Interesting post 😁 but i have a series questions that may follow from the one i'm about to ask, it would depend on your response.

2, 15/10, 142/100 ...

Are these set of values for r² belonging to Q or r belonging to Q ?

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Interesting post 😁 but i have a series questions that may follow from the one i'm about to ask, it would depend on your response.

2, 15/10, 142/100 ...

Are these set of values for r² belonging to Q or r belonging to Q ?

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These are values in Q. They are not contained in B. But they do bound B from above ^^

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 : r^{2} < 2 } only takes elements from Q. Indeed r^{2} = 1.5 or r^{2}=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.

2, 15/10, 142/100 are in Q but not in B. But they do bound B from above.

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.