How can I find the maximum, minimum, supremum, and infimum of a given set?

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In summary, the following is true:-There are no max or min of the set.-The supremum is $1$, and the infimum is $-1$.
  • #1
evinda
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Hey! :)
I am given the following exercise:
Find max,min,sup,inf of the set:
$$B=\{\frac{(-1)^{n}m}{n+m},n,m=1,2,...\}$$
I thought the following:
$$supB=\frac{1}{3},maxB=\frac{1}{3},infB=0, \nexists minB$$

Are the above right?? If yes,how can I prove that it is actually like that??
 
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  • #2
evinda said:
Hey! :)
I am given the following exercise:
Find max,min,sup,inf of the set:
$$B=\{\frac{(-1)^{n}m}{n+m},n,m=1,2,...\}$$
I thought the following:
$$supB=\frac{1}{3},maxB=\frac{1}{3},infB=0, \nexists minB$$

Are the above right?? If yes,how can I prove that it is actually like that??

Hola! :D

Suppose we pick $m=n=2$, then I get \(\displaystyle B=\frac 2 4\).
Isn't that greater than your $\sup B$?

And suppose we pick $m=n=1$, isn't $B$ smaller than your $\inf B$ then?
 
  • #3
I like Serena said:
Hola! :D

Suppose we pick $m=n=2$, then I get \(\displaystyle B=\frac 2 4\).
Isn't that greater than your $\sup B$?

And suppose we pick $m=n=1$, isn't $B$ smaller than your $\inf B$ then?

Oh,yes!Right! :eek: So:
$$ supB=\frac{1}{2},maxB=\frac{1}{2},infB=\frac{-1}{2},minB=\frac{-1}{2}$$

Is there a way to prove that it is like that? (Thinking)
 
  • #4
evinda said:
Oh,yes!Right! :eek: So:
$$ supB=\frac{1}{2},maxB=\frac{1}{2},infB=\frac{-1}{2},minB=\frac{-1}{2}$$

Is there a way to prove that it is like that? (Thinking)

Good!

... but suppose we pick $m=3, n=2$, then I get \(\displaystyle B=\frac 3 5\).
Isn't that greater than your new $\sup B$?

Let's take a look at B.
$$B=\frac{(-1)^n m}{n+m}$$

To find the largest value, we need a numerator that is a big as possible, and a denominator that is as small as possible.
Oh, and we have the additional condition that the fraction should be positive.

Suppose we pick a "large" value for the number in the numerator. Say $m=1000$.
What should you pick for $n$ to make $B$ as big as possible?
 
  • #5
I like Serena said:
Good!

... but suppose we pick $m=3, n=2$, then I get \(\displaystyle B=\frac 3 5\).
Isn't that greater than your new $\sup B$?

Let's take a look at B.
$$B=\frac{(-1)^n m}{n+m}$$

To find the largest value, we need a numerator that is a big as possible, and a denominator that is as small as possible.
Oh, and we have the additional condition that the fraction should be positive.

Suppose we pick a "large" value for the number in the numerator. Say $m=1000$.
What should you pick for $n$ to make $B$ as big as possible?

I understand..I think that we should pick then $n=2$..Right?
 
  • #6
evinda said:
I understand..I think that we should pick then $n=2$..Right?

Yep. :)
Care to make a new guess about the largest value?
And what about the smallest value?
 
  • #7
I like Serena said:
Yep. :)
Care to make a new guess about the largest value?
And what about the smallest value?

Hmm..I don't really know.. (Thinking) If we want the numerator to be as big as possible,we have to take $ m\to \infty$ and $n=2$..But then we will have $\frac{\infty}{\infty}$ (Worried).. Could you give me a hint how I can find it? :eek:
 
  • #8
evinda said:
Hmm..I don't really know.. (Thinking) If we want the numerator to be as big as possible,we have to take $ m\to \infty$ and $n=2$..But then we will have $\frac{\infty}{\infty}$ (Worried).. Could you give me a hint how I can find it? :eek:

What is the following limit?
$$\lim_{m \to \infty} \frac m {2+m} =\lim_{m \to \infty} \frac {1} {\frac 2 m + 1}$$
 
  • #9
I like Serena said:
What is the following limit?
$$\lim_{m \to \infty} \frac m {2+m} =\lim_{m \to \infty} \frac {1} {\frac 2 m + 1}$$

It is equal to $1$! So,this is the supremum,right?? :rolleyes:
And,to find the infimum,we pick $n=1$ and calculate the limit $\lim_{m \to \infty}\frac{-m}{1+m}=-1$,so the infimum is $-1$,or am I wrong?
So,$supB=1$ and $infA=-1$ ?
 
  • #10
evinda said:
It is equal to $1$! So,this is the supremum,right?? :rolleyes:
And,to find the infimum,we pick $n=1$ and calculate the limit $\lim_{m \to \infty}\frac{-m}{1+m}=-1$,so the infimum is $-1$,or am I wrong?
So,$supB=1$ and $infA=-1$ ?

Yes! (Mmm)
 
  • #11
I like Serena said:
Yes! (Mmm)

And... there is no min and max of the set,right? (Thinking)
 
  • #12
evinda said:
And... there is no min and max of the set,right? (Thinking)

Right!
 
  • #13
I like Serena said:
Right!

And how can I prove it?
 
  • #14
evinda said:
And how can I prove it?

Note that since $m,n \ge 1$, we have that:
$$-1 < \frac{(-1)^n m}{n+m} < 1$$
This is true because the absolute value of the numerator is always smaller than the denominator.

Furthermore, we have already found sequences approaching both $-1$ and $1$.

Therefore $\sup B = 1, \inf B = -1$, and $\min B, \max B$ do not exist. $\qquad \blacksquare$
 
  • #15
I like Serena said:
Note that since $m,n \ge 1$, we have that:
$$-1 < \frac{(-1)^n m}{n+m} < 1$$
This is true because the absolute value of the numerator is always smaller than the denominator.

Furthermore, we have already found sequences approaching both $-1$ and $1$.

Therefore $\sup B = 1, \inf B = -1$, and $\min B, \max B$ do not exist. $\qquad \blacksquare$

Great!Thank you very much! :)
 

FAQ: How can I find the maximum, minimum, supremum, and infimum of a given set?

What is the difference between max and sup?

Max refers to the maximum value in a set of numbers, while sup (supremum) is the least upper bound of a set. This means that sup may not necessarily be a member of the set, but it is greater than or equal to all the values in the set.

How do you find the maximum or minimum value in a data set?

To find the maximum or minimum value in a data set, you can arrange the numbers in ascending or descending order and then select the first or last number, respectively. Alternatively, you can use a mathematical function or algorithm to determine the maximum or minimum value.

What is the difference between min and inf?

Min refers to the minimum value in a set of numbers, while inf (infimum) is the greatest lower bound of a set. This means that inf may not necessarily be a member of the set, but it is less than or equal to all the values in the set.

Can you have multiple maximum or minimum values in a data set?

Yes, it is possible to have multiple maximum or minimum values in a data set. This can occur when there are ties in the data, meaning that two or more values are equally the maximum or minimum.

Why is finding the sup and inf important in mathematics?

Finding the sup and inf is important in mathematics because it helps to define the upper and lower limits of a data set. This can be useful in various applications, such as optimization problems, where the goal is to find the best possible solution within a given range of values.

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