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If you understand this, you should do the second half of the proof for the case where ##s^2 > 2##.
The discussion revolves around the concept of supremum in the context of rational numbers and the properties of upper bounds. Participants explore definitions and properties of relations, total orders, and partial orders, while also analyzing a proof related to the supremum of a specific set of rational numbers.
Participants express disagreement regarding the validity of the original proof and its conclusions. Multiple competing views are presented on the nature of upper bounds and the existence of a supremum for the set of rational numbers defined.
Participants note limitations in the original proof, including the need to consider both positive and negative rational numbers and the implications of the Archimedean principle. There is also mention of the necessity to avoid assumptions about the properties of real numbers in the proof.
is 2s/n+1/n^2 in S tho?PeroK said:That's the contradiction. And it all started by assuming that ##s = \sup(S)##.
##s## is rational and ##n## is a natural number, so yes, that is definitely a rational number.lys04 said:is 2s/n+1/n^2 in S tho?
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.PeroK said:##s## is rational and ##n## is a natural number, so yes, that is definitely a rational number.
That question makes no sense.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 number is arbitrarily small for increasing ##n##.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?
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##.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
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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?