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.

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    $\begingroup$ 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
    Commented Jul 30, 2018 at 12:10
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    $\begingroup$ 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. $\endgroup$ Commented Jul 30, 2018 at 12:11

3 Answers 3


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.

  • $\begingroup$ 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
    Commented Jul 30, 2018 at 15:19
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    $\begingroup$ @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
    Commented Jul 30, 2018 at 16:11

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.

  • $\begingroup$ 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
    Commented Jul 30, 2018 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
    Commented Jul 31, 2018 at 9:30

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!


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