it starts out okay but 6/0 is not inf.
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mexicancartel@lemmy.dbzer0.com 7 months agoAdding 3 on both sides
3-3=3+3
0 = 6
1•0 = 6
1 = 6/0
1 = inf
Multiplying e^(iπ) on both sides,
e^(iπ) = - inf
iπ = ln|-inf|
π = ln|-inf| ÷ i
huf@hexbear.net 7 months ago
mexicancartel@lemmy.dbzer0.com 7 months ago
Its just the joke i take watever i want. There is | | inside log which is just made up. 6/x as x->0 is infinity though so i’mma use that
hemko@lemmy.dbzer0.com 7 months ago
I gotta say half of that goes over my head, but I raise my hat to you
mexicancartel@lemmy.dbzer0.com 7 months ago
I remember you. You saved me from rickroll on xkcd
hemko@lemmy.dbzer0.com 7 months ago
Damn me
Maalus@lemmy.world 7 months ago
1/0 isn’t infinity though.
mexicancartel@lemmy.dbzer0.com 7 months ago
Like if everything else is true.
1/0 can be infinity as a limiting case but not always. So imma use what i like since i also took ln|x| (took mod out of nowhere just to make it positive)
emergencyfood@sh.itjust.works 7 months ago
Division is repeated subtraction. How many times can you subtract 0 from 1?