Proving LUB and GLB Properties in Ordered Fields

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1. Prove that an ordered field has the LUB property iff it has the GLB property.


I know that I need to prove that if the ordered field has the GLB property, then it has the LUB property, and that if the ordered field does NOT has the GLB property, then it also does not have the LUB property. I'm just really stuck on how to start the proof - do you assume the ordered field is bounded?

Any help would be appreciated!
 
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Hint: if a set S has a lower bound, what can you say about the set (-1)*S?
 
That it has an upper bound?...
 
ragnes said:
That it has an upper bound?...

Yes. And how is the lower bound of S related to the upper bound of (-1)*S?
 
ragnes said:
That it has an upper bound?...
Don't guess! If U is the LUB, it is, first, of all, an upper bound. In other words, for any x in the field, x\le U. Multiplying both sides by -1, -x\ge -U. But if y is any member of the field, x= -y is also in the field and so y= -x\ge -U. That is, -U is a lower bound. Now you need to show it is the greatest lower bound.

Don't forget that this is an "ordered field", not necessarily the field of real numbers. What is meant by "-1"? Have you proven or can you prove that "if a< b, then -a> -b"?
 
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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