find supremum and infimum of {((2p-1)/(2pq+1))| p,q ∈ Z}

[erblina]

find the supremum and infimum of the following set :

{((2p-1)/(2pq+1))| p,q ∈ Z}

I just do not know how to start this
please help me !

 

[stapel]

find the supremum and infimum of the following set :

{((2p-1)/(2pq+1))| p,q ∈ Z}

I just do not know how to start this

A good place to start would be with the definitions. What is the "supremum" of a set? What is the "infimum" of a set? (In this case, does "Z" stand for "the set of integers"?)

What are your thoughts? What have you found, when working with the sorts of elements in this set? What are those elements? Can you figure out any particular characteristics of them?

Please be complete. Thank you!

 

[erblina]

A good place to start would be with the definitions. What is the "supremum" of a set? What is the "infimum" of a set? (In this case, does "Z" stand for "the set of integers"?)

What are your thoughts? What have you found, when working with the sorts of elements in this set? What are those elements? Can you figure out any particular characteristics of them?

Please be complete. Thank you!

well Z is the set of all integers and i tried firstly taking p=q>0 and i found this set of elements {1/3 ; 3/9 ; 5/19 ; 7/33 ; 9/51 ...} so i have this set
{(2n-1)/(2n^2+1)| n∈N(set of all natural numbers)} and now does it mean that the supremum is 0 while n raises the elements get smaller and infimum 1 ? Now i need to see when p>q or p<q and when both of them are negative ? do i have to prove all these cases?

 

[HallsofIvy]

Stapel said "A good place to start would be with the definitions. What is the "supremum" of a set? What is the "infimum" of a set?" and you did NOT answer those questions. In fact, since you asked "it mean that the supremum is 0 while n raises the elements get smaller and infimum 1?", it is clear that you do NOT know those definitions! If you knew those definitions then you would know that the supremum of a set is never less than the infimum.

The "supremum" of a set of numbers is a number larger than or equal to any number in the set (an "upper bound") such that there is no upper bound less (the "least upper bound"). In particular, the supremum is larger than or equal to any number in the set (if the supremum is equal to a number in the set, it must be the largest number in the set and, conversely, if the set has a largest number, it is the supremum).

The "infimum" of a set of numbers is a number less than or equal to any number in the set (a "lower bound") such that there is no upper bound larger (the "greatest lower bound"). In particular, the supremum is less than or equal to any number in the set (if the infimum is equal to a number in the set, it must be the smallest number in the set, and, conversely, if a set has a smallest number, it is the infimum.)

Okay, I will accept that this was a typo and you accidently reversed the words "supremum" and "infimum". You can make a fraction large by making the numerator as large as possible and the denominator as small as possible. Here, the numerator is 2p- 1 and you can make that as large as possible by making p as large as possible. The denominator is 2pq+ 1 as small as possible by taking q= 0. In that case you get the fraction is (2p-1)/1= 2p-1.

 

[erblina]

Stapel said "A good place to start would be with the definitions. What is the "supremum" of a set? What is the "infimum" of a set?" and you did NOT answer those questions. In fact, since you asked "it mean that the supremum is 0 while n raises the elements get smaller and infimum 1?", it is clear that you do NOT know those definitions! If you knew those definitions then you would know that the supremum of a set is never less than the infimum.

The "supremum" of a set of numbers is a number larger than or equal to any number in the set (an "upper bound") such that there is no upper bound less (the "least upper bound"). In particular, the supremum is larger than or equal to any number in the set (if the supremum is equal to a number in the set, it must be the largest number in the set and, conversely, if the set has a largest number, it is the supremum).

The "infimum" of a set of numbers is a number less than or equal to any number in the set (a "lower bound") such that there is no upper bound larger (the "greatest lower bound"). In particular, the supremum is less than or equal to any number in the set (if the infimum is equal to a number in the set, it must be the smallest number in the set, and, conversely, if a set has a smallest number, it is the infimum.)

Okay, I will accept that this was a typo and you accidently reversed the words "supremum" and "infimum". You can make a fraction large by making the numerator as large as possible and the denominator as small as possible. Here, the numerator is 2p- 1 and you can make that as large as possible by making p as large as possible. The denominator is 2pq+ 1 as small as possible by taking q= 0. In that case you get the fraction is (2p-1)/1= 2p-1.

