this post was submitted on 21 Aug 2025
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Sqrt(-1) is still wrong tho. I'm commuting a sin by writting it. Correct expression is i^2=-1
Wolfram tells me
sqrt(-1) = iand it hasn't lied to me yet.In what meaningful way is
i^2 = -1different fromsqrt(-1) = i?sqrt(-1) = ±i. The negative answer is also valid.
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.
Ah, good point; I'd forgotten that part.
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.
In my case, I did prep courses for entrance exam to engineering schools (something like in dead poet society but more modern), using sqrt(-1) somewhere would be an instant 0 mark. Like forgetting a unit in a physics test answer.
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
So what’s the square root of your mom then?
Damn, hoisted by my own petard.
√∞ = ∞
You mean limit of sqrt(x) when x->+inf = +inf
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 https://lemmy.world/comment/18924227
Except that yes, math is about precision and rigor, and yes, the square root of negative numbers is totally allowed (https://en.wikipedia.org/wiki/Square_root#Square_roots_of_negative_and_complex_numbers) and yes, math totally allows for statements with multiple solutions.
sin(x) = 0also has infinite solutions and yet it's totally legal for x to make an equation like that.I'm very sorry you had to go through such stupid school tests, welcome to the real world.
sqrt(x)is just a shorthand forx^(1/2).Wouldn't the square root just give plus/minus i? Seems correct enough.
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 y, you mustn't write y=√x, but rather y=±√x to be formally correct.
But (-i)^2=-1 as well. So we still need a convention to distinguish i from -i.
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
They’re the same thing. You just take the square root of both sides to get i = sqrt(-1).
Indeed, usually you would want to avoid a notation of
sqrt(-1)or(-1)^(1/2). You would usee^(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 ase^z = xhas infinitely many complex solutionsz. Mathematicians have already decided on a default branch of the logarithm you would usually use. This matters because depending on the branch you choosesqrt(-1)either givesior-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 suresqrtis 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 giveior-ibut writinge^(1/2 log(-1))then everyone would just assume you use the default logarithm branch and the solution isi.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.
Simply define it as i = e^(iπ/2) 🤣