematics articles, not just this one. Todd W. Carter 23:59, 20 November 2010 (UTC) Todd W. Carter 00:03, 21 November 2010 (UTC)

You're probably right. it occurs so often and in so many circumstances in maths it's had to realize someone wouldn't see why it is important. Probably the first major use is in finding a distance when a place is given by two distances at right angles, the distance given by the long side of a right angled triangle, walk four steps to the east and three steps north, how far away ar you now. Something should be added to the leader about the basics liek this okay. Dmcq (talk) 10:24, 21 November 2010 (UTC)
I think Wikipedia is probably doing thinga a little better with square roots than a certain well known venerable competitor which offers this, "in mathematics, a factor of a number that, when multiplied by itself, gives the original number", which happens to be nonsense (the square root of 2 is not a 'factor' of 2).
I do sympathise to some extent with your comments, especially when I struggle myself with mathematical articles outside or beyond my own expertise in mathematics or notice a contributor I suspect more intent on showing off his expertise than instructing his reader.
However the fact of the matter is that an encyclopedia is not a text-book and in particular (in this case) not a remedial textbook for students challenged by mathematics. This article does give a clear definition of what a square root is. If a student has a difficulty with the definition then he should seek instruction from a textbook. You don't go to an encylopedia to learn how to play the piano, neither should you to learn mathematics. You do go to an encyclopedia to learn about playing the piano, about mathematics. The emphasis is different. FightingMac (talk) 03:29, 16 April 2011 (UTC)

## New sections that I think should be included

[quote]

#### Square roots in abstract algebra

For an abstract element a of (say) a semigroup, there may be any number of elements b so that b2 = a. So one could say there could potentially be many square roots of a.

However, in an integral domain, suppose the element a has some square root b, so b2 = a. Then this square root is unique in the following sense: If x too is a square root of a, then x2 = a = b2. So x2b2 = 0, or (x + b)(xb) = 0. Because there are no zero divisors in the integral domain, we conclude that one factor is zero, and x = ±b. The square root of a, if it exists, is therefore unique up to a sign, in integral domains.

[/quote]

I think it is relevant to mention that one can consider square roots in more general settings. I also think it is relevant to mention that in order for the number of square roots of a given element to be limited to at most two, it is sufficient to restrict oneself two integral domains. This is pretty obvious to a mathematician, but readers of Wikipeadia might not realize it if we don't cover that.

But User:DVdm undid my change both times because of missing/poor sources. However, most of the article is without sources, as it stands. Will someone else please contribute with information like that above, sources or not? /79.138.236.143 (talk) 18:38, 14 January 2011 (UTC)

It seems to me that the material ought to be in here (but should be merged with the short preceding section, "Square roots of matrices and operators"). The link to square root of a matrix is appropriate, although I think it should appear in a sentence in the text, not in a footnote. The external link to Project Crazy Project is not appropriate, since that site says "No doubt these pages are riddled with typos and errors in logic."
I think that absolutely no reference is needed for these two assertions ((1) there can be more than two square roots, but (2) not in an integral domain). The first assertion is self-evidently true once the reader sees the example in the linked wikipedia page (and in fact maybe that example should be put in here as well). The second assertion is extremely obviously true and relevant. Neither assertion should be removed, as they are uncontroversial. Duoduoduo (talk) 19:13, 14 January 2011 (UTC)
Anon79 and Duoduoduo, "However, most of the article is without sources, as it stands" is not a reason to add more content without sources. Quite on the contrary.

I know it is pretty obvious to a mathematician, but wikipedia not written for mathematicians only. It is written for everyone and the policies demand sources. See wp:RS, see wp:CALC, see wp:BURDEN. If this is really extremely obviously true and relevant, then it should be really extremely easy to find a source for it. Taking it over from another WP-article is no good either, since WP is not considered to be a reliable source, and specially since that article Square root of a matrix is tagged as poorly sourced, and it doesn't even mention semigroups or integral domains, which immediately turns the first sentence into wp:SYNT and thus wp:OR. So come on, this is good content, so just find a good source and you're in business. DVdm (talk) 19:47, 14 January 2011 (UTC)

Hi DVdm: "So come on, this is good content...." If you agree that it is good content, you should not be reverting it -- you should be labeling it [citation needed]. If it's not controversial, it should stay in even without citations. As for the link to the other article, I view it not as Wikipedia being used as a source -- it's just a wikilink to more information.
The decision whether to revert something should be based on whether the article is better with or without the edit. In this case, when you say that it is good content, that means the article is better with it than without it, so a [citation needed] tag is better than a reversion.
In any event, I'll put in a cite for the multiple square roots of matrices part. Duoduoduo (talk) 22:15, 14 January 2011 (UTC).
Yes, please do so, because whether you and I think it is good content is entirely irrelevant. When a source is requested it really should be provided. Cheers and thanks. DVdm (talk) 23:35, 14 January 2011 (UTC)

## Square roots of negative and complex numbers

Can I suggest someone do a rewrite of this section? It looks dreadfully old fashioned, doesn't actually mention what a complex number is and it surely cannot be a plus that it contrives to fail to mention that i can be thought of a rotation through 90 degrees. I question whether the algebraic derivation of the square root of i serves any purpose, equally the formal derivation using de Moivre's theorem as presented here. In plain what I'm suggesting is not that the treatment is too technical but that it's too pedantic. FightingMac (talk) 16:09, 16 April 2011 (UTC)

1. I find that the article puts entirely too much emphasis on the "principal value" of the square root, without making it clear enough that this is not a natural mathematical concept but rather a convention.

