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\noindent {\bf Question}

\noindent Show that the rationals ${\bf Q}$ with their usual order
$<$ form an ordered field but not a complete ordered field.

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\noindent {\bf Answer}

\noindent To see that ${\bf Q}$ is not a complete ordered field,
note that the subset $A =\{ a\in {\bf Q}\: |\: a < \sqrt{2}\}$ is
bounded above, for instance by $s = 2$, but has no supremum in
${\bf Q}$: that is, for every rational number $s$ so that $a \le
s$ for every $a\in A$, we have that there exists another rational
number $t$ so that $t < s$ and $a\le t$ for every $a\in A$.  (One
way to see this is to use decimal expansions, and to recall that a
number is rational if and only if its decimal expansion is either
repeating or terminating.)


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