Recent content by pessimist

Thanks for the info.

CERN anyways extended the LHC run to search for Higgs to mid 2012, so curtailing it by a quarter doesn't seem nonsense to me. I may be wrong ofcourse. http://www.redorbit.com/news/science/1987868/cern_extends_lhcs_run_until_2012/index.html

This is an extremely speculative post. I came to know through a friend of mine, that his professor told him that CERN is going to announce by spring 2012 the closure of all Higgs and Supersymmetry related programs. I do not have any citation for this claim. More knowledgeable people can tell me...

The adjacency matrix is as named, a matrix. After all when you want to find the number of walks from one vertex to another you multiply the matrix to itself using matrix multiplication.

Yes it is. Use the definition of limit and choose your epsilon to be a certain constant (say 1). Then work your magic.

Do you mean Paul Dirac? http://www.dirac.ch/PaulDirac.html

I only meant to do it so that a better understanding was possible but I erred.

I just realized there is an error in my proof :) sorry about that. |x| = |x + a - a| <= |x - a| + |a| < 1 + |a| This one posted above is abolutely correct.

|x - a| < 1 -1 + a < x < 1 + a |x| < |1 + a| <= 1 + |a|

In case this wasn't clear I meant we have to prove that cardinality of the set of Fibonacci primes is aleph null. How do you say that each new term is a next prime? It could be that it has a prime factor that is a new prime but may not be prime itself?

Dedekind did indeed prove that an infinite set can be set up in a bijection with a proper subset of itself. But that just proves the existence of A proper subset (say in N) to which it is bijective (say even natural numbers). It doesn't mean that "Given any proper subset of an infinite set...

That is exactly what one would like to prove isn't it? But how do we?

well I've got it, thanks anyways.

Homework Statement h: X x Y --> R is a function from X x Y to R. X,Y nonempty. If range is bounded in R. then let f : X --> R st f(x) = sup{h(x,y): y belongs to Y} (call this set A) g :Y --> R st g(y) = inf{h(x,y) : x belongs to X} (call this set B) Then prove that sup{g(y) : y belongs to...

The cantor set is indeed one of the most remarkable things one encounters in mathematics. In my opinion it is the height of non intuitiveness.