Give an example of a non-empty subset of rational numbers which is bounded but doesn't have greatest lower bound.
Answers
Answer:
) The Archimedian property of the real-number system was deduced as a consequence of the least-upper-bound axiom. Prove that the set of rational numbers satisfies the Archimedian property but not the least-upper-bound property. This shows that the Archimedian property does not imply the least-upper-bound axiom.
I don't get what I have to do, and what it means -- to satisfy the least-upper-bound-property. I know it is an axiom, I didn't know it is a property. My closest guess is that if I have two rational numbers x<y, I can find an integer n such that nx>y. And I don't need to refer to that axiom, because I can find this n by playing with the integers inside the rational numbers.
Step-by-step explanation:
) The Archimedian property of the real-number system was deduced as a consequence of the least-upper-bound axiom. Prove that the set of rational numbers satisfies the Archimedian property but not the least-upper-bound property. This shows that the Archimedian property does not imply the least-upper-bound axiom.
I don't get what I have to do, and what it means -- to satisfy the least-upper-bound-property. I know it is an axiom, I didn't know it is a property. My closest guess is that if I have two rational numbers x<y, I can find an integer n such that nx>y. And I don't need to refer to that axiom, because I can find this n by playing with the integers inside the rational numbers.