subiprime

joined 3 months ago
[–] subiprime@lemmy.blahaj.zone 7 points 1 week ago (1 children)

petition to make an okbuddyrosalyn community here

[–] subiprime@lemmy.blahaj.zone 14 points 1 week ago (1 children)

6 articles?

amateurs...

meanwhile the polish words for "this" and "that":

[–] subiprime@lemmy.blahaj.zone 6 points 1 month ago

man i hate it when i walk by a nuclear power plant and get hit with crippling amounts of ionizing radiation and my immune system stops working

...they should really figure out some way to like, keep the radiation inside or something...

[–] subiprime@lemmy.blahaj.zone 1 points 1 month ago

room-temperature is safe mf's when they are doomed to feel an eternal nothing on their skin

[–] subiprime@lemmy.blahaj.zone 2 points 1 month ago (1 children)

is that saddam hussein

[–] subiprime@lemmy.blahaj.zone 2 points 2 months ago

I think this could use a bit more elaboration, since if x-y+y-z < -(|x-y|+|y-z|), then ||x-y|+|y-z|| >= |x-y+y-z| wouldnt be true. This is impossible though since q >= -|q|

[–] subiprime@lemmy.blahaj.zone 2 points 2 months ago

I see, thanks! :3

[–] subiprime@lemmy.blahaj.zone 3 points 2 months ago (3 children)

I'm confused about this step in the final condition's proof:

|🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)| >=|🍎(x) -🍌(x) +🍌(x) - 🍇(x)| = |🍎(x) - 🍇(x)| since |q| >= q forall q

I can see how it's true by proving that |p| + |q| >= |p + q|, but that's not stated anywhere and I can't figure out how |q| >= q forall q is relevant.

Also, thanks a lot for making/showing a proof :D

[–] subiprime@lemmy.blahaj.zone 2 points 2 months ago

i saw this before going to bed and i dreampt mojang actually added this and i was watching hbg highlights videos in the new update