I Supremum of a set, relations and order

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The discussion revolves around the proof of the supremum of the set S = {x ∈ Q: x² < 2}, questioning the validity of certain definitions and assumptions. Participants clarify that an upper bound cannot be defined solely by the condition r² > 2, as negative values can also satisfy this inequality. They emphasize the importance of excluding negative rationals and correctly applying the Archimedean principle to demonstrate that no rational number can equal √2. The conversation highlights the need for a more formal proof that adheres to the properties of rational numbers without invoking irrational numbers. Overall, the thread underscores the complexities involved in establishing the supremum of a set within rational numbers.
  • #31
If you understand this, you should do the second half of the proof for the case where ##s^2 > 2##.
 
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  • #32
PeroK said:
That's the contradiction. And it all started by assuming that ##s = \sup(S)##.
is 2s/n+1/n^2 in S tho?
 
  • #33
lys04 said:
is 2s/n+1/n^2 in S tho?
##s## is rational and ##n## is a natural number, so yes, that is definitely a rational number.
 
  • #34
PeroK said:
##s## is rational and ##n## is a natural number, so yes, that is definitely a rational number.
Ok so you’re supposing s is the supremum of S, which when squared should be a number that is extremely close to 2, but not exactly since there is no sqrt 2 in Q.
And since s is the supremum, there should be no number whose squared is bigger than it?
But how is 2s/n+1/n^2 in S? Even though it’s a rational number. (2s/n+1/n^2)^2=4s^2/n^2+4s/n^3+1/n^4. How is this less than 2?
 
  • #35
lys04 said:
Ok so you’re supposing s is the supremum of S, which when squared should be a number that is extremely close to 2, but not exactly since there is no sqrt 2 in Q.
And since s is the supremum, there should be no number whose squared is bigger than it?
But how is 2s/n+1/n^2 in S?
That question makes no sense.
lys04 said:
Even though it’s a rational number. (2s/n+1/n^2)^2=4s^2/n^2+4s/n^3+1/n^4. How is this less than 2?
That number is arbitrarily small for increasing ##n##.

To be honest, I don't think you understand what I'm doing at all. I'm not sure how much more I can help.

Have you done any subject that requires proofs?
 
  • #36
lys04 said:
TL;DR Summary: prove that a supremum for a set doesn't exist; relations, total order and partial order

Hello, found this proof online, I was wondering why they defined r_2=r_1-(r_1^2-2)/(r_1+2)? i understand the numerator, because if i did r_1^2-4 then there might be a chance that this becomes negative. But for the denominator, instead of plus 2, can i do plus 10 as well? or some other number thats positive
View attachment 342079

I also did some reading on Cartesian product, relations, total order and partial order.
So a Cartesian product AxB is just ordered pair (a,b) where a is an element of A and b is an element of B right.
And a relation is just a subset of the Cartesian product.
Now total and partial orders.
Total order is denoted by < and partial order is denoted by <=? I’m a bit unsure about these, please correct me if I’m wrong.
And a total order relation requires four things:
Reflexive, anti-symmetric, transitive and comparability? I’m a bit unsure what comparability is though.
And for partial order relation I think it just needs to be reflexive, anti-symmetric and transitive?
The product ## A \times B## is not a pair ##(a,b)## but rather the collection of _all_ pairs ##(a,b)## for ##a \in A, b\in B##.
 

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