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

[evinda]

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??

 

[I like Serena]

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 $$B=\frac 2 4$$.
Isn't that greater than your $\sup B$?

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

 

[evinda]

Hola! :D

Suppose we pick $m=n=2$, then I get $$B=\frac 2 4$$.
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! 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)

 

[I like Serena]

Oh,yes!Right! 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 $$B=\frac 3 5$$.
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?

 

[evinda]

Good!

... but suppose we pick $m=3, n=2$, then I get $$B=\frac 3 5$$.
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?

 

[I like Serena]

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?

 

[evinda]

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?

 

[I like Serena]

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?

What is the following limit?
$$\lim_{m \to \infty} \frac m {2+m} =\lim_{m \to \infty} \frac {1} {\frac 2 m + 1}$$

 

[evinda]

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??
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 $infA=-1$ ?

 

[I like Serena]

It is equal to $1$! So,this is the supremum,right??
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)

 

[evinda]

Yes! (Mmm)

And... there is no min and max of the set,right? (Thinking)

 

[I like Serena]

And... there is no min and max of the set,right? (Thinking)

Right!

 

[evinda]

Right!

And how can I prove it?

 

[I like Serena]

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. $\qquad \blacksquare$

 

[evinda]

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! :)