Let f : Q ⟶ Q be a continuous function. Is there necessarily an interval ( a , b ), with a , b ∈ Q such that a &lt; b, such that the restriction of f to ( a , b ) ∩ Q is a rational function?My guess is that the answer is negative. I tried to prove it as follows: I took a countable and dense set { r n ∣ n ∈ N } of irrational numbers and defined, for each x ∈ Q , f ( x ) = ∑ r n &lt; x 2 − n . This will work, in the sense that f is continuous and that the restriction of f to any interval ( a , b ) is never a rational function. The problem is that, of course, in general, f ( x ) ∉ Q .

Question

Let   f  :      Q    ⟶      Q   be a continuous function. Is there necessarily an interval   (  a  ,  b  ), with   a  ,  b  ∈      Q   such that   a  &amp;lt;  b, such that the restriction of   f to   (  a  ,  b  )  ∩      Q   is a rational function?My guess is that the answer is negative. I tried to prove it as follows: I took a countable and dense set   {      r    n    ∣  n  ∈      N    } of irrational numbers and defined, for each   x  ∈      Q  ,   f  (  x  )  =      ∑                  r        n            &amp;lt;      x            2          −      n      . This will work, in the sense that   f is continuous and that the restriction of   f to any interval   (  a  ,  b  ) is never a rational function. The problem is that, of course, in general,   f  (  x  )  ∉      Q  .

Savanah Hernandez · Accepted Answer

Let&#039;s show the existence of a continuous function   f  :      R    →      R   such that   f  (      Q    )  ⊂      Q   and   f is a not a rational function on any interval.For this, consider   A  =  {      a    1    ,      a    2    ,  …  ,      a    n    …  },   B  =  {      b    1    ,      b    2    ,  …  } two countable dense subsets of       R  , labelled by       N  . Fact: there exists an increasing bijection   f  :  A  →  B. Indeed, consider the binary search tree corresponding to   A with its labelling, and to   B with its labelling. Because   A,   B are both dense, the associated search tree will be complete binary trees. Now map the elements by considering the obvious bijection isomorphism between the two trees. Note that   f extends (uniquely) to a homeomorphism of       R  .Now take   A  =      Q   with some labelling, and   B a subset of rationals such that the denominators of its elements grow very fast ( faster than any polynomial). To be explicit, take   B so that      |    {  b  ∈  B         |    denom  ⁡  (  b  )  ≤  n  }      |    =  o  (      n          ϵ        )for every   ϵ  &amp;gt;  0.Now, assume that   f  (      m    n    )  =            P      (      m      ,      n      )              Q      (      m      ,      n      )       for all       m    n    ∈  (  α  ,  β  ), where   P,   Q are homogenous of degree   d with integer coefficients. There exists for every   n at least   c  n elements in   (  α  ,  β  ) with numerator and denumerator   ≤  n ( a poor estimate, but enough for our purposes). Applying   f we conclude   B contains at least   c  n elements with denominator   ≤  k      n    d  , for all   n. We got a contradiction.

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