Does the completeness axiom apply to bounded subsets of the rationals?

[annoymage]

Homework Statement

let A be a set of all positive rational number such that $p^2<2$

B be a set of all positive rational number such that $p^2>2$

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 realized 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

 

[fzero]

The completeness axiom applies to the reals, there is no contradiction. As you say, the set A demonstrates that Q does not have the least-upper-bound property, so Q is not complete. The completeness axiom guarantees that a subset of Q has a least-upper-bound in R. However that least-upper-bound is not in Q.

 

[annoymage]

AAAAAAAAAAAAAAAAAAA in R, not Q

Thank YOUUUUUUU, I'm so stupid