
#1
Oct310, 05:40 PM

P: 26

Find the supremum and infimum of S, where S is the set
S = {√n − [√n] : n belongs to N} . Justify your claims. (Recall that if x belongs to R, then [x] := n where n is the largest integer less than or equal to x. For example, [7.6] = 7 and [8] = 8) I found my infimum to be 0 and my supremum to be 1, but how do i go about proving them? Help please. 



#2
Oct310, 06:52 PM

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It's pretty obvious that all of the elements in S are in [0,1), right? And you shouldn't have any trouble showing the infimum is 0. Just find a n where f(n)=sqrt(n)[sqrt(n)] is 0. Showing the supremum is 1 is a little harder. You want to find a sequence of integers a_n such that f(a_n) approaches 1.




#3
Oct310, 07:06 PM

P: 851





#4
Oct310, 07:10 PM

P: 617

Analysis Question: Find the supremum and infimum of S,where S is the set S = {√n − [√
I actually don’t think an ε proof will work for this, since n must be a natural number, unless you restrict ε to naturals too.
I’m not sure on this by any means but this is the approach I would take. First because S is a subset of the reals it must have a LUB. Arguing that 1 is an upper bound is easy. I would try to show that if √n − [√n]<1 you can find an √m − [√m] that’s even closer to 1 which would make 1 the smallest upper bound. These three facts together ill show that 1 is the LUB. 



#5
Oct310, 08:05 PM

P: 851

Originally, I was going to use density of R. But since there are countably many irrationals in the set proposed it is obvious that I can't use it. Since the set proposed is a subset of all irrational numbers between (0,1). Your approach is similar to the epsilon argument and I doubt it would work. Even the sequence approach suggested is a little hairy as it requires an epsilon argument to show convergence. And one cannot guarantee there are no "jumps". Eg. sqrt(1023) = 31.98437118..... and then sqrt(1024)=32. 



#6
Oct310, 08:46 PM

P: 617

But I got it, man did this take me awhile, but my idea can work. I don't wanna give it away, but here's the general idea. For contradiction sake, assume that there is a n in N, and for all other m in N that: 1 –(√n − [√n]) < 1 – (√m − [√m]). Simplify this, then pick a clever m in terms of n that will get rid of the radicals that cause problems. 



#7
Oct310, 09:14 PM

P: 851

I said "Since the set proposed a subset of all irrational numbers between (0,1)." Transcendentals are irrationals right ? I don't see where I went wrong. What exactly are you getting a contradiction from ? I don't follow your argument. Using your argument, I suceeded in showing that (√n − [√n]) is not bounded above by any number of the form (√m − [√m]). Maybe I am missing something but that doesn't prove suprema. 



#8
Oct310, 09:34 PM

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Uh, pick a_n=n^21. That's the worst case in some sense. What is [n^21]? What's the limit as n>infinity of the difference? And for Simkate, please don't double post again, ok?




#9
Oct310, 09:41 PM

P: 851

Did you not say f(n) = sqrt(n)  [sqrt(n)] Then f(a_n) = sqrt(n^2 1)  [ sqrt(n^2 1)] Seems to me like this may not even converge. For some large n we could find f(a_n) =0 



#10
Oct310, 09:54 PM

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#11
Oct310, 10:02 PM

P: 851

It is defintely too late for me to be thinking :(. Your solution works. Hopefully OP can use it. In the analysis books I have seen limits appear after suprema and the like. Mine is certainly like that. 



#12
Oct310, 10:31 PM

P: 129

 sqrt(n^2 + 1) = n1 wooot? I think he meant that it is never an integer because for it to be an integer it need to be a quadratic number which n^2 + 1 never is. 


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