An injection $F$ from the set of real numbers to itself with $F(x)-F(y)\not\in\mathbb Q$ for $x\neq y.$ Please help me find an injection, $F$, from the set of real numbers into itself such that $F(x) - F(y)$ is an irrational number for any two distinct real numbers $x$ and $y$.
Thank you.
 A: Consider $\mathbb R$ as vector space over $\mathbb Q$. Complete the set $\{1\}$ to a basis, call it $B$. There is a bijection $f \colon B \to B - \{1\}$ which extends to a linear transformation $F \colon \mathbb R \to \mathbb R$ that satisfies the requirement, because it is irrational everywhere except 0.

A constructive answer without AC:
Lemma:
There is an absolutely convergent sequence of real numbers $x_i$ such that $\sum a_i x_i$ is irrational for every $a_i \in \{-1,0,1\}$ except all $a_i = 0$.
Proof assuming the lemma:
Since $\mathbb R$ and $\{0,1\}^{\mathbb N}$ have equal cardinality, it is enough to show $f \colon \{0,1\}^{\mathbb N} \to \mathbb R$.
Define $f(a_1 a_2 \dots) = \sum a_i x_i$. The difference $f(x)-f(y)$ is always irrational by the lemma.
Proof of the lemma:
Parition $\mathbb N$ into infinitely many infinite subsets $A_i$ and let $x_i = \sum_{k \in A_i} 10^{-k!}$. The series $\sum a_i x_i$ is, except when all $a_i = 0$, a Liouville number and therefore irrational.
(Alternatively, the lemma can be shown with precise but boring diagonalization.)
A: Let us pull this question off from unanwered questions' hell, following Leonid's idea. First, taking additive groups, we look at the quotient $\,\mathbb R/\mathbb Q\,$ . By Lagrange's theorem, $$|x+\mathbb Q|=|\mathbb Q|=\aleph_0\,\,,\,\,\forall x\in \mathbb R,$$and since the different cosets are in fact a partition of the set $\,\mathbb R\,$ , their (disjoint, of course) union is $\,\mathbb R\,$ . It follows that there are uncountable many different cosets as above (why?).
Now, we can make use of the Axiom of Choice in its form called Zermelo's Well Ordering Theorem, so we can introduce a well ordering in $\,\mathbb R\,$ which in turn induces a well ordering in $\,\mathbb R/\mathbb Q\,$, and then form any index (out of an uncoutable set) $\,\alpha\,$ , we cand define $$f:\mathbb R\longrightarrow \mathbb R\,\,,\,\,f(x_\alpha):=y_\alpha$$where $\,y_\alpha\,$ belongs to the $\,\alpha$-th coset in $\,\mathbb R/\mathbb Q\,$ (also here we use AC...)
A: Under AC there is a well-order of $\mathbb R$ with each element having $\lt \mathfrak c$ predecessors.  So just go along the well-order assigning an allowable value to each real.  Each previous element excludes $(\lt \mathfrak c) \times \aleph_0$ values, so you will have at least one available.
