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Prove that if $x<y$ then there is a rational number $r$ with a finite decimal expansion and an integer $k$ such that $x<r<r+10^{-k}<y$

So far. Let $x=a_0.a_1a_2...$, $y=b_0.b_1b_2...$,then there is an index $k$ such that $a_k<b_k$ Now I am having trouble choosing a suitable $r$ and $k$ to satisfy the property.

I was thinking of choosing an $r$ with a decimal expansion $r=c_0.c_1c_2...$ with $c_i=a_i$ for all $i$ and $c_k=a_k$ and then take $c_l>a_l$ for some $l>k$. Do I need to provide a specific choice of $r$? Can someone help with choosing this $r$ and $k$? I am having serious trouble with these decimal expansion proofs.

  • ",then ai=bi for all i<k and ak<bk for k,since x<y" Not quit if $x= 2$ and $y= \pi$ you do not have $a_i = b_i$. But if $x < y$ there must be some $k$ where $a_k \ne b_k$. Let $k$ but that and let $r$ be any decimal between $a_0.a_1..........a_k$ and $b_0.b_2.....b_k$ If they are off by only $10^{-k}$ go one decimal place further and let $k$ be the next value. – fleablood Aug 27 '20 at 01:29
  • So leave that part in quotes out and just suggest there is some index $k$ for which $a_k \neq b_k$? Also would I need to be specific in the number I choose for $r$ between $x$ and $y$? –  Aug 27 '20 at 01:33
  • no,... if more than one exist just pick one. But now this is important. Are you allowed to assume real numbers do have decimal expansions? I think this is an excercise to lead to showing that decimal expansions are valid. – fleablood Aug 27 '20 at 02:21

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$y- x > 0$. So let $k$ be so that $10^{-k} < y -x$. The archmedian principal says that there is a natural $n$ so that $n*10^{k-1} < y \le (n+1)*10^{k-1}$.

So $y-10^{-k} < n*10^{-k}$. But $y-10^{-k} > x$ so

$x < n*10^{-k} < y$.

That's it. We're done. $\frac n{10^{k}}$ is the rational we want.

And Yes, $n*10^{k}$ will agree with the decimal expansions of $x$ and $y$ so long as they agree but ... we don't have to do all that calculating and manipulating

fleablood
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