Talk:Square root

Semi-protected edit request on 2 July 2023

It’s written that -4 and 4 are square root of 16, but that’s false. 4 is the square root of 16, but not -4. -4 is a solution of x^2=16 because x=±sqrt(16) 2A02:8440:7136:43EB:F01A:F1AE:DCCA:271C (talk) 20:36, 2 July 2023 (UTC)

not done: If you read the next three paragraphs it does explain it fully. Moons of Io (talk) 20:39, 2 July 2023 (UTC)

It does not matter. the initial statement is FALSE

square root is FUNCTION, it cannot have 2 results or tell me the result of this sqrt(2) + sqrt(4) + sqrt (6). what is the result ? according to you and that definition you have 9 results of that sum !!!

i remind you that sqrt(2) is NUMBER. real number 109.110.224.32 (talk) 21:16, 7 December 2023 (UTC)

 Not done: it's not clear what changes you want to be made. Please mention the specific changes in a "change X to Y" format and provide a reliable source if appropriate. Shadow311 (talk) 21:32, 7 December 2023 (UTC)

The definition of square root is false

Hello everyone,

In the English Wikipedia article “square root” the square root is defined as follows:

“In mathematics, a square root of a number x is a number y such that y² = x; in other words, a number y whose square (the result of multiplying the number by itself, or y ‧ y) is x. For example, 4 and −4 are square roots of 16 because 4² = (−4)² = 16.”

That’s wrong. What is being described here are the solutions to the quadratic equation x² = 16. But this is not the square root of 16. The square root of 16 is 4, not 4 or −4. The result of a square root is always nonnegative. More precisely: an absolute number. The square root is defined as follows:

√(x²) = |x|

Everything else is not square root. This eliminates the notion of “principal square root”. The “principal square root” is simply the square root.

It should also be noted that the so-called “roots of a quadratic equation” are not square roots, and are not actually roots at all, but the solutions of a quadratic equation. A quadratic equation like x² = 16 is solved as follows (with row numbers):

(1) x² = 16   |√
(2) |x| = 4
(3) x₁ = 4
(4) x₂ = −4

Only line (2) shows a square root, which is the square root of 16. What is in lines (3) and (4) are not square roots. In English, what is in lines (3) and (4) is called “roots of a quadratic equation”. That is an extremely misleading expression. It can be confused with the notion of the square root. Therefore, the term “solutions to a quadratic equation” should always be used instead.

I would be in favor of replacing the quoted incorrect definition of a square root with the correct definition found in the German article “Quadratwurzel”. There it says:

"Die Quadratwurzel (umgangssprachlich Wurzel; englisch square root, kurz sqrt) einer nichtnegativen Zahl y ist jene (eindeutig bestimmte) nichtnegative Zahl, deren Quadrat gleich der gegebenen Zahl y ist."

Translated to English:

“The square root of a nonnegative number y is that (uniquely determined) nonnegative number whose square is equal to the given number y.”

This corresponds exactly to the mathematical definition of the square root as I gave it above. The rest of the article would need to be revised accordingly.

Best regards
Jake2042 (talk) 15:47, 3 September 2023 (UTC)

Firstly, a definition cannot be wrong, as it is the result of an arbitrary choice. It is often the case in mathematics, that there are several different definitions of the same phrase that are used in different contexts. This is the case here. Your prefered definition is commonly used in very elementary mathematics, but it does not work in more advanced contexts, especially in the case of square roots of complex numbers (note that, in the first sentence, it is not said that x must be a real number). The case of nonnegative real numbers is considered in the second paragraph, and you must note that although "a square root" is defined for any number, the notation ${\sqrt {x}}$ is reserved for nonnegative real numbers. In other word, your definition is that of the notation of ${\sqrt {x}}.$

However, your post convinced me that, in the second paragraph, it must be said that the principal square root is often called simply "the square root". I'll add this to the article. D.Lazard (talk) 16:40, 3 September 2023 (UTC)

I agree with D.Lazard adding the clarification to the article as suggested by you. But one thing to keep in mind on math Wikipedia is that you're not supposed to use your own logic and reasoning to decide what goes in articles, you're supposed to cite sources such as textbooks, peer-reviewed publications, and other reliable sources. You may be logical and correct, but the next editor might be a crackpot, so that's why original research is frowned on and citing sources is favored.Brirush (talk) 23:43, 3 September 2023 (UTC)

Guys. square root in complex numbers is ok. BUT normal people understand from your phrase that the square root of 4 is +/- 2 !!! So yes the definition is WRONG because is NOT the definition. The definition is that the sqre root is the POSITIVE number that raised to the power of 2 is the original number. IS NOT +/- square root of X is ABS(square root of X)

You also state that clearly below as well.

