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## Homework Statement

let A be a set of all positive rational number such that [itex]p^2<2[/itex]

B be a set of all positive rational number such that [itex]p^2>2[/itex]

## Homework Equations

n/a

## The Attempt at a Solution

Set A is clearly non empty, and is a subset of real number, anyway i can choose 3 is upperbound, therefore upperbound exist, so by completeness axiom, supremum exist.

But the book here said

"Set A is bounded above, in fact every element in B a the upperbound of A. Since B has no smallest element, A has no least upper bound/ supremum in Q."

i'm really sure i'm not wrong. But am i wrong?

p/s; i just realised that this book define least-upper-bound property(more general case from completeness axiom), and also above example are the counterexample that proves Q does not have least-upper-bound property(follows from what the book have shown, not mine).

But aren't this contradicting the completeness axiom?

since Q is a subset of R, and any non-empty subset of Q that bounded from above has supremum(from completeness axiom), therefore Q has the least-upper-bound property.

help, where i gone wrong T_T

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