= Bounded rational sequences can converge outside the rationals
{title2=$q_n=\lfloor10^n\sqrt2\rfloor/10^n$}
The <rational numbers> $q_n=\lfloor10^n\sqrt2\rfloor/10^n$ lie in $[1,2]$ and obey $0\leq\sqrt2-q_n<10^{-n}$. They form a <convergent sequence> in the <real numbers>, with <limit of a sequence> $\sqrt2$. Every <subsequence> has that same real limit. Because $\sqrt2$ is an <irrational number>, no <subsequence> converges to a point in the <rational numbers> with their usual <metric>. Boundedness alone therefore does not give the corresponding theorem in every <metric space>; <completeness of the real numbers> is doing real work here. The <floor function> simply truncates the decimal expansion in this example.
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