So please do rephrase that because as it is is wrong and confusing. 109.110.224.32 (talk) 07:32, 1 November 2023 (UTC)

I'm not sure what the purpose is in repeating what has already been said and answered, but the expression "square root" is used by mathematicians to refer to either root. In contexts where it is clear that postive numbers are concerned, it is usual not to redundantly include the word "positive", so "square root" is effectively shorthand for "positive square root". "Contexts where it is clear that only postive numbers are concerned" include most elementary contexts, and are also very common in more advanced contexts, so the use of the expression "square root" to refer only to the positive use is very common, and therefore in a wide range of contexts it is effectively the default use. It may also be the only usage that is encountered at all by many people whose knowledge of mathematics does not go beyond the elementary, so to them it appears to be the only meaning. However, that doesn't mean that the other usage doesn't exist or is in some sense wrong. A word or expression means precisely what it is used to mean and understood to mean, and "square root" is in some contexts used to mean and understood to mean only the positive root, and in some contexts to mean either root: that is to say both meanings exist. The article quite correctly gives both the meanings. To omit either would be misleading. JBW (talk) 12:21, 1 November 2023 (UTC)

A definition can be wrong if is a wrong definition lol. Just because you call it definition it does not mean it is one.

First of all square root is FUNCTION.

A function can have only one value for a given input not 2

Think otherwise: if sqrt(4) = +/-2 then pls tell me how much is this : sqrt(2) + sqrt(3) + sqrt(10) ?

You do realize that sqrt(2) is a NUMBER so it cannot have 2 values

That definition is simply WRONG because is NOT the definition. 109.110.224.32 (talk) 21:12, 7 December 2023 (UTC)

Is this a difference between UK and US usage? Obviously mathematically the square roots of 25 are + and - 5. But there are no end of YouTube videos that say this is wrong and √25 (e.g.) refers only to the positive square root +5 and other usage (-5) is confusing or even wrong. I understand how with 8 year old students you might want to keep things simple and, for the time being, ignore the negative roots, but Wikipedia is not written for 8 year olds.Cross Reference (talk) 00:02, 11 March 2024 (UTC)

In some contexts, including elementary arithmetics, only non-negative real numbers have a square root and that square root is always non-negative. In other contexts, including the study of complex numbers, any number other than zero has n n-th roots, for instance there are 5 "fifth roots of unity" located in the complex plane at the vertices of a regular pentagon centered at zero and one of them is 1. Saying that what the other guy says is "wrong" leads nowhere. He's just standing in a different context and talking to other people. To reconcile "both kinds" of square roots (or of nth roots in general), the one (always non-negative) used in elementary arithmetic is called "the principal square root" when there is a risk of ambiguity; other times one knows what is talked about — for instance when talking about the five fifth roots of unity it is obviously not the single non-negative one, but when asking what is the square root of ten the adjective "principal" is left implicit. — Tonymec (talk) 12:54, 13 April 2024 (UTC)

"Think otherwise: if sqrt(4) = +/-2 then pls tell me how much is this : sqrt(2) + sqrt(3) + sqrt(10) ?" Taking √2 = +/- 1.4; √3 = +/- 1.7; and √10 = +/- 3.2 (all approximations of course) then your expression can have any one of 8 values from -6.3 to +6.3 depending on which root you take for each element. For many (most?) purposes the negative square root can safely be ignored as a mathematical artifact with no practical meaning in the circumstance - e.g., a square of area 10 square units has a side of 3.2 units and the negative value can be ignored.

