The set of rational numbers is not order complete. Let us consider, for example, a set :
"A:=\\{x\\in\\mathbb{Q} : x^2<2\\}"
We can clearly see that "A" is bounded in "\\mathbb{Q}" , for example we have "\\forall x\\in A, x<2" , as we know that for all "y\\geq2, y^2\\geq2^2=4>2". But the set "A" does not have a precise upper bound (supremum), as if we had "x_0 = \\sup A" , then (we could prove that) "x_0^2=2". But we know that there is no such rational number for arithmetical reasons. Therefore "\\mathbb{Q}" is not order complete.
In fact the set of real numbers "\\mathbb{R}" is obtained from "\\mathbb{Q}" by order-completing it.
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