Given the Point set S - Part (a)

[nasi112]

Given the point set [MATH]S: \{i, \frac{i}{2}, \frac{i}{3}, \frac{i}{4}, . . .\}[/MATH]
(a) Is [MATH]S[/MATH] bounded?

 

[Deleted member 4993]

Given the point set [MATH]S: \{i, \frac{i}{2}, \frac{i}{3}, \frac{i}{4}, . . .\}[/MATH]
(a) Is [MATH]S[/MATH] bounded?

Do you know the definition of a bounded set? Do you know the characteristics of a bounded set?

I have to repeat myself!!

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[HallsofIvy]

First, what is i? Is i a fixed number or is this to include all possible values (integers?)?

 

[nasi112]

Do you know the definition of a bounded set? Do you know the characteristics of a bounded set?

No, I don't know.

First, what is i? Is i a fixed number or is this to include all possible values (integers?)?

[MATH]i = \sqrt{-1}[/MATH]

Please show us what you have tried and exactly where you are stuck.

Can I consider the set [MATH]S[/MATH] as a sequence [MATH]\frac{i}{n}[/MATH]?
Then, I take the limit[MATH] \lim_{n\to\infty} \frac{i}{n} = 0[/MATH], then I conclude [MATH]S[/MATH] is bounded!

Or
Can I take an integer n = 100, for example, then I say since input [MATH]= 100[/MATH] (bounded), output [MATH]= \frac{i}{100}[/MATH] (bounded), then [MATH]S[/MATH] is bounded?

 

[Steven G]

No, I don't know.

I for one will not help you until you supply the definition the Sir Subhotosh Khan asked for.

 

[Deleted member 4993]

No, I don't know.

You need to read your textbook or use Google to learn that (definition of bounded set) first.

 

[pka]

Given the point set [MATH]S: \{i, \frac{i}{2}, \frac{i}{3}, \frac{i}{4}, . . .\}[/MATH](a) Is [MATH]S[/MATH] bounded?

Once we know that \(i\) is the solution to the equation \(x^2+1=0\)
then we know that \(|i|=1\) so \((\forall n)\left[\left|\dfrac{i}{n}\right|<\dfrac{1}{n}\right]\).
Is the set \(S\) bounded?

 

[nasi112]

Thanks guys for replying. I think that there is no need for any definition. If lower bound is [MATH]\frac{i}{\infty} = 0[/MATH] and upper bound is [MATH]i[/MATH], then the set [MATH]S[/MATH] is bounded.

Can I consider the set [MATH]S[/MATH] as a sequence [MATH]\frac{i}{n}[/MATH]?
Then, I take the limit[MATH] \lim_{n\to\infty} \frac{i}{n} = 0[/MATH], then I conclude [MATH]S[/MATH] is bounded!

It is funny that you are asking for my working and where I am struggling, but then all of you ignore my working. HaHa. It would be nice if anyone of you said yes your answer is correct and this is one approach to solve the problem, and then suggesting reading the definition as another approach.

Once we know that \(i\) is the solution to the equation \(x^2+1=0\)
then we know that \(|i|=1\) so \((\forall n)\left[\left|\dfrac{i}{n}\right|<\dfrac{1}{n}\right]\).
Is the set \(S\) bounded?

I consider this approach as brilliant as its owner (yes it is bounded)

 

[Steven G]

Is it not clear that the least upper bound is i and the greatest lower bound is 0?