11
$\begingroup$

I'm slightly confused on the subject of conjugates and how to define them.

I know that for a complex number $ a - bi $ the conjugate is $ a + bi $ and similarly for $ 1 + \sqrt 2 $ the conjugate is $ 1 - \sqrt2 $ because when multiplied it gives a rational answer.

But how about for just a simple real number like 1 or 2, what would be the conjugate for this? Does a conjugate exist for a real number?

I'm new to this topic and have tried searching Maths SE and Google in vain; any help would be appreciated.

$\endgroup$
  • 1
    $\begingroup$ Since $\mathbb{R} \subset \mathbb{C}$, any $a \in \mathbb{R}$ can be written in the following way: $a=a+i\cdot 0$. $\endgroup$ – Iuli Jul 30 '18 at 12:10
  • 14
    $\begingroup$ There is complex conjugate for complex numbers, and there is also quadratic conjugate for algebraic numbers of the form $p+q\sqrt{d}$ with $p,q\in\mathbb{Q}$ and $d\in\mathbb{Z}$ is not a square. $\endgroup$ – Batominovski Jul 30 '18 at 12:11
42
$\begingroup$

Careful! These are two different notions of conjugate.

First we have the complex conjugate, given by $\overline{a+bi} = a-bi$. Then, since we can write a real number $x$ as $x+0i$, the complex conjugate of a real number is itself.

There is also a second idea of a rational conjugate, where as in your example, if $a,b$ are rational and $d$ is squarefree, the conjugate of $a+b\sqrt{d}$ is $a-b\sqrt{d}$.

There is a connection between these two ideas. In general, given a field extension $E/F$, take an algebraic element $\alpha$ of $E$, and let $m(x)$ be it's minimal polynomial over $F$. Then we call the other roots of $m$ in $E$ the conjugates of $\alpha$.

In the case of the extensions $\mathbb{C}/\mathbb{R}$ and $\mathbb{Q}(\sqrt{d})/ \mathbb{Q}$ this agrees with the above.

$\endgroup$
  • $\begingroup$ I can see that those are two different kinds of conjugates, but by treating $i$ and $\sqrt{d}$ as (mere) indeterminate variables, don't we get a unified notion of conjugation, as in mapping from $\alpha+\beta y$ to $\alpha-\beta y$ when this mapping makes sense (as in well defined and preserves algebraic structure)? In both cases, $\alpha$ gets mapped to $\alpha$ and we declare that the conjugate of $\alpha$ is itself. $\endgroup$ – Frenzy Li Jul 30 '18 at 15:19
  • 6
    $\begingroup$ @FrenzyLi In fact, that's what Daniel Mroz is hinting at with his part on field extensions. Conjugation is an automorphism of the bigger field, leaving the smaller field fixed; an element of the Galois group. $\endgroup$ – Babelfish Jul 30 '18 at 16:11
6
$\begingroup$

The notion of a conjugate comes up when you have a naturally defined function $c$ from a set to itself such that for all $x$ you have $c(c(x)) = x$. Then $c(x)$ is the conjugate of $x$.

That happens naturally when the numbers you are thinking about are of the form $a+b\sqrt{d}$ for some $d$. Then you define the conjugating function by switching the sign of $b$.

In this case $2$ will be its own conjugate.

The conjugate of $1 + \sqrt{2}$ is tricky. It's sometimes $1 - \sqrt{2}$ but it's just $1 + \sqrt{2}$ (itself) in the complex numbers, since it's a real number.

$\endgroup$
  • $\begingroup$ Perhaps outside the scope of the question, but I would say a conjugate must be with respect to some subset, i.e. $c(x)=x$ for every $x$ in the sub-set as well as $c(c(x))=x$ for every $x$ in the whole set. Assuming only members of the sub-set satisfy the first condition, we have $c(x \cdot c(x))=c(x)\cdot c(c(x))=x\cdot c(x)$ which means $x\cdot c(x)$ is always in our "special subset" (in other words - a conjugate of $L$ over $K$ is a non-trivial involution in $\mathrm{Aut}(L/K)$) $\endgroup$ – Itai Jul 30 '18 at 13:27
  • $\begingroup$ I think you should clarify "numbers of the form a + b sqrt (d)", it's evidently not well defined when considering real numbers, or when d is a perfect square. Also the "It's sometimes..." sentence could be clarified, it's not like it is non-deterministic, depending on the context and object considered it will be clear what notion of conjugation one is referring to. $\endgroup$ – Maxim Jul 31 '18 at 9:30
2
$\begingroup$

Usually our definition of "conjugate" refers to complex numbers: the conjugate of $a+bi$ is $a-bi$. You could say "complex conjugate" be be extra specific.

Note that $1+\sqrt{2}$ is a real number, so its conjugate is $1+\sqrt{2}$.

A nice way of thinking about conjugates is how they are related in the complex plane (on an Argand diagram). Given a complex number, reflect it across the horizontal (real) axis to get its conjugate. Since $1$, $2$ and $1+\sqrt{2}$ all lie on the real line, they are their own conjugate.

Getting a little bit more advanced, you can define conjugates in different fields. The complex numbers are an extension of the real numbers with the number $i$. We can write this "field extension" as $\mathbb{C}=\mathbb{R}(i)$.

Working instead in $\mathbb{Q}(\sqrt{2})$ (the rationals extended with $\sqrt{2}$), you can define the conjugate of $a+b\sqrt{2}$ as $a-b\sqrt{2}$. But don't worry so much about this if you are new to the topic!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.