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In several online high school teaching resources (I do not want to single out any) I see that $a-b$ is referred to as the conjugate of $a+b$ with no restriction on $a$ and $b$.

I can understand that $a-bi$ is the conjugate of $a+bi$ when we consider complex numbers.This is a standard terminology.

After asking a question, I was directed to Galois extensions, so now I understand why we call $a-\sqrt{b}$ the conjugate of $a+\sqrt{b}$ (where $a$ and $b$ are rational and $\sqrt{b}$ is irrational).

However, I still don't understand why it is justified to use the term "conjugate" when we simply change the sign in a binomial. Is it really necessary to give names to everything? Is a formal similarity enough to use the same expression as a name? In the previous two examples the similarity is more than formal. The two type of conjugation have similar properties (since they are both special cases of a general theory). However, I do not see how changing the sign in a binomial without any restriction on the terms in the binomial fits in this general theory. enter image description here

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    $\begingroup$ The conjugate of $5+x$ is $5-x$. No, I would not say that. $\endgroup$ – Gerald Edgar Feb 27 at 12:43
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    $\begingroup$ So is the conjugate of a+b equal to a-b, while the conjugate of b+a is b-a? $\endgroup$ – Ben Crowell Feb 27 at 15:03
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    $\begingroup$ To get a more general view, you need to involve the language of Galois theory. $\endgroup$ – Bill Cook Feb 27 at 18:42
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    $\begingroup$ The OP (and one of the answers below) say that this usage is found in many online resources. Has anybody seen this in any textbooks? I don’t recall ever having seen it. $\endgroup$ – mweiss Feb 28 at 1:27
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    $\begingroup$ @FerencBeleznay I ought to have been more clear; I am aware of the widespread use of “conjugate” in the context of expressions of the form $a \pm \sqrt{b}$, and think that is both reasonable and useful. It’s in the context of $a+b$ and $a-b$ in general that I have never seen this. Are there really sources that claim the conjugate of $5+3$ is $5-3$? That seems utterly incoherent to me. $\endgroup$ – mweiss Mar 1 at 3:27
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Calling both $a+ib \mapsto a-ib$ and $a+b \sqrt{2} \mapsto a-b \sqrt{2}$ conjugations is reasonable because of the general understanding given by Galois Theory (which has been known much longer than our lifetimes even if the educational establishment can't see fit to require it in our education). Using the same name is in part reasonable beyond the abstract union of the concepts in that similar calculational strategies follow for both: $$ \frac{1}{2+3i} = \frac{1}{2+3i} \cdot \frac{2-3i}{2-3i} = \frac{2-3i}{13} $$ and $$ \frac{1}{2+3\sqrt{2}} = \frac{1}{2+3\sqrt{2}}\cdot \frac{2-3\sqrt{2}}{2-3\sqrt{2}} = \frac{2-3\sqrt{2}}{-14} $$ However, there is danger in using something without understanding. If we try to naively extend this idea of conjugation to other numbers such as $1+\sqrt[3]{7} \mapsto 1-\sqrt[3]{7}$ then we'll find the rationalization scheme above falls flat: $$ \frac{1}{1+\sqrt[3]{17}} = \frac{1}{1+\sqrt[3]{17}}\cdot \frac{1-\sqrt[3]{17}}{1-\sqrt[3]{17}} = \frac{1-\sqrt[3]{17}}{1+(\sqrt[3]{17})^2} $$ the denominator is still ugly. To find the magic number to rationalize $1+\sqrt[3]{17}$ we can consider the polynomial $x^3-17$ which clearly takes $\sqrt[3]{17}$ as a root. I'll use my brother's handy-dandy Sage-based polynomial Euclidean algorithm calculator to see that the $gcd(x^3-17,x+1) = 1$ and in fact: enter image description here $$ \frac{-1}{18}(x^3-17)+\left(\frac{1}{18}x^2-\frac{1}{18}x+\frac{1}{18} \right)(x+1) = 1. $$ Now, set $x = \sqrt[3]{17}$ in the identity above to see that: $$ \left(\frac{1}{18}(\sqrt[3]{17})^2-\frac{1}{18}\sqrt[3]{17}+\frac{1}{18} \right)(1+\sqrt[3]{17}) = 1. $$ Which goes to show: $$ \frac{1}{1+\sqrt[3]{17}} = \frac{(\sqrt[3]{17})^2-\sqrt[3]{17}+1}{18} $$ To calculate the same using Galois conjugates is more difficult since the splitting field of $x^3-17$ has more than one conjugation and essentially you have to multiply by the product of all the Galois conjugates to produce an integer.

