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I’m a fan of the Eilenberg–Mazur swindle. It is based on the “proof” that [math][/math][math] 1 = 0 [/math][math][/math][math][/math] by[math][/math][math] 1 = 1 [/math][math][/math][math]+ (-1 [/math][math][/math][math]+ 1) [/math][math][/math][math]+ (-1 [/math][math][/math][math]+ 1) [/math][math][/math][math]+ \ldots = (1 [/math][math][/math][math]+ -1) [/math][math][/math][math]+ (1 [/math][math][/math][math]+ -1) [/math][math][/math][math]+ (1 [/math][math][/math][math]+ -1) [/math][math][/math][math]+ \ldots = 0 [/math][math][/math][math][/math].Now, this “proof” doesn’t work, because the series [math][/math][math] 1 [/math][math][/math][math]+ -1 [/math][math][/math][math]+ 1 [/math][math][/math][math]+ -1 [/math][math][/math][math]+ \ldots [/math][math][/math][math][/math] doesn’t converge and so it just doesn’t make sense to manipulate it algebraically like this.However, there are other places in mathematics where it makes sense to write [math][/math][math] [/math][math]A[/math][math] [/math][math][/math][math]+ [/math][math]B[/math][math] = [/math][math]B[/math][math] [/math][math][/math][math]+ [/math][math]A[/math][math] = 0 [/math][math][/math][math][/math] and to conclude from this that[math][/math][math] [/math][math]A[/math][math] = [/math][math]A[/math][math] [/math][math][/math][math]+ ([/math][math]B[/math][math] [/math][math][/math][math]+ [/math][math]A[/math][math]) [/math][math][/math][math]+ ([/math][math]B[/math][math] [/math][math][/math][math]+ [/math][math]A[/math][math]) [/math][math][/math][math]+ \ldots = ([/math][math]A[/math][math] [/math][math][/math][math]+ [/math][math]B[/math][math]) [/math][math][/math][math]+ ([/math][math]A[/math][math] [/math][math][/math][math]+ [/math][math]B[/math][math]) [/math][math][/math][math]+ ([/math][math]A[/math][math] [/math][math][/math][math]+ [/math][math]B[/math][math]) [/math][math][/math][math]+ \ldots = 0 [/math][math][/math][math][/math].Mazur, for example, used this to prove results about knots. Knots are loops drawn in 3D space—the thing of interest is whether a given knot can be moved around so that it becomes the trivial knot (i.e. just a circle). On the left, I have drawn an example of a knot that is equivalent to the trivial knot. On the right is the trefoil knot, which isn’t.You can “add” knots by taking out small chunks out of them and connecting them. That looks as follows:Note that if [math][/math][math] [/math][math]A[/math][math], [/math][math]B[/math][math] [/math][math][/math][math][/math] are knots, then [math][/math][math] [/math][math]A[/math][math] [/math][math][/math][math]+ [/math][math]B[/math][math] = [/math][math]B[/math][math] [/math][math][/math][math]+ [/math][math]A[/math][math] [/math][math][/math][math][/math] (as long as you are willing to think of the knots as being equivalent if you can move one around into another). Furthermore, it makes sense to write the trivial knot as [math][/math][math] 0 [/math][math][/math][math][/math], since if you add it to any other knot, you get the same knot back.One can ask: is it possible to take two knots [math][/math][math] [/math][math]A[/math][math], [/math][math]B[/math][math] [/math][math][/math][math][/math] that are not equivalent to the trivial knot, but when you add them, you do get the trivial knot? Or, to put it another way, is it possible that [math][/math][math] [/math][math]A[/math][math] [/math][math][/math][math]+ [/math][math]B[/math][math] = 0 [/math][math][/math][math][/math] if [math][/math][math] [/math][math]A[/math][math], [/math][math]B[/math][math] \neq 0 [/math][math][/math][math][/math]?Enter the Mazur swindle. In this case, we can absolutely think about adding together an infinite amount of knots (if you like, you can always draw each successive knot smaller), and so we see that if [math][/math][math] [/math][math]A[/math][math] [/math][math][/math][math]+ [/math][math]B[/math][math] = 0 [/math][math][/math][math][/math], then[math][/math][math] [/math][math]A[/math][math] = [/math][math]A[/math][math] [/math][math][/math][math]+ 0 [/math][math][/math][math]+ 0 [/math][math][/math][math]+ \ldots = [/math][math]A[/math][math] [/math][math][/math][math]+ ([/math][math]B[/math][math] [/math][math][/math][math]+ [/math][math]A[/math][math]) [/math][math][/math][math]+ ([/math][math]B[/math][math] [/math][math][/math][math]+ [/math][math]A[/math][math]) [/math][math][/math][math]+ \ldots [/math][math][/math][math][/math][math][/math][math] = ([/math][math]A[/math][math] [/math][math][/math][math]+ [/math][math]B[/math][math]) [/math][math][/math][math]+ ([/math][math]A[/math][math] [/math][math][/math][math]+ [/math][math]B[/math][math]) [/math][math][/math][math]+ ([/math][math]A[/math][math] +[/math][math]B[/math][math]) [/math][math][/math][math]+ \ldots [/math][math][/math][math][/math][math][/math][math] = 0 [/math][math][/math][math]+ 0 [/math][math][/math][math]+ 0 [/math][math][/math][math]+ \ldots [/math][math][/math][math][/math][math][/math][math] = 0 [/math][math][/math][math][/math].Therefore, if you take any two non-trivial knots and you add them, you get another non-trivial knot.

I find that it is best to not speculate and instead RTFM.Steam Subscriber Agreement2. LICENSES⏶A. General Content and Services LicenseSteam and your Subscription(s) require the download and installation of Content and Services onto your computer. Valve hereby grants, and you accept, a non-exclusive license and right, to use the Content and Services for your personal, non-commercial use (except where commercial use is expressly allowed herein or in the applicable Subscription Terms). This license ends upon termination of (a) this Agreement or (b) a Subscription that includes the license. The Content and Services are licensed, not sold. Your license confers no title or ownership in the Content and Services. To make use of the Content and Services, you must have a Steam Account and you may be required to be running the Steam client and maintaining a connection to the Internet.You are paying for a license.Here’s the thing - thats how nearly all games sold are. Buy some digital playstation games and get your account banned - games are gone. Sony goes out of business - games are gone. Auth servers go down - games are gone for a while.Even physical media won’t save you if the game requires online components.Pretending this is just a Steam thing is a bit disingenuous.

The latest issue of the New England Journal of Medicine published a case study of a man with uncontrolled HIV and multiple tumors in various organs. When pathologists looked at biopsies of the tumors, they noticed some weird things. The cells were extremely small and they were syncytial–they had multiple nuclei sharing a connected cytoplasmic space–which isn't unheard of, but is still a bit odd.So it was suspected that the nodules were some sort of weird fungal infection and researchers sequenced the DNA of the tumor cells. They definitely weren't human; they were tapeworm cells.Hymenolepis nana scolex. It looks happy to see you.The man also had an infection of Hymenolepis nana, a common tapeworm that usually causes little pathology in adults. As near as anyone can tell, tapeworm stem cells with mutations similar to those seen in human cancer somehow traveled to his organs and proliferated.Intraspecies transfer of cancer is well documented, if rare. As far as I know, this is the first and only known case of interspecies transfer of cancerous cells. And it absolutely blows my mind. I may update this post as this information digests.The paper is short and free, so you can read it yourself! PDFReference:Muehlenbachs, A., Bhatnagar, J., Agudelo, C. A., Hidron, A., Eberhard, M. L., Mathison, B. A., … Zaki, S. R. (2015). Malignant Transformation of Hymenolepis nana in a Human Host. New England Journal of Medicine, 373(19), 1845–1852. http://doi.org/10.1056/NEJMoa1505892