We only have the epsilon-delta definition to work with for these.
Prove that f is integrable and verify the value. On [0,1] f(x)=1 if x=1/2 else 0. $$\int_{0}^{1} \,f$$ =0
Prove: If f is integrable on [0,1] then $$\lim_{{n}\to{\infty}}\ \frac{1}{n} \sum_{k=1}^{n} f(\frac{k}{n})$$ =...
Would this be reasonable: Since a<b mean b-a is positive then x(b-a)<y(b-a) by axiom. then I get xb-xa<yb-ya then because of this definition of < and the face all the terms are positive get xa<xb<ya<yb and from there that xa<yb.
We have but since we are working in only the positive natural numbers for these proofs there are no multiplicative inverses. So far I know the relations ax<ay and bx<by hold. As does ax+bx<ay+by. I've been trying to find a way to add/multiply them so that I can find one that will show ax<by.
We have not defined division so I don't know for sure I could use y/x. I know that ax>0 and by>0 by closure under addition for positive numbers. But I still can't find a way to link ax and yb so that ax<yb.
Do you mean for (1) If x<y then x+z<y+z?
If so I've been using both of those and we were told we can use anything else we've already proved as well. I still don't see a way to get the ax<by using just those two axioms or the other properties we have.
I think I'm supposed to be able to do it off the definition we have of x<y mean y-x is positive. So I was able to easily do 0<x<y and 0<a<b mean x+a<y+b. I've tried writing everything I can think of that follows from multiplying various ways but I can't seem to get just ax or yb without...