Conjugate of real number

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.

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.

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!