Recent content by aortizmena

Thanks chiro, yes i know alternatives of proving the statement, i was just trying to give a proof with this alternative aproach. I know i can prove it by matrix properties, by linear independence of the vectors, and by proving <{ u , v , w }>=V<{ u+v+w , v+w , w }>. Thanks though.

I thought it was implicit, that was my intention

Thanks Hurkyl but isn't saying that for for any arbitrary x∈V=<{ u , v , w }> with d=a, d+e=b and d+e+f=c it implies the unique solution d=a, e=b-d=b-a , f=c-d-e=c-a-(b-a)=c-a-b+a=c-b ?

Let u,v,w\in V a vector space over a field F such that u≠v≠w. If { u , v , w } is a basis for V. Prove that { u+v+w , v+w , w } is also a basis for V. Proof Let u,v,w\in V a vector space over a field F such that u≠v≠w. Let { u , v , w } be a basis for V. Because { u , v , w } its a basis...

you should try ramsey's model its an aplication to economics

If you really want to learn calculus, you shoud first learn logic and how to do proofs. Then read and work on any of these authors calculus books by Spivak, Apostol or Courant for a deep formal approach (depending on which one you like the most, you can't go wrong with this ones). And maybe...

Homework Statement Prove that D={\frac{m}{2^{n}} : n\in N , m=0,1,2,...,2^{n}} (dyatic rationals set) is dense on [0,1] , i.e. if (a,b) \subset [0,1] then (a,b) \bigcap D \neq emptysetHomework Equations The Attempt at a Solution Is it wrong if I just state that because a,b\in\Re we know that...