Talk:Square root

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To-do list for Square root: edit · history · watch · refresh
  • 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.

Contents

[edit] 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)

[edit] 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:
>>And what do you think about this :
>>
>>
>> -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.

[edit] 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.

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)

[edit] 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".

[edit] 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 xn + 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}.

[edit] 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 xn + 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 ).

[edit] 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

[edit] 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)

[edit] 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

[edit] continued fraction rewrite

[edit] 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.

[edit] 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 N0 using some other method. The best method is N0 = intsqrt(r) using some other algorithmn to determine the integer square root.

N0 = 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 X0 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 X0 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 N1 from the numeric values of X0LowerBound and X0UpperBound because N_1 \quad \in \quad \mathbb{Z}

Step 5. Once we know the value of N1, we can rework the equation for X1

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 X1 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 X1 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 N2 from the numeric values of X1LowerBound and X1UpperBound because N_2 \quad \in \quad \mathbb{Z}

Step 8. Once we know the value of N2, we can rework the equation for X2

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 .

[edit] 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)

[edit] 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)

[edit] 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 definately support moving all the square root computation portion to Methods of calculating square roots (or something similar). Eric119 06:09, 14 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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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? —Mets501 (talk) 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. —Mets501 (talk) 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. —Mets501 (talk) 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)

[edit] 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. —Mets501 (talk) 13:02, 20 July 2006 (UTC)

[edit] 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. —Mets501 (talk) 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. :-) —Mets501 (talk) 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. —Mets501 (talk) 11:45, 25 July 2006 (UTC)
Actually, that's already in the square root of 2 article. I wikilinked to it in the intro. —Mets501 (talk) 15:08, 10 August 2006 (UTC)

[edit] 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)
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