Least Upper Bound Principle for Q?

Can anyone give me an example showing how Q (rationals), in general,
don't follow the Least Upper Bound Principle (assuming the reals have
already been constructed as a superset of rationals)?

Using purely intuition, it seems like every bounded subset of Q has a
least upper bound. Even if we have a subset like this (defined using
irrationals):

S = { x rational : x^2 = 3 }

sup S = root(3)
inf S = -root(3) correct?

Can someone build a subset of Q bounded above that does not have a
least upper bound?
Or is my definition of Least Upper Bound wrong?

please and thank you

.

Relevant Pages

  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... The fact that the reals are complete is used in many places - analysis ... S does not have a rational least upper bound. ... You _said_ it was not clear to you why the rationals were not ... suprema and infima assumes the real numbers to be defined already. ...
    (sci.math)
  • Order Complete
    ... We remark that a set A is referred to as order complete relative to a ... an upper bound also has a least upper bound. ... How is it the reals are order complete and rationals are not? ...
    (sci.math)
  • Re: Sets and Real Numbers
    ... Fields and Completenes Axioms. ... subset of Q which is bounded above has a least upper bound *in Q*. ... So to complete the rationals, you need to construct the reals, and the ...
    (sci.math)
  • Re: Orlow cardinality question
    ... Tony Orlow wrote: ... as there is no upper bound. ... Does the set of rationals have an ... It's an infinite ...
    (sci.math)
  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... David C. Ullrich wrote: ... The fact that the reals are complete is used in many places - analysis ... S does not have a rational least upper bound. ... rationals are not. ...
    (sci.math)