Indeed, it is a useful convention, one that enables people to speak unambiguously about square roots of real and complex numbers by turning what is naturally a multi-valued function into an unnaturally single-valued one. But still.

2. There is not even a graph of the relation y = x2, i.e., a picture of the set {(x,y) ∈ $\mathbb{R}$2 | y = x2}, which is an astonishing omission for an article on the square root.

3. Another thing: What is the point of showing that the square root is not unique in $\mathbb{Z}/8\mathbb{Z}$ as a demonstration of its non-uniqueness in "general rings" when this has already been made clear for the simplest ring, $\mathbb{Z}$ ???

A better idea might be to use integers mod 8 to show that square roots in general rings can be of multiplicity greater than 2. And the integers mod 5 could be used to show that for some elements in a general ring (in this case, ±2), a square root need not exist.Daqu (talk) 02:32, 6 July 2011 (UTC)

I think you have raised some good points there. In the context, I think the person who wrote about square roots not being "unique" in $\mathbb{Z}/8\mathbb{Z}$ actually meant not "unique", but rather "unique up to sign", in which case it makes more sense. However, the use of the word "unique" in that section was most unhelpful, so I have rewritten it. Oddly enough your suggestion of using integers mod 5 to show that a square root need not exist in a general ring suffers from exactly the same fault that you criticised the example $\mathbb{Z}/8\mathbb{Z}$ for: the ring $\mathbb{Z}$ gives a simpler illustration of the same fact. Nevertheless, I have added a brief mention of the fact that 2 has no square root in the integers mod 8. It seems simpler and clearer to me to use integers mod 8 to illustrate both points, rather than mod 8 for one and mod 5 for the other. I don't have time now to address your other points, but maybe I will come back and do so, unless someone else does first. JamesBWatson (talk) 09:28, 6 July 2011 (UTC)

## √

Could someone add something about √ being the sqrt character? I don't know where it would fit, but I think a lot of people use Wikipedia to get special characters easily (just go to the article and you got it, for almost everything), but this article still lacks a copy/paste-able character for those people. It would be nice if someone could add that at a fitting place on the article.Joeytje50 (talk) 20:18, 4 April 2012 (UTC)

## Why Taylor series of sqrt(x + ..)?

Isn't it possible to also give a Taylor series of sqrt(x) itself, in which the terms are integer (and non-negative) powers of x? The article doesn't really explain why sqrt(x+1) is notable if sqrt(f(x)), for any f(x)!=x+1, is less so. Cesiumfrog (talk) 01:21, 13 October 2012 (UTC)

It appears to be sqrt (1+x) that is claimed to be notable in the article, not sqrt (x+1). That seems trivial but in reality it isn't. -- Glenn L (talk) 03:45, 13 October 2012 (UTC)
How so? Cesiumfrog (talk) 10:08, 13 October 2012 (UTC)
Re Cesiumfrog: If you try to calculate the Taylor-coefficients for sqrt(x), you will notice that you fail, as they're all (except the 0th order) infinite, so to speak. So you need some simple function that doesn't have this property. As Taylor is used to calculate function values, the idea is to take as simple a function as possible, hence f(x)=x+1. DVdm (talk) 09:14, 13 October 2012 (UTC)
Sure, forget Taylor then, but isn't it still possible to have a power series for sqrt(x) that works for small positive values of x? For example, can't the terms of the sqrt(1+x) series be modified and rearranged to construct a power series in x converging to sqrt(x) for 0<x<2? Cesiumfrog (talk) 10:08, 13 October 2012 (UTC)
The wp:Reference desk/Math is perhaps more appropriate for this — see wp:TPG. - DVdm (talk) 10:43, 13 October 2012 (UTC)
Is this really the wrong place to point out that the inclusion criteria of this article's section on properties is unclear (Should I just go ahead and mark elements of the article with inline importance templates?) and to discuss the content for a short explanatory sentence designed to resolve that and improve the accessibility of this article's treatment of its topic? (Something along the lines of "It is (not) possible to create a series for sqrt(x) because Y but most references instead frequently list the series of sqrt(W) which is as follows[ref]"?)
Looking at the article more broadly, I notice the problem is more serious and that several entire sections do not even contain a single reference. Please let me hereby WP:CHALLENGE all of it, to give a little time for anyone to try to provide specific references. Cesiumfrog (talk) 13:21, 13 October 2012 (UTC)
I agree that large parts lack inline references. Feel free to improve upon that. I already made this little cosmetic change. - DVdm (talk) 13:29, 13 October 2012 (UTC)

By the way, 1 + x does converge (absolutely) in the closed segment [−1, 1]. This fact, namely the convergence in x = −1, allows for a definition of the square root in Banach algebras without resorting to any kind of the spectral theorem. Incnis Mrsi (talk) 09:38, 16 March 2014 (UTC)

## Lede image, self-publishing, original research, and the "square roots of natural numbers as square" image.

Hi. I just replaced the lede image, which was changed to this in January:

to something a bit more introductory, a simple "square root of x" symbol.

My concerns are twofold: First, while as an amateur mathematician I find the 'square roots of natural numbers as squares" to be a nifty image, it's not an appropriate lede image. It's confusing, it's not well-labelled, and the overall effect is going to be to make the topic seem even more confusing to general interest readers. Second, I'm worried there's a bit of original research going on here; not in the sense that the facts in the image are in dispute, but in the sense that we should prefer visualizations either created by or at least endorsed by reliable third party sources.

Thoughts? Nandesuka (talk) 00:08, 5 April 2014 (UTC)