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.
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...
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 aproach (depending on which one you like the most, you cant go wrong with this ones). And maybe (just...
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 emptyset
Homework Equations
The Attempt at a Solution
Is it wrong if I just state that because a,b\in\Re we know...