TITLE: Why is the difference of distinct roots of irreducible $f(x)\in\mathbb{Q}[x]$ never rational?
QUESTION [17 upvotes]: The way I understand it, is that if $f(x)$ is an irreducible polynomial in $\mathbb{Q}[x]$ of degree at least 2, then a difference of distinct roots $a_i-a_j$ is never rational for any of the $a_1,\dots,a_n$ which are the roots of $f(x)$ in $\mathbb{C}$. 
Why is this? If $a_i-a_j\in\mathbb{Q}$ for some distinct roots, what goes wrong? Would it follow somehow that $f(x)$ is reducible over $\mathbb{Q}$? Or perhaps there's a more direct explanation?

REPLY [11 votes]: As a footnote to Pierre-Yves's unimprovable answer, let me add that the theorem becomes false over a field $k$ of characteristic   $p\gt 0 $ : an irreducible polynomial $f(X)\in k[X]$ may very well have distinct roots differing by an element of $k$.  
For example Artin-Schreier's theorem (Lang, Algebra, Ch.6,Th.6.4) implies  that if $char.(k)=p$ and if the polynomial $f(X)=X^p-X-q\in k[X]$ has no zero in  $k $ , then $f(X) $ is actually irreducible over $k$.  Moreover, if $a\in K$ is one of its roots in some  splitting field, the other roots are $a+1,a+2,...,a+p-1$ and are thus obtained by adding elements from the ground field to one of the zeros.   
And how does Pierre-Yves's wonderful argument break down?
It breaks down because $g(X)=f(X+b)-f(X)$ is actually the zero polynomial $g(X)=0\in k[X]$ for any $b\in \mathbb F_p$ so that you no longer have the contradiction present in the characteristic zero case .
(Pierre-Yves's argument is of course valid if you replace the ground field $\mathbb Q$ by any field of characteristic zero)