These are values in Q. They are not contained in B.
This looks convincing. In fact, that response alone cancels the other questions i would have asked.
But they do bound B from above ^^
But then this seems to be complicating your response.
The values you presented don't follow the rules of B ( for example, r = √1.5 or √(15/10) isn't a rational number). So my question, bound B in what context ?
B:={ r ∊ Q : r2 < 2 } only takes elements from Q. Indeed r2 = 1.5 or r2=15/10 are not contained in it. We are viewing B as subset of Q. And looking for a least upper bound in Q (it is the same what we did for the R story but now R is replaced by Q). For B we cannot apply axiom of completeness and (you can prove that) B has no least upper bound in Q
I have seen the Wikipedia article on this topic, it looks more explanatory. You didn't answer my first question well, probably why i was still confused.
The set of values you presented (2, 1.5, 1.42....) are actually for r belonging to Q. You said they belonged to Q but never stated if it was for r or r².
Sorry I don't understand the question. r can only exist within the definition of A or B. So we can only view it as something contained in a set.
2, 15/10, 142/100 are in Q but not in B. But they do bound B from above.
I pretty much understand you, ok.
please don't repeat this again, it's making you sound like a robot. 😂
How about using the Cauchy sequence approach, I like that one, it looks more understandable.
:P