Comment on nooo my genderinos

m8052@lemmy.world ⁨2⁩ ⁨months⁩ ago

• rudyharrelson@lemmy.radio ⁨2⁩ ⁨months⁩ ago

  Wolfram tells me sqrt(-1) = i and it hasn’t lied to me yet.

  In what meaningful way is i^2 = -1 different from sqrt(-1) = i?

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  • lvxferre@mander.xyz ⁨2⁩ ⁨months⁩ ago

    sqrt(-1) = ±i. The negative answer is also valid.

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    • excral@feddit.org ⁨2⁩ ⁨months⁩ ago

      This is technically incorrect. While the square root is both the positive and the negative solution, the sqrt (√) operator results in the principal square root. For nonnegative numbers this is the nonnegative square root and more generalised for complex numbers it’s the square root that halves the complex phase.

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    • rudyharrelson@lemmy.radio ⁨2⁩ ⁨months⁩ ago

      Ah, good point; I’d forgotten that part.

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  • m8052@lemmy.world ⁨2⁩ ⁨months⁩ ago

    Square root definition does not allow a negative number as an input. Only positives and zero. Although it is possible to expand the definition to negative numbers, complex numbers, matrices… So unless you followed a course where you thoroughly defined your expansion of sqrt, it only applies to real, positives number and zero. Its the thing with math, you have to define what you work with.

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    • rudyharrelson@lemmy.radio ⁨2⁩ ⁨months⁩ ago

      Sounds to me like this is exactly what the OP meme is referencing.

      Basic math: square root only of positive numbers and 0. Advanced math: square root of anything you want

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      • LibertyLizard@slrpnk.net ⁨2⁩ ⁨months⁩ ago

        So what’s the square root of your mom then?

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      • m8052@lemmy.world ⁨2⁩ ⁨months⁩ ago

        You’re missing the point. Math (especially advanced) is about precision and rigor. Writing sqrt of something negative is ambiguous. There are better ways of writing it as explained here lemmy.world/comment/18924227

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    • TwilightKiddy@programming.dev ⁨2⁩ ⁨months⁩ ago

      I’m very sorry you had to go through such stupid school tests, welcome to the real world. sqrt(x) is just a shorthand for x^(1/2).

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• Khanzarate@lemmy.world ⁨2⁩ ⁨months⁩ ago

  Wouldn’t the square root just give plus/minus i? Seems correct enough.

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  • Opisek@lemmy.world ⁨2⁩ ⁨months⁩ ago

    No. The symbol √ signifies the principal square root of a number. Therefore, √x is always positive. The two roots of x, however, are ±√x. If you therefore have y²=x and you want to find x, you mustn’t write y=√x, but rather y=±√x to be formally correct.

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• msfroh@lemmy.ca ⁨2⁩ ⁨months⁩ ago

  But (-i)^2=-1 as well. So we still need a convention to distinguish i from -i.

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  • excral@feddit.org ⁨2⁩ ⁨months⁩ ago

    That’s fairly simple: we restrict the complex phase to the range (-pi, pi] and the principal square root halves the complex phase. -1 has the phase value pi, so the principal square root has the the complex phase pi/2, so it’s i, while -i has a phase of -pi/2

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• moobythegoldensock@infosec.pub ⁨2⁩ ⁨months⁩ ago

  They’re the same thing. You just take the square root of both sides to get i = sqrt(-1).

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• Ethanol@pawb.social ⁨2⁩ ⁨months⁩ ago

  Indeed, usually you would want to avoid a notation of sqrt(-1) or (-1)^(1/2). You would use e^(1/2 log(-1)) instead because mathematicians have already decided on a “natural” way to define the logarithm of complex numbers. The problem here lies with choosing a branch of the logarithm as e^z = x has infinitely many complex solutions z. Mathematicians have already decided on a default branch of the logarithm you would usually use. This matters because depending on the branch you choose sqrt(-1) either gives i or -i. A square-root is usually defined to only give the positive solution (if it had multiple values it wouldn’t fit the definition of a function anymore) but on the complex plane there isn’t really a “positive” direction. You would have to choose that first to make sure sqrt is defined as a function and you do that via the logarithm branch.
  So, just writing sqrt(-1) leaves ambiguity as you could either define it to give i or -i but writing e^(1/2 log(-1)) then everyone would just assume you use the default logarithm branch and the solution is i.

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• NoneOfUrBusiness@fedia.io ⁨2⁩ ⁨months⁩ ago

  Nah, sqrt(x) is the principal branch (the one with a positive real part) of x^½, and you can do (-1)^½ because it's just exponentiation.

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• silasmariner@programming.dev ⁨2⁩ ⁨months⁩ ago

  Simply define it as i = e^(iπ/2) 🤣

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