Rational numbers - bounded subset with no least upper bound

ciarax
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Homework Statement




Give an example of a bounded subset of Q which has no least upper
bound in Q. Explain why your answer has this property.


Homework Equations





The Attempt at a Solution



[1/8, 1/4, 3/8, 1/2, 5/8, 3/4...infinity]
is this correct?
 
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ciarax said:

Homework Statement




Give an example of a bounded subset of Q which has no least upper
bound in Q. Explain why your answer has this property.


Homework Equations





The Attempt at a Solution



[1/8, 1/4, 3/8, 1/2, 5/8, 3/4...infinity]
is this correct?

No. Hint: What kind of numbers are missing from Q? Find a bounded set that has one of them for its lub.
 
ciarax said:
Give an example of a bounded subset of Q which has no least upper
bound in Q. Explain why your answer has this property.

[1/8, 1/4, 3/8, 1/2, 5/8, 3/4...infinity]
Your set appears to be integer multiples of 1/8. This set is not bounded, so doesn't qualify as an example in this problem.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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