Is there a way to prove that a set is bounded using calculus techniques?

[Anne5632]

Homework Statement

Prove that S which is a subset of R is bounded

Relevant Equations

S=(n^2-1)/(n^3-1)
S is an element of natural numbers

I know that for a set to be bounded it is bounded above and below, for the bound below is it 0 and n cannot equal 1 and u paper bound is inf but how do I prove that it is bounded?

 

[PeroK]

Homework Statement:: Prove that S which is a subset of R is bounded
Relevant Equations:: S=(n^2-1)/(n^3-1)
S is an element of natural numbers

I know that for a set to be bounded it is bounded above and below, for the bound below is it 0 and n cannot equal 1 and u paper bound is inf but how do I prove that it is bounded?

Did you try writing down the first few elements in the set to get an idea? I assume we are taking about $n > 1$ here?

 

[Anne5632]

Did you try writing down the first few elements in the set to get an idea? I assume we are taking about $n > 1$ here?

the inf of the set is 0 so does that count in the interval?

 

[PeroK]

the inf of the set is 0 so does that count in the interval?

Yes, it should be clear that for all $n > 1$ we have $s_n > 0$. $S$ is, therefore, bounded below by $0$. What about an upper bound?

 

[Anne5632]

Yes, it should be clear that for all $n > 1$ we have $s_n > 0$. $S$ is, therefore, bounded below by $0$. What about an upper bound?

1

 

[PeroK]

1

How do you prove that is the question?

Have you done many proofs?

 

[Anne5632]

How do you prove that is the question?

Have you done many proofs?

No I haven't done proofs
I simplified the fraction in the equation given and factorised out (x-1) then got a polynomial to degree one in top and a polynomial to degree 2 on the bottom.
If i want to show it's smaller than 1 should I rewrite 1 into a polynomial with degree 2÷polynomial to degree 2

 

[PeroK]

No I haven't done proofs
I simplified the fraction in the equation given and factorised out (x-1) then got a polynomial to degree one in top and a polynomial to degree 2 on the bottom.
If i want to show it's smaller than 1 should I rewrite 1 into a polynomial with degree 2÷polynomial to degree 2

You can either do a direct proof, which is to show directly that $\frac{n^2 -1}{n^3 - 1} \le 1$. Or, you can do a proof by contradiction by assuming that for some $n$ we have $\frac{n^2 -1}{n^3 - 1} > 1$ and reaching a contradiction.

Cancelling the common factor of $n - 1$ is an option.

 

[Anne5632]

You can either do a direct proof, which is to show directly that $\frac{n^2 -1}{n^3 - 1} \le 1$. Or, you can do a proof by contradiction by assuming that for some $n$ we have $\frac{n^2 -1}{n^3 - 1} > 1$ and reaching a contradiction.

Cancelling the common factor of $n - 1$ is an option.

Got the bounds I think , thanks.
Part 4 of that q asked:
Let B be a bounded subset of R. Prove -B + S is bounded from below.

How would I know the bounds of B? Does it have no lower bound?

 

[PeroK]

Got the bounds I think , thanks.
Part 4 of that q asked:
Let B be a bounded subset of R. Prove -B + A is bounded from below.

How would I know the bounds of B? Does it have no lower bound?

What's A?

 

[Anne5632]

What's A?

Sorry meant to write S,
A is the set S described in the question above

 

[PeroK]

Sorry meant to write S,
A is the set S described in the question above

First, if $X$ is a bounded subset of $\mathbb R$, then the set $-X \equiv \{-x: x \in X\}$ is also a bounded subset of $\mathbb R$.

And, more generally $X$ is bounded above iff $-X$ is bounded below.

Also, if $X$ and $Y$ are bounded above, then the set $X + Y \equiv \{x+y: x \in X, y \in Y\}$ is bounded above. Similarly, if $X$ and $Y$ are bounded below, then so is $X + Y$.

These are things that I assume you will be shown in your course or asked to prove as an exercise. You need to learn some of the techniques that are used to prove things like this. The proofs are straighforward, but not always easy to see for someone new to formal proofs.

Have you seen anything like this on your course? It seems to me that you are working in the dark here.

 

[Anne5632]

First, if $X$ is a bounded subset of $\mathbb R$, then the set $-X \equiv \{-x: x \in X\}$ is also a bounded subset of $\mathbb R$.

And, more generally $X$ is bounded above iff $-X$ is bounded below.

Also, if $X$ and $Y$ are bounded above, then the set $X + Y \equiv \{x+y: x \in X, y \in Y\}$ is bounded above. Similarly, if $X$ and $Y$ are bounded below, then so is $X + Y$.

These are things that I assume you will be shown in your course or asked to prove as an exercise. You need to learn some of the techniques that are used to prove things like this. The proofs are straighforward, but not always easy to see for someone new to formal proofs.

Have you seen anything like this on your course? It seems to me that you are working in the dark here.

No I haven't, and none of the other exercises are similar.
But I'll looks throught the theorems in that topic

 

[PeroK]

No I haven't, and none of the other exercises are similar.
But I'll looks throught the theorems in that topic

If we take the first one.

First, if $X$ is a bounded subset of $\mathbb R$, then the set $-X \equiv \{-x: x \in X\}$ is also a bounded subset of $\mathbb R$.

A simple proof of that would be;

Let $U$ be an upper bound for $X$.

Let $y \in -X$. Then $y = -x$ for some $x \in X$. Now $x \le U \ \Rightarrow \ -x \ge -U$. Hence $y \ge -U$.

As $y$ was any member of $-X$ we see that $-U$ is a lower bound for $-X$ and so $-X$ is bounded below.

Then you would do something similar for a lower bound $L$.

That's the style of proof that I assume is required for these problems.

 

[Anne5632]

If we take the first one.

A simple proof of that would be;

Let $U$ be an upper bound for $X$.

Let $y \in -X$. Then $y = -x$ for some $x \in X$. Now $x \le U \ \Rightarrow \ -x \ge -U$. Hence $y \ge -U$.

As $y$ was any member of $-X$ we see that $-U$ is a lower bound for $-X$ and so $-X$ is bounded below.

Then you would do something similar for a lower bound $L$.

That's the style of proof that I assume is required for these problems.

Would the formula
Inf(N)=-sup(-N)
be useful?

 

[PeroK]

Would the formula
Inf(N)=-sup(-N)
be useful?

That result is similar and has the same sort of proof as the one I gave. But, ultimately, pure mathematics isn't about plugging numbers into formulas; it's about being able to construct logically valid proofs.

 

[Anne5632]

That result is similar and has the same sort of proof as the one I gave. But, ultimately, pure mathematics isn't about plugging numbers into formulas; it's about being able to construct logically valid proofs.

True,
To find the lower bound of -B I thought I could use that theorem

 

[PeroK]

True,
To find the lower bound of -B I thought I could use that theorem

Well, let's see you attempt to prove it. You haven't actually posted anything yet that gives me any indication of what you can do!

 

[WWGD]

@Anne5632 :
If you've done a bit of Calculus or advanced pre-Calculus, you can consider $f(x)=\frac {x^2-1}{x^3-1}$ and use standard optimization techniques: Take derivative f'(x), set to 0, etc.