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UwU brat mathematician behavior
(mander.xyz)
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this post was submitted on 14 Jul 2025
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Fake and gay.
No way the engineer corrects the mathematician for using j instead of i.
As an engineer I fully agree. Engineers¹ aren't even able to do basic arithmetics. I even cannot count to 10.
¹ Except maybe Electrical engineers. They seem to be quite smart.
Engineer here, I can definitely count to 10 tho
0 1 10
0 1 everything that comes after is simply summarizes as "many"
Electrical engineers are the ones that use j though (because i is used for current)
I am used for current
10? That’s the name some put to 1e1, right?
Yup, I can count just fine to the 10th number in a zero-indexed counting system: black, brown, red, orange, yellow, green, blue, violet, gray, white.
https://xkcd.com/227/
The inner machinations of an electrical engineer is too complicated for me to understand, I think they might be thinking on a higher order to understand these circuits
Thats why I barely passed my electrical engineering class lol
Having worked with electrical engineers, some of them are quite smart, the rest have lead poisoning.
How do we know it's gay though? OP could be a girl (male)
Because it's 4chan. And there are no women on the Internet on 4chan
Sure OP is a girl. Guy In Real Life
Newfag.
(sorry! seemed like the appropriate 4chan reply)
Right? They got that shit backwards. Op is a fraud. i is used in pure math, j is used in engineering.
The mathematician also used "operative" instead of, uh, something else, and "associative" instead of "commutative"
I think they meant "operand". As in, in the way dy/dx can sometimes be treated as a fraction and dx treated as a value.
I think you mean operator. The operand is the target of an operator.
Correct. Thus, dx is an operand. It's a thing by which you multiply the rest of the equation (or, in the case of dy/dx, by which you divide the dy).
I'd say the $\int dx$ is the operator and the integrand is the operand.
You're misunderstanding the post. Yes, the reality of maths is that the integral is an operator. But the post talks about how "dx can be treated as an [operand]". And this is true, in many (but not all) circumstances.
∫(dy/dx)dx = ∫dy = y
Or the chain rule:
(dz/dy)(dy/dx) = dz/dx
In both of these cases, dx or dy behave like operands, since we can "cancel" them through division. This isn't rigorous maths, but it's a frequently-useful shorthand.
I do understand it differently, but I don't think I misunderstood. I think what they meant is the physicist notation I'm (as a physicist) all too familiar with:
∫ f(x) dx = ∫ dx f(x)
In this case, because f(x) is the operand and ∫ dx the operator, it's still uniquely defined.
Ok that's some really interesting context I didn't know. I've only ever seen it done the mathematician's way with dx at the end. Learning physicists do it differently explains why the person in the post would want to discuss moving it around.
But I still think they have to mean "if dx can be treated as an operand". Because "if dx can be treated as an operator" doesn't make sense. It is an operator; there's no need to comment on something being what it objectively is, and even less reason to pretend OOP's partner was angry at this idea.
My thoughts exactly lol