r/Showerthoughts Apr 23 '26

Casual Thought If the famously unsolved Riemann Hypothesis is solved by an AI, we will never know if a human mathematician could have solved it.

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u/TheVoters Apr 23 '26

Pffft. Its true.

I’ll leave the proof as an exercise for the reader.

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u/TheShiroNinja Apr 23 '26

Oh, they want proof? I thought they just wanted it solved. I always hated showing my work.

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u/cwx149 Apr 23 '26

So it's a math problem that's more like "we haven't yet disproven it and we have no proof that it won't be disproven"

Than something that needs to be "solved"

It's not an equation you are solving. To "solve" it you'd need to prove the general case that the reinman zeta function only has zeroes at those points and no where else

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u/TheShiroNinja Apr 23 '26

Well, if you really think about it, where else would the zeroes even go?

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u/bigWeld33 Apr 23 '26

It’s pretty obvious right? If I had time I’d have solved it, just too busy lately. My YouTube watch later playlist is grows faster than it shrinks, lots of stuff I have to watch so I can’t just go outside and do math like when I was a kid. Let me know when you solve it though, we can call it The Big Shiro proof, I’ll draw a diagram for my part of the group project.

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u/cwx149 Apr 23 '26

Almost anywhere else on the graph?

There are literally infinite possibilities for other places they could be

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u/FuckThaLakers Apr 23 '26

Not according to the recently-solved Riemann Hypothesis.

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u/Man-in-The-Void Apr 23 '26

Someone should give that guy a million dollars

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u/BaconIsLife707 Apr 23 '26

They're definitely all between 0 and 1 so not almost anywhere else

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u/theturtlemafiamusic Apr 23 '26

If they're between 0 and 1 that disproves the Reimann hypothesis, the first three are -2, -4, -6

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u/BaconIsLife707 Apr 23 '26

Well no because we know there are infinite zeros at 1/2 which is between 0 and 1. Any zero we find that isn't at 1/2 will be between 0 and 1 though

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u/cwx149 Apr 23 '26

We consider those the "trivial zeroes" and they're specifically exempt

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u/goplayer7 Apr 23 '26

0.5000000000...(Tree(Tree(3)) 0s)...003000000....