this post was submitted on 14 Jul 2025
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[–] Zagorath@aussie.zone 296 points 2 weeks ago (5 children)

Fake and gay.

No way the engineer corrects the mathematician for using j instead of i.

[–] LeFrog@discuss.tchncs.de 53 points 2 weeks ago (6 children)

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.

[–] boonhet@sopuli.xyz 46 points 2 weeks ago (2 children)

Engineer here, I can definitely count to 10 tho

0 1 10

0 1 everything that comes after is simply summarizes as "many"

[–] gnutrino@programming.dev 33 points 2 weeks ago (1 children)

Electrical engineers are the ones that use j though (because i is used for current)

[–] Klear@sh.itjust.works 2 points 2 weeks ago

I am used for current

[–] thomasloven@lemmy.world 12 points 2 weeks ago

10? That’s the name some put to 1e1, right?

[–] exasperation@lemmy.dbzer0.com 10 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

Except maybe Electrical engineers.

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.

[–] Baguette@lemmy.blahaj.zone 6 points 2 weeks ago

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

[–] Fedizen@lemmy.world 5 points 2 weeks ago

Having worked with electrical engineers, some of them are quite smart, the rest have lead poisoning.

[–] Hoimo@ani.social 39 points 2 weeks ago (3 children)

How do we know it's gay though? OP could be a girl (male)

[–] SippyCup@feddit.nl 59 points 2 weeks ago

Because it's 4chan. And there are no women on the Internet on 4chan

[–] floquant@lemmy.dbzer0.com 19 points 2 weeks ago

Sure OP is a girl. Guy In Real Life

[–] ByteJunk@lemmy.world 11 points 2 weeks ago

Newfag.

(sorry! seemed like the appropriate 4chan reply)

[–] TheSlad@sh.itjust.works 20 points 2 weeks ago (1 children)

Right? They got that shit backwards. Op is a fraud. i is used in pure math, j is used in engineering.

[–] Chakravanti@monero.town 2 points 2 weeks ago (1 children)

That's hilarious. You're not seeing what's going on backwards just like that (as I point at the point going nowhere shitty) in an equation that is finding as many clAEver ways to say something you actually not caring about talking about.

That's like, "How many time van express the only thing that van't be done until the 'verse itself tries to do what can't be done and sever your...

...Oh, I see...you don't have ([of course, because you can't have to give {is}) nothing)] to give.

Unable to sea time doesn't mean we can't see(k)ER the mAETh.ac(k).cc(k).08

The only thin(g):(k) that doesn't ever be never, is not at alla hack(g)in(g).G your lackthereof to divi...

[–] TheSlad@sh.itjust.works 3 points 2 weeks ago (1 children)

Is this a copypasta or are you having a stroke?

[–] Chakravanti@monero.town 1 points 2 weeks ago

Wah, wah, wah...

Not my problem.

Keep trying. /s

[–] kogasa@programming.dev 20 points 2 weeks ago (1 children)

The mathematician also used "operative" instead of, uh, something else, and "associative" instead of "commutative"

[–] Zagorath@aussie.zone 3 points 2 weeks ago (1 children)

"operative" instead of, uh, something else

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.

[–] Chrobin@discuss.tchncs.de 1 points 2 weeks ago (1 children)

I think you mean operator. The operand is the target of an operator.

[–] Zagorath@aussie.zone 2 points 2 weeks ago (1 children)

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).

[–] Chrobin@discuss.tchncs.de 2 points 2 weeks ago (1 children)

I'd say the $\int dx$ is the operator and the integrand is the operand.

[–] Zagorath@aussie.zone 1 points 2 weeks ago (1 children)

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.

[–] Chrobin@discuss.tchncs.de 2 points 2 weeks ago (1 children)

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.

[–] Zagorath@aussie.zone 0 points 1 week ago

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.

[–] NeatNit@discuss.tchncs.de 12 points 2 weeks ago

My thoughts exactly lol