Show that each real number is the supremum of a set of a rational numbers and also the supremum of a set of irrational numbers. He is what i got so far. Let $$Q= [m/n | m,n \in Z, where n \not=0]$$ The set $Q$ is not empty since $0 \in Q$. Since Q is not empty by the completeness axiom $Q$ is bounded above. So by definition there is a number $a \in R$ such that $a=SupQ$, Now the question is how do how show that $a$ is the supremum. I know that it is supposed to simple but i am having a hard time showing a is the least upper bound
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Not every non-empty set is bounded from above. – Amitai Yuval Sep 16 '15 at 18:06
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What is your model of a "real number"? – Christian Blatter Sep 16 '15 at 18:16
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Yeah in this particular problem the teacher told me that the set is bounded above so this part is true – user146269 Sep 16 '15 at 18:18
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possible duplicate of Show that $\mathbb{Q}$ is dense in the real numbers. (Using Supremum) – Deliasaghi Sep 16 '15 at 18:26