I was reading this answer explaining why all real number have a decimal representation. I think it is really a nice explanation but I don't really see were (I think it is a little hidden) we use the density of $\mathbb Q$ into $\mathbb R$ (which I think somehow we should use).
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It seems to me the linked answer is virtually a proof that $\mathbb Q$ is dense in $\mathbb R$. It shows that a subset of $\mathbb Q$—fractions with denominators which are powers of $10$—has elements arbitrarily close to any real number. Which makes the same true of $\mathbb Q$, thereby proving that $\mathbb Q$ is dense in $\mathbb R$.
So the reason it doesn't use density of $\mathbb Q$ in $\mathbb R$ is that it proves it instead.
timtfj
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Oh, l see. I could use the same reasonning for base b, where $b \ge 2$ but instead of dividing the interval into $10$ pieces I divide in $b$ pieces right? For example for base $2$ I just choose if the number is in the left or right of the interval. – roi_saumon Jan 22 '19 at 20:35
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1Yes, you'd be doing essentially the same thing but end up with a representation in base $b$. – timtfj Jan 22 '19 at 20:52
