# Real number which is different from all rationals [closed]

By diagonalization, it is possible to construct a real number $$r \in [0,1]$$ such that for every rational $$q \in [0,1]$$, there exists an index $$i \in \mathbb{N}$$ such that $$r_i \neq q_i$$ (where $$x_i$$ is the $$i$$'ith digit in $$x$$).

Can we make a stronger claim, and construct a real number $$r \in [0,1]$$ such that for every rational $$q \in [0,1]$$ there exists an index $$i \in \mathbb{N}$$ such that for every $$j \geq i$$, $$r_j \neq q_j$$?

• This is not a research-level question, and as such it is off-topic for this site. It might be more appropriate for math.stackexchange.com . Anyway, a hint: think about $q=k/9$ (assuming decimal digits). – Emil Jeřábek supports Monica Feb 2 at 9:35
• Sorry! I didn't know this website is for research level questions. Should I delete the question? – Larry Feb 2 at 9:40