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I've searched for a question, and it has been already asked here Baby Rudin: Chapter 1, Problem 6{d}. How to complete this proof?, but it hasn't yet a satisfactory answer. Can you help? If you go there, you'll see I've done the case $x+y\notin \Bbb{Q}$. Can you guees something about the case $x+y\in \Bbb{Q}$?
Thanks!

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Derso
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  • C'mon guys. Open bounty +100 for that question... – Derso Jan 10 '15 at 12:53
  • in the original question a comment has been added pointing to the solution of the exercises in the book... – Thomas Jan 10 '15 at 13:06
  • Yes. I've checked it, but it's not satisfactory. If you read it, that solution "cheats". It consider $B(x)={b^t|t<x;t\in \Bbb{Q}}$. While the book (Rudin) defines $B(x)={b^t|t\leq x;t\in \Bbb{Q}}$. Is exactly this equality that is causing troubles. @Thomas – Derso Jan 10 '15 at 13:09
  • See this solution, pp. 2-3 . – Tony Piccolo Jan 10 '15 at 13:56

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