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...
I'm fine with the proof from the notes, its just that since you said it wasn't from first principles I was just wondering on how your proof, which comes from first principle works.
properties of the rational numbers as in the distance d is irrational? (They used the fact that the difference between a rational and an irrational number is also irrational?) So does first principle refer to only using properties of the rational numbers?
I had a read over your proof for this...
Sorry for the late reply. I was looking through my universitys notes and this is what they did, although it’s kind of a different example. Is this kind of similar to what I did?
Yes your right.
So there is no q in Q that is equal to sqrt(2) since its irrational.
For q>sqrt(2) I can do this:
q=m/n >sqrt(2) for it to be an upper bound.
Then using Archimedean principle, that for every epsilon>0 (Take this to be the distance between q and sqrt(2), this is an irrational...
q<sqrt(2) cannot be an upper bound based on the definition of the set though, because again q is rational here, and using Archimedean Principle the distance between q<sqrt(2) and sqrt(2) is irrational, so I can always find a bigger number that is in the set?
I'm not sure how I'm meant to find a contradiction for this. As in q=m/n is not an upper bound?
Could I do this instead? Let q=m/n be an upper bound. Then q=m/n >=sqrt(2)
But there is no q in Q that is equal to sqrt(2) since its irrational. so q=m/n >sqrt(2) for it to be an upper bound.
Then...
Yes I know, I was referring to the original post where they concluded that r is an upper bound if and only if r^2>2. But r^2 for r=-3 is greater than 2, but it's definitely not an upper bound.
The original post is from here...