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[–] 274 points 1 year ago (27 children)

Fake and gay.

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

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  • [–] 53 points 1 year ago (9 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.

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  • [–] 6 points 1 year 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

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  • [–] 19 points 1 year ago (3 children)

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

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  • [–] 2 points 1 year ago (2 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...

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  • [–] 37 points 1 year ago (3 children)

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

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  • [–] 18 points 1 year ago (7 children)

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

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  • [–] 3 points 1 year ago (6 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.

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  • [–] 1 point 1 year ago (5 children)

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

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  • [–] 2 points 1 year ago (4 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).

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  • [–] 2 points 1 year ago (3 children)

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

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  • [–] 1 point 1 year ago (2 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.

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  • [–] 2 points 1 year ago (1 child)

    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.

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  • [–] 0 points 1 year 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.

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