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#1 2010-10-30 14:06:28
- xsw001
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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.
#2 2010-10-30 14:27:08
- xsw001
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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.
#3 2011-03-16 19:08:36
- rstarling
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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.
#4 2011-03-16 20:10:06
- bobbym
- Administrator
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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.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.