Cirno proves that every Boolean ring is commutative
Cirno has a PhD in mathematics.
Here's an educational slideshow. Don't ask me why I made this; all you gotta know is that this won't become a math channel. Well, maybe.
My good friend Emiko generously let me borrow her voice for this skit; I told her to record everything in one take.
Anyway, the pictures were taken in MikuMikuDance using models by arlvit and Maddoktor2. I wanted to animate Cirno properly but it would've taken too long for a shitpost so fuck it.
Here's an educational slideshow. Don't ask me why I made this; all you gotta know is that this won't become a math channel. Well, maybe.
My good friend Emiko generously let me borrow her voice for this skit; I told her to record everything in one take.
Anyway, the pictures were taken in MikuMikuDance using models by arlvit and Maddoktor2. I wanted to animate Cirno properly but it would've taken too long for a shitpost so fuck it.


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@RedTheNEET Not necessarily. The simplest example of a boolean ring is just the field R = {0,1} with sum and product modulo 2 and 0 != 1 (aka Z/2Z, the integers mod 2). You can easily check that it's a boolean ring because 0^2 = 0 and 1^2 = 1. In R it happens that 1 + 1 = 0, so 1 is its own additive inverse (1 = -1), but 1 is distinct from 0. The boolean ring R is used in logic, with 0 interpreted as FALSE, 1 as TRUE, the product as AND and the sum as XOR.
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