The number of solutions/roots is equal to the highest power x is raised to (there are other forms with different rules and this applies to R and C not higher order systems)
Some roots can be complex and some can be duplicates but when it comes to the real and complex roots, that rule generally holds.
To translate: As a child learning math this equates to “ignore math, the explanations don’t explain anything real, they only explain more math.“
“The only explanation is more abstraction with no real world application as far as math class is concerned. Frankly, there’s more application to your own life experience if you focus on language and the arts.”
I was one of those students who asked how it would be used, the teachers didn’t do the whole real world application part, and I never needed to go past trig.
I work with engineers and use math like any other human on the planet but really wish mathematics was taught differently to make it more interesting. You hear a PHD candidate talk about the hairy ball problem and the math is interesting. Math class never was.
I think you can make arbitrarily complicated roots if you move over to G^n^ which includes the R and C roots…
For example the grade 4 blade (3e1e2e3e4)^2 = 9 in G^4^
Complex roots are covered because the grade 2 blade (e1e2)^2 = -1 making it identical to i so G^n^(n>=2) includes C.
G^n^ also includes all the scalars (grade 0 blades) so all the real roots are included.
G^n^ also includes all the vectors (grade 1 blades) so any vector with length 3 will square to 9 because u^2 = u dot u = |u|^2 where u is a vector.
All blades will square to a scalar but blades are not the only thing in G^n^ so things get weird with the multivectors(sums of different grades). Any blade with grade n%4 < 2 will square to a positive scalar and the other grades will square to a negative, with the abs of the scalar equal to the norm^2^ of the blade. Can pretty much just make as many roots as you want if you are willing to move into higher dimensional spaces and use a way cooler product.
I thought this would be related to quaternions, octonions etc. but no, it’s multivectors and wedge products. Very neat, I didn’t know you could use them like that.
Oh no, you were right on the money. In G^2^ you have two basis vectors e1 and e2. The geometric product of vectors specifically is equivalent to uv = u dot v + u wedge v… the dot returns a scalar, the wedge returns a bivector. When you have two vectors be orthonormal like the basis vectors, the dot goes to 0 and you are just left with u wedge v. So e1e2 returns a bivector with norm 1, its the only basis bivector for G^2^.
e1e2^2 = (e1e2)*(e1e2) = e1e2e1e2
A nice thing about the geometric product is its associative so you can rewrite as e1*(e2e1)*e2… again that middle product is still just a wedge but the wedge product is anti commutative so e2e1 = -e1e2. Meaning you can rewrite the above as e1*(-e1e2)*e2 = -(e1e1)*(e2e2) = -(e1 dot e1)*(e2 dot e2) = -(1)*(1) = -1… Thus e1e2 squares to -1 and is the same as i. And now you can think of the geometric product of two vectors as uv = u dot v + u wedge v = a + bi which is just a complex number.
In G^3^ you can do the same but now you have 3 basis vectors to work with, e1, e2, e3. Meaning you can construct 3 new basis bivectors e1e2, e2e3, e3e1. You can flip them to be e2e1, e3e2, e1e3 without any issues its just convention and then its the same as quaternions. They all square to -1 and e2e1*e3e2*e1e3 = -e2e1e2e3e1e3 = e2e1e2e1e3e3 = e2e1e2e1 = -1 which is the same as i,j,k of quaternions. So just like in G^2^ the bivectors + scalars form C you get the quaternions in G^3^. Both of them are just bivectors and they work the same way. Octonions and beyond can be made in higher dimensions. Geometric algebra is truly some cool shit.
i think this is a really clean explanation of why (-3) * (-3) should equal 9. i wanted to point out that with a little more work, it’s possible to see why (-3) * (-3) must equal 9. and this is basically a consequence of the distributive law:
the first equality uses 0 * anything = 0. the second equality uses (3 + -3) = 0. the third equality uses the distribute law, and the fourth equality uses 3 * (-3) = -9, which was shown in the previous comment.
in other words, 9 = (-3) * (-3). this basically says that if we want the distribute law to be true, then we need to have (-3) * (-3) = 9.
it’s also worth mentioning that this is a specific instance of a proof that shows (-a) * (-b) = a * b is true for arbitrary rings. (a ring is basically a fancy name for a structure with addition and distribute multiplication.) so, any time you want to have any kind of multiplication that satisfies the distribute law, you need (-a) * (-b) = a * b.
in particular, (-A) * (-B) = A * B is also true when A and B are matrices. and you can prove this using the same argument that was used above.
1/0 can be infinity as a limiting case but not always. So imma use what i like since i also took ln|x| (took mod out of nowhere just to make it positive)
Eh, not really. It’s been a while, but I’m pretty sure the rule in algebra when solving for a squared variable like this is to use ± for exactly that reason.
Afaik sqrt only returns positive numbers, but if you’re searching for X you should do more logic, as both -3 and 3 squared is 9, but sqrt(9) is just 3.
If I’m wrong please correct me, caz I don’t really know how to properly write this down in a proof, so I might be wrong here. :p
(ps: I fact checked with wolfram, but I still donno how to split the equation formally)
So I checked this on my smartphone first, and thought maybe the software is just shit… So then I checked it on a Casio scientific calc, and both agree.
-3^2 = -9… And 9 != -9
… Are all the calculators somehow wrong? What’s the math rule I’m forgetting here…
Add comment