
#1
Mar912, 07:06 AM

P: 8

1. The problem statement, all variables and given/known data
Find the least upper bound and greatest lower bound (if they exist) of the following sets and state whether they belong to the set: a. {1/n:n[itex]\in[/itex]"Natural Number"} b. {x[itex]\in[/itex]"Rational Number":0≤x≤√5 c. {x irrational:√2≤x2} d. {(1/n)+(1)^{n}:n[itex]\in[/itex]"Natural Number"} 2. Relevant equations Not applicable. 3. The attempt at a solution a. least upper bound does not exist; greatest lower bound is 1 and does belong to the set b. least upper bound is √5 and does not belong to the set; greatest lower bound is 0 and does belong to the set. c. least upper bound is 2 and does not belong to the set; greatest lower bound is √2 and does belong to the set. d. I am not sure about this one, I don't know what the graph would look like. Am I getting the right idea here? Any ideas for d.? Thanks! 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
Mar912, 08:19 AM

HW Helper
P: 3,225

For a. Is 1 <= x, for all x in your set? Revise the definitions of "greatest lower bound", i.e. "infimum".




#3
Mar912, 08:45 AM

P: 8

Well, I thought it was, because x has to be a natural number.




#4
Mar912, 09:09 AM

HW Helper
P: 3,225

Least Upper Bounds 



#5
Mar912, 12:22 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

if n= 1, 1/n= 1 if n= 2, 1/n= 1/2 if n= 3, 1/n= 1/3 if n= 4, 1/n= 1/4 ... I strongly recommend that you write out at least a few of the numbers in each problem. 



#6
Mar912, 01:13 PM

P: 8

This is why I thought that the greatest lower bound was one.
Because n=1, 1/n=1. as far as I know n cannot be smaller than one as a natural number. Doesn't this make 1 the greatest lower bound?? 



#7
Mar912, 02:05 PM

HW Helper
Thanks
PF Gold
P: 7,187





#8
Mar912, 03:11 PM

P: 8

Okay! Thanks!



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