Ok , thank you very much for these explanations i can understand much better now and i'm gonna write all of my combinations that i did because i don't know if i'm doing it right . Now like i see i need to prove for more values of p and q because in each case there are different sets , i began as it follows ,

1. For p=q=0 i have this set {-1,-1,-1,...}
2. For p>0 and q=0 its {(2p-1)| p>0 , p is integer(while its bigger than 0 its a natural )} this is equal to {1,3,5,7,...} so the inf=1 and sup= does not exist
3. For p=0 and q>0 or q<0 its again {-1,-1, ...}
4. For p,q>0 and p=q its {(2p-1)/(2p^2+1)| p>0 , p integer } its equal to {1/3 ; 3/9 ; 5/19; ...} inf =0 and sup=1/3
5. For p,q>0 and p different from q {(2p-1)/(2pq+1)| p,q>0 } while p,q>0 2p<2pq follows 2p-1<2pq+1 0<(2p-1)/(2pq+1)<1 inf=0 and sup=1
so its this right till now and can you please tell me what can i conclude in the 1. and 3. step for inf and sup ?
thank you very much btw.

 

[HallsofIvy]

Ok , thank you very much for these explanations i can understand much better now and i'm gonna write all of my combinations that i did because i don't know if i'm doing it right . Now like i see i need to prove for more values of p and q because in each case there are different sets , i began as it follows ,

1. For p=q=0 i have this set {-1,-1,-1,...}

Either you are misunderstanding the problem or you are writing this wrong. "For p= q= 0", you do NOT get any set at all You get the single number -1. For any specific values of p and q, you get a single member of the set. Perhaps you mean the subset {-1}. (A set cannot contain more than one copy of a single item.)

2. For p>0 and q=0 its {(2p-1)| p>0 , p is integer(while its bigger than 0 its a natural )} this is equal to {1,3,5,7,...} so the inf=1 and sup= does not exist

This tells you, since these numbers are in the set that the original set does not have a supremum. It does NOT tell you anything about the infimum of the original set.

3. For p=0 and q>0 or q<0 its again {-1,-1, ...}

Again, not a set- p= 0 and q any positive number gives the single number, -1. (or the set {-1})

4. For p,q>0 and p=q its {(2p-1)/(2p^2+1)| p>0 , p integer } its equal to {1/3 ; 3/9 ; 5/19; ...} inf =0 and sup=1/3

The infimum of that subset is 0 which tells you that the infimum of the original set is less than or equal to 0.

5. For p,q>0 and p different from q {(2p-1)/(2pq+1)| p,q>0 } while p,q>0 2p<2pq follows 2p-1<2pq+1 0<(2p-1)/(2pq+1)<1 inf=0 and sup=1
so its this right till now and can you please tell me what can i conclude in the 1. and 3. step for inf and sup ?
thank you very much btw.

You have, in 1, shown that a subset, and therefore the entire set, contains unboundedly large numbers so has no supremum. You have not yet concluded anything about the infimum. But what if you take q= 0 and p< 0?

 

[erblina]

Either you are misunderstanding the problem or you are writing this wrong. "For p= q= 0", you do NOT get any set at all You get the single number -1. For any specific values of p and q, you get a single member of the set.


This tells you, since these numbers are in the set that the original set does not have a supremum. It does NOT tell you anything about the infimum of the original set.


Again, not a set- p= 0 and q any positive number gives the single number, -1.


The infimum of that subset is 0 which tells you that the infimum of the original set is less than or equal to 0.


You have, in 1, shown that a subset, and therefore the entire set, contains unboundedly large numbers so has no supremum. You have not yet concluded anything about the infimum. But what is you take q= 0 and p< 0?

so if i take this q= 0 and p< 0 ill have another subset which contains {-3 ,-5,-7,...} like it seems the biggest element is -3 but i can either tell nothing about supremum and infimum of the original set is it right ? So what can i actually say about this problem ? Can i writethe supremum and infimum of all the subsets that i got because how it seems it is impossible to give an answer in general ?

 

[Ishuda]

find the supremum and infimum of the following set :

{((2p-1)/(2pq+1))| p,q ∈ Z}

I just do not know how to start this
please help me !

As a different way of saying what HallsofIvy said we have
(1) Supremum: Is there a number so that it is larger than every number in the set [larger than the number for every combination of p and q]. If so, there is a maximum for the set and the smallest maximum is the supremum. If not, there is no supremum.

(2) Infimum: Is there a number so that it is smaller than every number in the set [smaller than the number for every combination of p and q]. If so, there is a minimum for the set and the largest minimum is the infinmum. If not, there is no infimum.

So, is there a number larger than the number for every combination of p and q.?

Is there a number smaller than the number for every combination of p and q.?