My apologies if this is slightly vague, I am trying to avoid getting further into the story of Galois theory here. You can look at Section 14.1, Example 5 in Dummit and Foote's 3rd edition where the Galois group of $x^3-2$ is shown to be $S_3$ which means there are five nontrivial automorphisms which we would call Galois conjugates.

All of this said, I would not personally say $a-b$ is the conjugate of $a+b$. I suspect such terminology is made because of similarity in the pattern discussed here for $a+bi$ or $a+b\sqrt{2}$. If I am going to do something with both $a+b$ and $a-b$ I'll likely say something about remembering the difference of squares formula $a^2-b^2 = (a+b)(a-b)$. But, that's just my personal preference.

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    $\begingroup$ In really old math (before Galois, but not too old such as before Viete's time) the term was "rationalizing factor" (actually, this was also used well after Galois . . .), and the form was whatever worked, not a "change sign" operation. Rationalizing factors could be found using factorizations such as $a^3 + b^3 = (a+b)(a^2 - ab + b^2),$ $a^4 - b^4 = (a-b)(a^3 - a^2b + ab^2 - b^3),$ $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ac),$ etc. $\endgroup$ – Dave L Renfro Feb 28 at 18:22
  • $\begingroup$ @DaveLRenfro Interesting, I was aware of the special form trinomial stuff, but I think it requires greater insight to apply to particular problems. Once we understand tthe technique I illustrate in this example it solves a wide swath of problems without too much thinking. I appreciate the historical remark. $\endgroup$ – James S. Cook Mar 1 at 15:28
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    $\begingroup$ FYI, I've amassed a huge literature on this kind of stuff (mostly 1800s; many hundreds of pages photocopied over a couple of decades from original library volumes), and one day I would like to write some kind of survey paper/book on the evolution and various applications of rationalization methods. I think the general procedure for rationalizing a finite sum of various-indexed radicals has been "known" since sometime in the 1700s, but I haven't tried to seriously look into this yet. I have an English paper from around 1800 (can't find it now, however), and some examples are: (continued) $\endgroup$ – Dave L Renfro Mar 1 at 17:01
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As James mentioned, to fully appreciate the term "conjugate" here we need to know a little about Galois theory.

The field $\mathbb{Q}[\sqrt{2}] =\{a+b\sqrt{2} \;|\; a,b \in \mathbb{Q}\}$ contains the field of rational numbers $\mathbb{Q}$. If we look for all automorphisms of $\mathbb{Q}[\sqrt{2}]$ that leave $\mathbb{Q}$ fixed pointwise, we get the Galois group: $\mathrm{Gal}(\mathbb{Q}[\sqrt{2}]/\mathbb{Q})$.

In particular, $\varphi \in \mathrm{Gal}(\mathbb{Q}[\sqrt{2}]/\mathbb{Q})$ if $\varphi$ is an invertible mapping from $\mathbb{Q}[\sqrt{2}]$ to itself and preserves addition and multiplication. It turns out that there are only two such mappings. The identity map: $\mathrm{id}(a+b\sqrt{2})=a+b\sqrt{2}$ and the conjugate map: $\varphi(a+b\sqrt{2})=a-b\sqrt{2}$.

What is going on here is that $\mathbb{Q}[\sqrt{2}]$ is obtained by attaching all of the roots of the irreducible (over $\mathbb{Q}$) polynomial $x^2-2$. These Galois automorphisms arise from the two ways we can permute the roots of $x^2-2$. Namely: $\pm\sqrt{2} \mapsto \pm\sqrt{2}$ (the identity) or $\pm\sqrt{2} \mapsto \mp\sqrt{2}$ (the conjugate).

