How to prove $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? How do I prove that $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? $\mathbb{Q}$ is the rational field.
I want to know the detail about the proof. Thanks in advance.
Actually I know any two of them and three of them are linearly independent.

I have no doubt that many slick solutions will be given here. I'll try to post an elementary one; which uses perhaps the most straightforward approach I can imagine. (The only things that are needed are some algebraic manipulation and basic properties of rational numbers; such as that the only solution of $x^2=6y^2$ in $\mathbb Q$ are $x=y=0$.)

This is equivalent to showing that if
$$a+b\sqrt6=c\sqrt3+d\sqrt2$$
then $a=b=c=d=0$.
By squaring both sides of the above equation we get
$$
\begin{align*}
a^2+6b^2+2ab\sqrt6&=3c^2+2d^2+2cd\sqrt6\\
2(ab-cd)\sqrt6=3c^2+2d^2-a^2-6b^2
\end{align*}
$$
Since $a,b,c,d\in\mathbb Q$, this implies
$$
\begin{align*}
ab-cd&=0\\
3c^2+2d^2-a^2-6b^2&=0
\end{align*}
$$
which is the same as
$$
\begin{align*}
ab&=cd\\
3c^2+2d^2&=a^2+6b^2
\end{align*}
$$
Suppose that $b\ne0$, $c\ne0$. (I'll leave the solution of these cases to the reader.) Then we can rewrite the first equation as $\frac ac = \frac db = x$, where $x\in\mathbb Q$. Now the second equation becomes
$$
\begin{align*}
3c^2+2x^2b^2&=x^2c^2+6b^2\\
x^2(2b^2-c^2)&=3(2b^2-c^2)\\
(x^2-3)(2b^2-c^2)&=0
\end{align*}$$
This implies that $x^2=3$ or $2b^2=c^2$. None of them has non-zero solutions in rational numbers.

Alternatively, we could solve the case $b=0$ first. Once we're looking only at $b\ne0$, we can assume w.l.o.g. that $b=1$. (After dividing both sides with $b$.)
So we're looking at $a+\sqrt6=c\sqrt2+d\sqrt2$. And squaring, as above, gives us now $a=cd$. Thus we get
\begin{align*}
cd-c\sqrt2-d\sqrt3+\sqrt6&=0\\
(c-\sqrt3)(d-\sqrt2)&=0
\end{align*}
which means that $c=\sqrt3$ or $d=\sqrt2$. In either case, it is the contradiction with the assumption $c,d\in\mathbb Q$.
