Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**xsw001****Member**- Registered: 2010-10-23
- Posts: 7

Suppose f:[a, b]-> R is bounded function

f(x)=0 for each rational number x in [a, b]

Prove Lower Integral <= 0 <= Upper Integral

Proof:

f(x) = 0 when x is rational

both L(f, p) = U(f, P) = 0

and L(f, p) <= Lower Integral <= Upper Integral <= U(f, p)

This function seems like discontinous even though there aren't any information of functional value when x is NOT rational. It looks like that the Intermediate Value Theorem needs to be appplied.

So I have to prove that the Lower Integral <=0, and the Upper Integral >=0. So the function itself has to cross f(x)=0 isn't it?

Any suggestions would be greatly appreciated.

Offline

**xsw001****Member**- Registered: 2010-10-23
- Posts: 7

Never mind, I got it.

Every interval of nonzero size contains a rational number.

So the min of f(x) on the interval MUST be <=0

and the max of f(x) on the interval is >=0.

Offline

**rstarling****Member**- Registered: 2011-03-08
- Posts: 1

cant quite get the whole indeger and order of opporation thing can u help make it a fun thing but easy.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,407

Hi rstarling;

Welcome to the forum.

cant quite get the whole indeger and order of opporation thing can u help make it a fun thing but easy.

I am not understanding what you need. If you show an example of what is unclear, then you can get help.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

Pages: **1**