CBenni
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 Jan 31 accepted Prime decomposition over the real numbers? Nov 3 awarded Yearling May 30 comment Showing that the rank of the complex projective space is 1 @mollyerin Thank you very much for your response. I have put some effort into double-checking this and I am almost certain that this proof is correct. (it, more or less, has to be) May 29 asked Showing that the rank of the complex projective space is 1 Nov 3 awarded Yearling Oct 15 comment TeX editor with instant preview @Moronplusplus I have probably deleted that file off the server, I can reupload it if you like. Sep 30 awarded Explainer Sep 24 awarded Autobiographer Jul 3 comment Why $2x$? Can't it be $x$? What about $x=2.5$ or $x=\pi$? Addition in $\mathbb R$ cannot be defined via a sum of $n$ summands easily. Jul 2 awarded Curious Feb 9 revised $\theta_1 + \theta_2 = -35.5$ how to find the values of those $\theta$s? fixed latex Feb 9 suggested approved edit on $\theta_1 + \theta_2 = -35.5$ how to find the values of those $\theta$s? Dec 15 comment Prime decomposition over the real numbers? Remember that I basically "choose" my prime elements and do not acquire them via divisibility rules - $\frac{1}{x}$ can just be defined as one, or else be the product of other elements in $P$. Dec 15 comment Prime decomposition over the real numbers? @edit - why would it give problems to have $\frac{1}{x_i}=x_i^{-1}\in P$? I really dont see that argument... Sorry if these objections are useless, but im trying to understand the matter. Dec 15 comment Prime decomposition over the real numbers? Sorry for taking away the accept for now, I would like to know: am I really trying to show that? The fact that each number is a unit gives problems usually, since each element $x$ can be written as $x y y^{-1}$ - I take out this case purposefully Dec 15 asked Prime decomposition over the real numbers? Nov 22 awarded Favorite Question Nov 9 awarded Notable Question Nov 3 awarded Yearling Sep 1 comment Jigsaw Puzzle Help $20''\times 27'' = 540$ sq inches Equally dividing that over 1000 pieces gives us $0.54$ sq inches per piece. Assuming that the pieces are square, we have a borderlength of $\sqrt{0.54}=0.735''$ per piece. The amount of pieces along one border is $27/0.735=36.73$ - This is not even close to an integer, meaning our assumption (pieces are square and equally in size) was incorrect - we need to know that in order to give qualified answers ;)