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Questions related from Andre Gontijo Campos
For instance, we know from Clifford algebra that the commutator of two vectors a and b is a bivector (i.e., an axial vector). That is, (ab-ba)/2=[a,b]/2=a^b. For space vectors (i.e., ordinary 3D...
06 June 2018 6,077 13 View
My answer is no. The reason is that we can never perform any measurement whose result is an irrational number. In this sense, perfect geometrical entities, such as spheres, squares, circles,...
05 May 2018 1,158 82 View
