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## #1 2010-10-29 15:06:28

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

### Prove Lower Integral <= 0 <= Upper Integral

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.

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## #2 2010-10-29 15:27:08

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

### Re: Prove Lower Integral <= 0 <= Upper Integral

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.

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## #3 2011-03-15 20:08:36

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

### Re: Prove Lower Integral <= 0 <= Upper Integral

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

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## #4 2011-03-15 21:10:06

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Prove Lower Integral <= 0 <= Upper Integral

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.

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