Prove that there does not exist any continuous function $f : \mathbb R → \mathbb R$ such that $f(x)$ is rational if and only if $f(x + 1)$ is irrational. What theorems can I use to prove the statement?

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    $\begingroup$ Isn't $f(x+1)-f(x)$ always irrational? $\endgroup$ – Lord Shark the Unknown Apr 23 '18 at 5:53
  • $\begingroup$ My first thought was a cardinality argument. My second thought is I'm about to crash, so maybe I'll look at this tomorrow and maybe not. $\endgroup$ – Michael Hardy Apr 23 '18 at 6:05

The two continuous functions $x\mapsto f(x)\pm f(x+1)$ take only irrational values, hence are both constant. Then their sum $2f$ is also constant - but the constant can neither be rational nor irrational.

  • $\begingroup$ What if we rephrase the question as follows: $f:\mathbb R \to L,$ where $L$ is a linearly ordered set, so ordered that between any two elements of $L$ there is another, and $K\subseteq L$ and between any two elements of $L$ there is an element of $K,$ and $K$ is countable. And $L$ has the least-upper-bound property. And let us assume $|L|>1.$ And $f(x)\in K$ if and only if $f(x+1)\notin K.$ Can that happen? $\endgroup$ – Michael Hardy Apr 23 '18 at 6:28
  • $\begingroup$ Thanks! but i still have one question about it, what has the question to do with the "continuous function" part? $\endgroup$ – Bella Apr 23 '18 at 6:38
  • $\begingroup$ It looks like the continuity condition is not necessary. The solution above does not use such a property of $f(x)$, does it? $\endgroup$ – pabodu Apr 23 '18 at 13:20
  • $\begingroup$ @pabodu Yes, it does. You can have non constant discontinuous functions with all its values irrational. $\endgroup$ – Julián Aguirre Apr 23 '18 at 14:02

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