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[-] pr06lefs@lemmy.ml 174 points 3 months ago* (last edited 3 months ago)

What about plain old x = -10?

-10 ^ 2 = 100
-10 ^ 3 = -1000
-10 ^ 5 = -100000

[-] Redacted@lemmy.world 121 points 3 months ago

Isn't that the joke?

[-] Enkers@sh.itjust.works 68 points 3 months ago

That's what he wrote, I imagine.

[-] Siethron@lemmy.world 39 points 3 months ago

It is, but with imaginary numbets

[-] JackbyDev@programming.dev 47 points 3 months ago

i² = -1 so...

[-] dwindling7373@feddit.it 27 points 3 months ago

10 * i^2 is -10.

[-] thedudeabides@lemmy.dbzer0.com 17 points 3 months ago
[-] pr06lefs@lemmy.ml 35 points 3 months ago

people being pedantic showoffs doesn't really register as humor for me, TBH

[-] thedudeabides@lemmy.dbzer0.com 13 points 3 months ago

That's true, the OOP is being quite snarky with their comment on a post where someone's had a genuine basic doubt

[-] Deconceptualist@lemm.ee 9 points 3 months ago

That was my immediate thought too.

[-] Sphks@lemmy.dbzer0.com 4 points 3 months ago
[-] Deebster@programming.dev 102 points 3 months ago

When all you have is an imaginary hammer, everything looks like a rotation around the imaginary unit circle.

Explanation of mathsx = -10, i = √-1 so i² = -1 and 10i²=-10

[-] MonkderVierte@lemmy.ml 15 points 3 months ago

Found the math but no explanation.

[-] JackbyDev@programming.dev 15 points 3 months ago

The squareroot of 100 is ±10.

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[-] fx3@beehaw.org 4 points 3 months ago

IIRC, your spoilery “so” is the other way round. The right side is the definition, and the left-hand side a layman’s shorthand, as the root operator isn’t defined on negative numbers.

I might very well be wrong. My being a mathematician has been over for a while now, my being a pedantic PITA not though.

[-] Deebster@programming.dev 5 points 3 months ago

I don't know enough to know how correct your pedantry is (technically or not), but to explain the meme it made sense to go through the symbols in the order you see them. I never got any points from the proof questions in exams anyway.

[-] NigelFrobisher@aussie.zone 37 points 3 months ago

Wait, isn’t x just -10 if x^3 is not 1000?

[-] bi_tux@lemmy.world 15 points 3 months ago
[-] iAvicenna@lemmy.world 27 points 3 months ago

that is a very long way to write -10

[-] southsamurai@sh.itjust.works 13 points 3 months ago
[-] Cryophilia@lemmy.world 45 points 3 months ago

That's because the explanation was about 10 times as complicated as it needs to be

[-] mexicancartel@lemmy.dbzer0.com 23 points 3 months ago

He is trolling with overcomplicating

[-] Yaysuz@lemm.ee 10 points 3 months ago

What an extremely unnecessary explanation. As a math teacher I would have deducted points for this answer.

[-] androogee@midwest.social 8 points 3 months ago

"show your work"

Malicious compliance intensifies

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[-] Ravi@feddit.org 9 points 3 months ago

No definition what values are suitable for x.

[-] quicksand@lemm.ee 21 points 3 months ago

x has to be -10, right? Or am I missing something?

[-] jacksilver@lemmy.world 7 points 3 months ago

Yeah, I think the point is that the person answering was wrong/over complicating. If x=10i, then x^2 would be -100 (or potentially -10 depending on what you think the ^2 is applied to).

[-] Khanzarate@lemmy.world 23 points 3 months ago

They said x=10i^2, not 10i. Difference is it equals -10, and they chose not to simplify.

[-] SoleInvictus@lemmy.blahaj.zone 5 points 3 months ago

They're correct, it's just overcomplicated as fuck in ways that are correct but completely irrelevant to the question.

[-] WolfLink@sh.itjust.works 4 points 3 months ago

The answer in the meme (10i^2) is -10

[-] Ravi@feddit.org 4 points 3 months ago

Depends on what are the allowed values for x are. Real numbers, complexe numbers, binary or I made up my own numbers ;)

[-] Malgas@beehaw.org 3 points 3 months ago

Probably what they were going for, but there are literally an infinite number of exotic arithmetic spaces you could ask this question in. For example, x=10 works in any ring with a modulus greater than 100 and less than 1000.

[-] Umbrias@beehaw.org 1 points 3 months ago

fortunately math problems are administered in the context of the class, so it will be pretty obvious that it's in the complex plane.

[-] Xavienth@lemmygrad.ml 5 points 3 months ago* (last edited 3 months ago)

Therefore i¹⁰ = ln(-1)¹⁰/pi¹⁰ = -1

This is true but does not follow from the preceding steps, specifically finding it to be equal to -1. You can obviously find it from i²=-1 but they didn't show that. I think they tried to equivocate this expression with the answer for e^iπ^ which you can't do, it doesn't follow because e^iπ^ and i¹⁰ = ln(-1)¹⁰/pi¹⁰ are different expressions and without external proof, could have different values.

[-] Dalvoron@lemm.ee 2 points 3 months ago

If we know the values of ln(-1)¹⁰ and pi¹⁰ we hypothetically could calculate their divided result as -1 instead of using strict logic, but it is missing a few steps. Moreover logs of negative numbers just end up with an imaginary component anyway so there isn't really any progress to be made on that front. Typing ln(-1)¹⁰ into my scientific calculator just yields i¹⁰pi¹⁰, (I'm guessing stored rather than calculated? Maybe calculated with built in Euler) so the result of division is just i¹⁰ anyway and we're back where we started.

[-] Xavienth@lemmygrad.ml 2 points 3 months ago

You can find the value of ln(-1)¹⁰ by examining the definition of ln(x): the result z satisfies eᶻ=x. For x=-1, that means the z that satisfies eᶻ=-1. Then we know z from euler's identity. Raise to the 10, and there's our answer. And like you pointed out, it's not a particularly helpful answer.

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this post was submitted on 24 Jul 2024
357 points (96.6% liked)

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