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Least Upper Bounds

by elizaburlap
Tags: least upper bound
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elizaburlap
#1
Mar9-12, 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
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radou
#2
Mar9-12, 08:19 AM
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For a. Is 1 <= x, for all x in your set? Revise the definitions of "greatest lower bound", i.e. "infimum".
elizaburlap
#3
Mar9-12, 08:45 AM
P: 8
Well, I thought it was, because x has to be a natural number.

radou
#4
Mar9-12, 09:09 AM
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Least Upper Bounds

Quote Quote by elizaburlap View Post
Well, I thought it was, because x has to be a natural number.
Unless I'm missing something, n has to be a natural number. Numbers of the form 1/n, where n is natural, are rational numbers.
HallsofIvy
#5
Mar9-12, 12:22 PM
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Quote Quote by elizaburlap View Post
Well, I thought it was, because x has to be a natural number.
??? There is NO mention of an "x" in problem (a). There is mention of a natural number, n, and the problem talks about the number 1/n for each n:
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.
elizaburlap
#6
Mar9-12, 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??
LCKurtz
#7
Mar9-12, 02:05 PM
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Quote Quote by elizaburlap View Post
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??
As Halls mentioned, ##S = \{1,1/2,1/3,\, ...\}##. Plot a few of those points on the ##x## axis. Then remember that a greatest lower bound of a set is at least a lower bound. Then ask yourself if 1 is a lower bound for ##S##.
elizaburlap
#8
Mar9-12, 03:11 PM
P: 8
Okay! Thanks!
LCKurtz
#9
Mar9-12, 05:57 PM
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Quote Quote by LCKurtz View Post
As Halls mentioned, ##S = \{1,1/2,1/3,\, ...\}##. Plot a few of those points on the ##x## axis. Then remember that a greatest lower bound of a set is at least a lower bound. Then ask yourself if 1 is a lower bound for ##S##.
Quote Quote by elizaburlap View Post
Okay! Thanks!
To whom are you replying? And what have you decided about this problem?


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