We can do the same thing for the complex numbers $\mathbb{C}$ extending the reals $\mathbb{R}$. Again, the Galois group only has two elements (called the identity map and the conjugate map). Again, $\mathbb{C}$ is generated over $\mathbb{R}$ by the roots of an irreducible (over $\mathbb{R}$) polynomial $x^2+1$ and these maps come from permuting the roots of this polynomial. Namely: $\pm i \mapsto \pm i$ (the identity) and $\pm i \mapsto \mp i$ (the conjugate).

Anytime (in characteristic 0) we have a quadratic extension (two dimensional extension) of one field over another, we'll get a Galois group with exactly two elements: an identity and a conjugate map.

The image of an element under a Galois automorphism is called a conjugate of that element. But in the context of a quadratic extension, if we say "conjugate" we usually mean the non-trivial one (not using the identity map and just getting our element itself right back).

Much more generally, if we take any Galois extension $\mathbb{K}/\mathbb{F}$ and consider some Galois automorphism $\varphi \in \mathrm{Gal}(\mathbb{K}/\mathbb{F})$ then given $a,b \in \mathbb{K}$ where $b=\varphi(a)$, we call $a$ and $b$ conjugates (relative to that Galois group).

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  • $\begingroup$ A generalized version of the "conjugate trick" from a Galois theory course I taught: billcookmath.com/courses/math4010-fall2019/… $\endgroup$ – Bill Cook Feb 27 at 19:06
  • $\begingroup$ The maple source code is also posted there under the date 11/08 on billcookmath.com/index.html?page=./courses/math4010-fall2019 $\endgroup$ – Bill Cook Feb 27 at 19:07
  • $\begingroup$ I have a question about the starting example from your link. If the quotient were given in the form $\frac{\sqrt{5}+1}{3\sqrt{5}+2}$, would you call $3\sqrt{5}-2$ the conjugate of the denominator? This would also work in rationalizing the denominator. $\endgroup$ – Ferenc Beleznay Feb 27 at 21:36
  • $\begingroup$ While using $3\sqrt{5}-2$ would allow you to rationalize, it is not a conjugate of $2+3\sqrt{5}$. A conjugate in this context replaces a root of $x^2-5$ with another root, so $\sqrt{5}$ and $-\sqrt{5}$ get swapped. The conjugate of $2+3\sqrt{5}$ is $2-3\sqrt{5}$. $\endgroup$ – Bill Cook Feb 28 at 0:32
  • $\begingroup$ I'm not sure if the Galois group of a quadratic extension is always non-trivial. $F_2(\sqrt{x})$ over $F_2(x)$ is a quadratic extension but there is only trivial homomorphism fixing underlying field. $\endgroup$ – Mihail Feb 28 at 2:21
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I have used this terminology for years. It helps students to see that being able to factor a square minus a square is powerful. There is one more context for this that I know of. When a or b is a trig function, multiplying by the conjugate gives us something that fits a Pythagorean identity. (I am now going to test whether that is always true, or just usually true.)

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    $\begingroup$ I do not see how using the word "conjugate" can help students more than saying "change the sign". It seems to me that in high school algebra it is used mostly when the $(a+b)(a-b)=a^2-b^2$ identity is involved. In other words, this is yet another technical term for students to remember. Instead of the procedural command "multiply by the conjugate", we could simply tell them "think of the factor form of the difference of squares". Probably less students would be able to rationalize a denominator (or simplify a trigonometric expression), but maybe more would understand the reason. $\endgroup$ – Ferenc Beleznay Feb 27 at 21:05
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    $\begingroup$ Sometimes language works as a framework for helping us understand things. I definitely put conceptual understanding first. You asked if it was a standard term, and part of the point of my answer was that it is. Clearly, math educators disagree on whether it should be. $\endgroup$ – Sue VanHattum Feb 28 at 3:04
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I don't think that this usage is standard, although it might be common. Still, I would argue that it is a reasonable enough usage that it is worth trying to fit it into the more standard theory.

