Supremum and infimum of specific sets

[haruspex]

For $C$, I am noticing that setting $n = 1$, and increasing the values of $m$, the numbers start to approach $0.9999...$ Is that right?

That is certainly true. Can you prove it gets arbitrarily close to 1, but no combination of m and n exceeds it?

 

[Bunny-chan]

That is certainly true. Can you prove it gets arbitrarily close to 1, but no combination of m and n exceeds it?

I don't know how to effectively prove it, but I can determine the possible values for $m$ and $n$ to see if it could possibly exceed $1$:

For $m = 0$, the values will be obviously $0$;
For $m \lt 0$, the values will be negative, which is inferior to $1$;
For $m > 0, m \geq n$, the denominator can't be less than the numerator (which implies in values smaller than $1$), seeing that $|m| \geq m \Rightarrow |m| + n \gt m$. The same can be said if $m \lt n$.

Is that right?

 

[haruspex]

I don't know how to effectively prove it, but I can determine the possible values for $m$ and $n$ to see if it could possibly exceed $1$:

For $m = 0$, the values will be obviously $0$;
For $m \lt 0$, the values will be negative, which is inferior to $1$;
For $m > 0, m \geq n$, the denominator can't be less than the numerator (which implies in values smaller than $1$, seeing that $|m| \geq m \Rightarrow |m| + n \gt m$. The same can be said if $m \lt n$.

Is that right?

Yes. To complete the proof, can you find a sequence of elements converging to 1?

 

[Bunny-chan]

Yes. To complete the proof, can you find a sequence of elements converging to 1?

Do I have to use limits for that?

 

[haruspex]

Do I have to use limits for that?

Yes, convergence is all to do with limits. Is that a problem? For the purposes of the question, I don't think you actually need to prove the convergence. Just show a sequence which pretty obviously converges to the extremum in your answer.

 

[Bunny-chan]

Yes, convergence is all to do with limits. Is that a problem? For the purposes of the question, I don't think you actually need to prove the convergence. Just show a sequence which pretty obviously converges to the extremum in your answer.

Let me see. I guess if we set $m > 0, n = 1$, we could say:$$C = \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \dots, \frac{9999}{10000}, \dots = 0.5, 0.6666..., 0.75, \dots, 0.9999, \dots$$which constantly approximates from $0,9999...$

Or I think we can also say$$\lim\limits_{m \rightarrow \infty} \frac{m}{m + 1}=\lim\limits_{m \rightarrow \infty} \frac{1}{1+\frac{1}{m}}=\frac{1}{1 + \lim\limits_{m \rightarrow \infty} \frac{1}{m}}=1$$Is that acceptable?

 

[haruspex]

Let me see. I guess if we set $m > 0, n = 1$, we could say:$$C = \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \dots, \frac{9999}{10000}, \dots = 0.5, 0.6666..., 0.75, \dots, 0.9999, \dots$$which constantly approximates from $0,9999...$

Or I think we can also say$$\lim\limits_{m \rightarrow \infty} \frac{m}{m + 1}=\lim\limits_{m \rightarrow \infty} \frac{1}{1+\frac{1}{m}}=\frac{1}{1 + \lim\limits_{m \rightarrow \infty} \frac{1}{m}}=1$$Is that acceptable?

Yes.

 

[Bunny-chan]

Yes.

Great! Thank you very much!

 

[haruspex]

Great! Thank you very much!

OK, that leaves B.
And don't forget it also asks for the infimum in each.

 

[Bunny-chan]

OK, that leaves B.
And don't forget it also asks for the infimum in each.

For $C$, I am assuming that the infimum is the opposite of $1$, assuming we take $m \lt 0$. Isn't that right?

 

[haruspex]

For $C$, I am assuming that the infimum is the opposite of $1$, assuming we take $m \lt 0$. Isn't that right?

Yes, swapping the sign of m swaps the sign of the result. I assume you mean -1 as the "opposite" of 1.

 

[Bunny-chan]

Yes, swapping the sign of m swaps the sign of the result. I assume you mean -1 as the "opposite" of 1.

