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Question  is this meaningful or is this retarded?
if
2*3 = 6
2*2 = 4
2*1 = 2
2*0 = 0
2*1 = 2
then why doesnt this work?
6/3 = 2
6/2 = 3
6/1 = 6
6/0 = 0
6/1 = 6
if n/0 is forbidden and 1/n returns the inverse of n, why shouldn't zero be its own inverse?
If we're talking "0" as in an infinitely precise definition of zero, then 1/n (where n is arbitrarily close to 0), then the result is an arbitrarily large answer, close to infinite, because any floating point number beneath zero (like an infinitely precise approximation of zero) when inverted, produces a number equal to or greater than 1.
If the multiplicative identity, 1, covers the entire set of integers, then why shouldn't division by zero be the inverse of the multiplicative identity, excluding the entire set? It ONLY returns 0, while anything n*1 ONLY returns n.
This puts even the multiplicative identity in the set covered by its inverse.
Ergo, division by zero produces either 0 or infinity. When theres an infinity in an formula, it sometimes indicates theres been
some misunderstanding or the system isn't fully understood. The simpler approach here would be to say therefore the answer is
not infinity, but zero. Now 'simpler' doesn't always mean "correct", only more elegant.
But if we represent the result of a division as BOTH an integer and mantissa
component, e.x
1.234567 or 0.1234567,
i.e. a float, we can say the integer component is the quotient, and the mantissa
is the remainder.
Logically it makes sense then that division by zero is equivalent to taking the numerator, and leaving it "undistributed".
I.e. shunting it to the remainder, and leaving the quotient as zero.
If we treat this as equivalent of an inversion, we can effectively represent the quotient from denominators of n/0 as 1/n
Meaning even 1/0 has a representation, it just happens to be 0.000...
Therefore
(n * (n/0)) = 1
the multiplicative identity
because
(n* (n/0)) == (n * ( 1/n ))
People who math. Is this a yea or nay in your book?26 
I spent the whole damn day trying to setup grpcweb, but this protocol is documented so damn poorly!
You manage to set grpc up for one language and it’s all cool, then you stupidly think that you are free to reuse the compiler you used for the nodejs version for your frontend part but nope! Our web module is now deprecated, please use this module instead!
“Ah yes just clone the repo and check out (…) and you can also check this link whic is in no way highlighted in the middle of a wall of text (…)”
*checking the other page*
Ah yes you need to install a package available only on your unix machine (great! Screw the devs in my team who use windows I guess, they’ll be happy to hear this!) and don’t forget to clone this repo to build your own plugin! And by that I ofc mean to compile it on your own!
 compiler error
After digging for an hour you find a requirement in an obscure issue opened and closed cause “ah yes we have a dependency not stated anywhere” *close issue and never add it to the project*
Fine, fine I can survive this bs
 another compiler error, no solution found after 2 hours
Honestly? Why the fuck do I need to compile this stuff? Just give me a damn npm package I can use? Goddamn it’s just transpiling, you don’t need access to my OS! (Aside for fs to save the files, and which btw is accessible via nodejs)
Now, I COULD download the latest realease as a precompiled, but… honestly?
I give up, I’ll do some shitty rest apis cause the customer’s not paying me enough for even THINKING to go trough this shit again when they’ll ask an iOS app. Or having colleagues asking me to help them understand how to do it.
Side note: also add typescript support to the webcodegeneration ffs! Why does node have it and web don’t?5 
"Fast" randomnumber/sample based estimation of logarithms:
https://pastebin.com/niVB57Ay
The result of rAvg(p) is usually "pretty close" to log(p)/2
rAvg(p, 1000) seems to be the golden number. 100 is a little low, but I've already pasted the code. Eh.
Don't know why it works, or if average results are actually considered "close" for logarithms of e.1