As observed in the comments, the $a+b\to a-b$ rule doesn't make much sense if $a$ and $b$ are concretely realized as numbers. However, it does make sense if we are looking at an expression which lives in the multivariate polynomial ring $\mathbb{R}[a,b]$. This ring also supports several other forms of conjugacy (i.e. ring automorphisms) such as changing the sign of $a$ and swapping $a$'s and $b$'s.

tl;dr: It is fine as long as you realize that what's being conjugated are formal expressions, not the numbers themselves.

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A Google search shows that this is common terminology.

Just because it's strange to you doesn't mean it is strange to the world.

Also a minus bi is often called the COMPLEX conjugate. That is to differentiate from the usual conjugate, I guess.

A comment asks if any books use this terminology. Well Google books shows several. Barron's pre calculus is a recent one. But also several algebra books from the late nineteenth and early twentieth century. Note that I'm excluding usage in terms of complex conjugated or binomial surds. But even with the very frequent usage of binomial surd, it would seem that the qualification surd differentiates from normal binomial conjugates.

I'm not sure the usage is most common. But it is certainly not uncommon. Really, what we are getting is a sounds strange to me from questioner. And Google is a helpful tool to show frequent usage. And I would caution questioners and answers to generalize from own personal experience. Sounds strange to me doesn't mean strange to the world.

For FB, well would a rose by any other name smell as sweet? We are not even kvetching about divided by zero. We are discussing terminology. And if people use the term such, how can that be wrong. Especially if common. It's just language.

Heck. My dictionary, lists one of the definitions as pairs having features in common but opposite in some particular. Further, the terms COMPLEX conjugate or binomial conjugate SURD each imply some need to further define the term conjugate...that is referring to binomials with imaginary and real or rational and sqrt irrational parts. I guess...as opposed to other sorts of conjugates.

So certainly a binomial conjugate, used as Sue and many other English speaking math teachers, for over a hundred years, is reasonable usage. It's not bastsrdizing the language. It fits fine with other uses if the word, is well defined in class, is supported in dictionaries, is common enough to be frequent, at least. Not saying it is majority though.

So anyhow if the usage is common, what is your problem with it. A rose is a rose by any other name. Oh...and a spade is a shovel. 😀

And as far as the opposite binomial surd, for rationalizing, so what? Standard method works as is and is easy to learn. Why complicate?

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    $\begingroup$ Yes, I can see that it is used in several online resources, but I think it is simply a result of the authors not being careful enough. I can even feel in examples where rationalizing the denominator is the goal, that "conjugate" is simply a synonym for "change the sign". When $1+\sqrt{2}$ is in the denominator, then multiplying by $1-\sqrt{2}$ and by $\sqrt{2}-1$ both work (and both are commonly referred to as conjugates), but as far as I know, only the first one is the conjugate of $1+\sqrt{2}$ in the field extension sense. $\endgroup$ – Ferenc Beleznay Feb 27 at 18:04
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    $\begingroup$ For those who have downvoted (6!), please consider whether your vote is because you have strong feelings on this issue. This seems like a good answer to me. $\endgroup$ – Sue VanHattum Feb 27 at 20:19
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    $\begingroup$ @guest I did not say in my post "it sounds strange to me". I wrote I don't understand the justification and the need of the use. I still don't understand it. I suspect that the authors who use it do it because "I heard it this way, let's pass it on". This is not a good enough justification for me. I am not generalizing (and did not downvote any answer), I am simply listening to the answers I get. $\endgroup$ – Ferenc Beleznay Feb 28 at 15:14
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    $\begingroup$ @guest You keep adding to your answer and it is getting to be quite aggressive. Can you please not do it? I did not say I have a problem with the usage. I observed that it is used frequently and asked help in understanding why. $\endgroup$ – Ferenc Beleznay Feb 28 at 18:01
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    $\begingroup$ Oh my. It does keep changing, doesn't it? The first 3 lines, I guess. I think that's what I saw when I posted my comment. But I am not used to answers changing like this. (I don't know if it's considered bad form, or not.) $\endgroup$ – Sue VanHattum Mar 1 at 18:31

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