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 2d accepted Calculus over integers for derivative/integral of factorial? 2d accepted Why do mathematicans care so much about the incompressible Navier-Stokes equations? 2d accepted If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$? 2d accepted Why is there only a complex conjugate, but no real conjugate? Feb 8 asked Is a random variate a functional? Jan 15 comment Is every square matrix a tensor of 2nd order? "The total number of indices required to uniquely select each component is equal to the dimension of the array, and is called the order, degree or rank of the tensor." (see en.wikipedia.org/wiki/Tensor#As_multidimensional_arrays). But searching through the article the term "order" seems to be the most used one. Jan 14 asked Is every square matrix a tensor of 2nd order? Jan 12 comment Why is there only a complex conjugate, but no real conjugate? Aren't the roots of $X^2 -1 = 0$ also symmetric, so $X = 1$ and $X=-1$ is an arbitrary choice? Jan 12 comment Why is there only a complex conjugate, but no real conjugate? @null Yes. I corrected the text. Jan 12 revised Why is there only a complex conjugate, but no real conjugate? added 2 characters in body Jan 12 asked Why is there only a complex conjugate, but no real conjugate? Dec 23 comment If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$? Could one not assume Positive × Negative == Zero and Negative × Positive == Zero ? Dec 23 comment If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$? @Pierre-GuyPlamondon The most symmetric (commutative) solution would be $(+1) \cdot (-1) = 0$. Dec 23 asked If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$? Dec 22 accepted How to generalize the Thue-Morse sequence to more than two symbols? Dec 18 awarded Yearling Dec 9 comment What is the advantage of the Fourier Transform over the Hartley Transform? I don't think so. A cosine transform is simply the real part of a Fourier transform. But the real part of the Fourier transform can be computed from a Hartley transform like $Re[F(\omega)] = (H(\omega) + H(-\omega))/2$. So only the even part of the Hartley transform is equivalent to a cosine transform. This is because the Hartley transform kernel is a shifted cosine function, which is not symmetric around the origin. Dec 7 asked What is the advantage of the Fourier Transform over the Hartley Transform? Nov 15 awarded Tumbleweed Nov 8 revised Why do mathematicans care so much about the incompressible Navier-Stokes equations? added 332 characters in body