You must log in or register to comment.
What about plain old
x = -10?-10 ^ 2 = 100
-10 ^ 3 = -1000
-10 ^ 5 = -100000Isn’t that the joke?
That’s what he wrote, I imagine.
It is, but with imaginary numbets
numbet
numberwang
i² = -1 so…
10 * i^2 is -10.
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
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
Found the math but no explanation.
The squareroot of 100 is ±10.
The square root is always positive, but you can plug it into the quadratic formula to get the two possible values.
Okay, fine the square roots of 100 are ±10.
Seems very inaccurate the we can only determine the square root to ±10.
That’s not how the square root is defined.
You’re confusing “square root of 100” with “the answer to x^2 = 100”. These are different things.
Which is why I differentiated between square roots and the principle square root by saying the square roots instead of the square root on the second comment.
so you came up with your own term to cover your mistake?
There’s no reason to bring the quadratic formula into this. Square roots can be negative, but when talking about the square root it’s normally assumed to be the principal square root, which is the positive one.
Nope. To clarify, square roots are the opposite of squaring.
Now ask yourself:
What is 10² ?
What is (-10)² ?
If you get the same answer, then they are both the roots of the answer. +10 and -10 then gets together called ±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
Unless I was in that clas where we had to write mathematical proofs. I HATED those. Sure, you solved the question but write out this complicated reason for why your answer is the correct answer.
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.
fortunately math problems are administered in the context of the class, so it will be pretty obvious that it’s in the complex plane.
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 eiπ which you can’t do, it doesn’t follow because eiπ 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.
deleted by creator
import math
