Is the function f(x)=1/x bounded on the interval (0,1)?

[k3k3]

Homework Statement

Let f be the function defined f(x)=1/x. Prove that f is not bounded on (0,1)

Homework Equations

The Attempt at a Solution

I think I should prove by contradiction.

Assume f is bounded on (0,1).
Since f is bounded, there exists a real number M such that |f(x)| ≤ M for all x in (0,1)
f(x) will never be negative since it is on the interval (0,1), hence |f(x)| = f(x)

This is where I begin to get unclear on where to go next. I want to show that M+1 ≤ M
Is it correct to use 1/(M+1) and plug it into f(x)?

 

[Dick]

I don't think you should prove it by contradiction. If n is an number greater than one then 1/n is in (0,1).

 

[k3k3]

Can I argue that since 1/n is an infinite sequence, then this function is not bounded?

 

[Dick]

Can I argue that since 1/n is an infinite sequence, then this function is not bounded?

You need a better argument than that. What is f(1/n)?

 

[k3k3]

f(1/n)=n

Then I could say for all n in the positive integers?

 

[Dick]

f(1/n)=n

Then I could say for all n in the positive integers?

You could say that, but it doesn't prove f is unbounded until you say why that proves f is unbounded.

 

[k3k3]

Since f(1/n)=n for all n in N. Since N has an infinite amount of elements, then the function is unbounded on (0,1)?

 

[Dick]

Since f(1/n)=n for all n in N. Since N has an infinite amount of elements, then the function is unbounded on (0,1)?

Having an infinite number of elements has little to do with being unbounded. What does unbounded mean?

 

[k3k3]

That there is no lower bound, no upper bound or both.

 

[Dick]

That there is no lower bound, no upper bound or both.

Ok, so give me an argument that f has no upper bound.

 

[k3k3]

There is no n such that 1/n is not in the interval (0,1), so there is no real number M that will satisfy |1/n|≤M.

 

[Dick]

There is no n such that 1/n is not in the interval (0,1), so there is no real number M that will satisfy |1/n|≤M.

You don't want to satisfy |1/n|<=M. You want to show that you can find a number in x in (0,1) such that f(x)>M.

 

[k3k3]

So if M is greater than one, 1/(M+1) is in (0,1) and M+1 < M is not true?

 

[Dick]

So if M is greater than one, 1/(M+1) is in (0,1) and M+1 < M is not true?

I really hope you meant f(1/(M+1))=M+1 > M.

 

[k3k3]

No, I was still thinking about the contradiction argument. Sorry.

 

[Dick]

No, I was still thinking about the contradiction argument. Sorry.

That's ok. But you've got it now, yes?

 

[k3k3]

Yep. Thank you again for your help!