Is there more than one opposite to $1$? o_o

 

[haruspex]

Is there more than one opposite to $1$? o_o

In relation to numbers, "opposite" is undefined. You can have "additive inverse" for x and -x, and "multiplicative inverse" for x and 1/x. For complex numbers there is the complex conjugate.

 

[Bunny-chan]

OK, that leaves B.
And don't forget it also asks for the infimum in each.

By the way, I just checked and I wrote set $B$ incorrectly, it should be $B = \left\{ {mn\over 4m^2+n^2} \mid m, n \in \mathbb N \right\}$. Anyway, setting $m = n = 1$, it results in $\frac{1}{5}$. Is there a way we could obtain a value greater than that considering $m, n \in \mathbb N$? Or is it safe to assume that that's the supremum?

 

[haruspex]

Is there a way we could obtain a value greater than that

A little greater, yes.
The quadratic is (after the correction) suggestive. It did look strange before.
Suppose the value for some m, n is a. Write that out as a quadratic equation.

 

[Bunny-chan]

A little greater, yes.
The quadratic is (after the correction) suggestive. It did look strange before.
Suppose the value for some m, n is a. Write that out as a quadratic equation.

Sorry, I don't follow what you mean. D:

 

[haruspex]

Sorry, I don't follow what you mean. D:

If a∈B then for some m, n, a=mn/(4m²+n²).
I.e. 4m²+n² = mn/a.
When you see an expression like that on the left, one thing to consider is 'completing the square'. There are two options for that here. What are they?

 

[Bunny-chan]

If a∈B then for some m, n, a=mn/(4m²+n²).
I.e. 4m²+n² = mn/a.
When you see an expression like that on the left, one thing to consider is 'completing the square'. There are two options for that here. What are they?

I'm clueless. I tried to complete the squares but I wasn't successful...

Should I divide the left side by four?

 

[haruspex]

I tried to complete the squares but I wasn't successful...

You cannot think what to add to or subtract from 4m²+n² to make a perfect square?

 

[Bunny-chan]

You cannot think what to add to or subtract from 4m²+n² to make a perfect square?

Oh.
$4m^2 + n^2 = (2m + n)^2 - 4mn$ or $(2m - n)^2 + 4mn$

 

[haruspex]

Oh.
$4m^2 + n^2 = (2m + n)^2 - 4mn$ or $(2m - n)^2 + 4mn$

Right. The 2m-n form is the useful one here.
Write the second equation in post #47 using that.
Can you get from that an upper bound on a?

 

[Bunny-chan]

Right. The 2m-n form is the useful one here.
Write the second equation in post #47 using that.
Can you get from that an upper bound on a?

Would that be $\frac{1}{4}$?

 

[haruspex]

Would that be $\frac{1}{4}$?

Yes.

 

[Bunny-chan]

Yes.

Excellent! And we have $0$ as infimum?

 

[haruspex]

Excellent! And we have $0$ as infimum?

As far as I am aware, there is not universal agreement on whether ℕ includes 0. What have you been taught?

 

[Bunny-chan]

As far as I am aware, there is not universal agreement on whether ℕ includes 0. What have you been taught?

Oh, I've forgot that. Particularly my professor doesn't include it either.

 

[haruspex]

Oh, I've forgot that. Particularly my professor doesn't include it either.

Maybe it does not matter. If we exclude 0, can you still get 0 as infimum?

 

[Bunny-chan]

Maybe it does not matter. If we exclude 0, can you still get 0 as infimum?

It's true that we can never get any value equal to zero, but the infimum doesn't need to be contained in the set, as you said. But I'm not very sure how to verify if $0$ is the infimum.

 

[haruspex]

It's true that we can never get any value equal to zero, but the infimum doesn't need to be contained in the set, as you said. But I'm not very sure how to verify if $0$ is the infimum.

Consider very unequal values of m, n.

 

[Bunny-chan]

Consider very unequal values of m, n.

The values indeed get more and more close to zero.