I understand that a function, by definition, returns a single value for a single input. What makes you think that a square root is a function and that therefore the radical sign √ returns only the principal (i.e. positive) square root? Cross Reference (talk) 21:49, 25 April 2024 (UTC)

Trouble understanding reference to sgn function

Under the heading "Algebraic formula", this remark in parentheses is confusing: "(except that, here, sgn(0) = 1)". First of all, what does "here" refer to? Secondly, according to the Wikipedia page for the Sign function, sgn(0) = 0, so how can it be legitimate to write sgn(0) = 1? 109.255.134.32 (talk) 18:56, 5 February 2024 (UTC)

This is written this way to be able to have a single RHS expression. The alternative is to replace the factor sgn(y) on the RHS with a plus/minus (±), and then explain with an elaborate string of words which root should be taken in which case. Indeed, this is not the standard definition of the sign function, which is why the exception is noted. Consider this a modified sign function. Much easier. - DVdm (talk) 19:26, 5 February 2024 (UTC)

I have fixed the confusing formulation. D.Lazard (talk) 19:37, 5 February 2024 (UTC)

Yes, correct. Try with x= ±2 and y=0.

Then

$${\sqrt {x+iy}}={\sqrt {{\tfrac {1}{2}}{\bigl (}{\sqrt {\textstyle x^{2}+y^{2}}}+x{\bigr )}}}+i\operatorname {sgn}(y){\sqrt {{\tfrac {1}{2}}{\bigl (}{\sqrt {\textstyle x^{2}+y^{2}}}-x{\bigr )}}}$$

becomes for x=2 and y=0

$${\begin{aligned}{\sqrt {2}}&={\sqrt {{\tfrac {1}{2}}{\bigl (}{\sqrt {\textstyle 2^{2}}}+2{\bigr )}}}+i{\sqrt {{\tfrac {1}{2}}{\bigl (}{\sqrt {\textstyle 2^{2}}}-2{\bigr )}}}\\&={\sqrt {{\tfrac {1}{2}}{\bigl (}2+2{\bigr )}}}+i{\sqrt {{\tfrac {1}{2}}{\bigl (}2-2{\bigr )}}}\\&={\sqrt {2}}\end{aligned}}$$

as it should be.

Similarly for x=-2 and y=0,

$${\begin{aligned}{\sqrt {-2}}&={\sqrt {{\tfrac {1}{2}}{\bigl (}{\sqrt {\textstyle (-2)^{2}}}+(-2){\bigr )}}}+i{\sqrt {{\tfrac {1}{2}}{\bigl (}{\sqrt {\textstyle (-2)^{2}}}-(-2){\bigr )}}}\\&={\sqrt {{\tfrac {1}{2}}{\bigl (}2+(-2){\bigr )}}}+i{\sqrt {{\tfrac {1}{2}}{\bigl (}2-(-2){\bigr )}}}\\&=i{\sqrt {2}}\end{aligned}}$$

as it should be. - DVdm (talk) 20:14, 5 February 2024 (UTC)

Semi-protected edit request on 11 April 2024

In "Computation" section, the following text...

"Furthermore, $(x+c)^{2}\approx x^{2}+2xc$ when c is close to 0, because the tangent line to the graph of $x^{2}+2xc+c^{2}$ at $c=0$, as a function of c alone, is $y=2xc+x^{2}$. Thus, small adjustments to x can be planned out by setting $2xc$ to a, or $c={\frac {a}{2x}}$."

Should be replaced with...

"Furthermore, $(x+c)^{2}\approx x^{2}+2xc$ when c is close to 0, because the tangent line to the graph of $x^{2}+2xc+c^{2}$ as a function of c alone, evaluated at $c=0$, is $y=2x$. Thus, small adjustments to x can be planned out by setting $2xc$ to $a-x^{2}$, or $c={\frac {a-x^{2}}{2x}}$."

...and this because the partial derivative of $f(x,c)=x^{2}+2xc+c^{2}$ w.r.t. c is ${\frac {df(x,c)}{dc}}=2x+2c$, and ${\frac {df(x,c)}{dc}}=2x$ at $c=0$. And that is why we can do the first order approximation $(x+c)^{2}\approx x^{2}+2xc$ when c is close to 0. Then, considering that x_n is the estimation at step n, we choose c such that the estimation x_n+1 at step $n+1$ is $(x_{n}+c)^{2}\approx a$, which leads to $c={\frac {a-x_{n}^{2}}{2x_{n}}}$. Nunoguedelha (talk) 19:33, 11 April 2024 (UTC)

 Not done. As neither the original nor the newly proposed version is sourced, I have removed ([1]) the challenged passage in the article. Feel free to add a correct version again, provided a proper source is added. - DVdm (talk) 22:09, 12 April 2024 (UTC)