What about plain old x = -10
?
-10 ^ 2 = 100
-10 ^ 3 = -1000
-10 ^ 5 = -100000
A place for majestic STEMLORD peacocking, as well as memes about the realities of working in a lab.
Rules
What about plain old x = -10
?
-10 ^ 2 = 100
-10 ^ 3 = -1000
-10 ^ 5 = -100000
Isn't that the joke?
That's what he wrote, I imagine.
i² = -1 so...
10 * i^2 is -10.
of course, but this problem is solvable without any understanding of complex numbers, and 10*i^2 is a really clunky, multi-operation expression whereas -10 is just an integer.
simplifying one's answers is standard practice and any grader who received the answer in the OP would be obligated to point out that while technically correct, they're missing the basic fact that the answer is -10.
The Rube Goldberg comment is apt as the solution is absurdly complicated and overengineered for the task it performs.
Yes that's what the thing you are looking at is built upon.
Nothing gets past you eh?
people being pedantic showoffs doesn't really register as humor for me, TBH
That's true, the OOP is being quite snarky with their comment on a post where someone's had a genuine basic doubt
Yeah exactly you're right, why overcomplicate the problem like the Reddit comment did? I guess that's just typical Reddit thinking that being pendantic and using lots of fancy words and long explanations makes you smart.
Nah just boiler plate autism.
no
That was my immediate thought too.
Boooooring
When all you have is an imaginary hammer, everything looks like a rotation around the imaginary unit circle.
Explanation of maths
x = -10, i = √-1 so i² = -1 and 10i²=-10
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.
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.
Wait, isn’t x just -10 if x^3 is not 1000?
yes, it is
that is a very long way to write -10
My brain
It hurts
That's because the explanation was about 10 times as complicated as it needs to be
Math pun intended?
He is trolling with overcomplicating
What an extremely unnecessary explanation. As a math teacher I would have deducted points for this answer.
"show your work"
Malicious compliance intensifies
No definition what values are suitable for x.
x has to be -10, right? Or am I missing something?
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).
They said x=10i^2, not 10i. Difference is it equals -10, and they chose not to simplify.
They're correct, it's just overcomplicated as fuck in ways that are correct but completely irrelevant to the question.
The answer in the meme (10i^2) is -10
Depends on what are the allowed values for x are. Real numbers, complexe numbers, binary or I made up my own numbers ;)
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.
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.
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.
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.