-3 feels like cheating.
this post was submitted on 17 Apr 2024
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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.
just wait for n-th roots of imaginary numbers :)
-3 id the hidden dark version character of the solution, like evil ryu or devil jin.
just a glimpse into my dark and twisted mind
Doesn’t x also equal -3?
Sqrt(9)
Uhm, actually 🤓☝️!
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)
You're correct. The square root operator only returns the principal root (the positive one).
So if x^2 = 9 then x = ±√9 = ±3
That's why in something like the quadratic formula we all had to memorize in school its got a "plus or minus" in it: -b ± √...(etc)
Thanks, I haven't connected the dots to that (+-) sign and this problem.
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x^2 = 9
<=>
|x| = sqrt(9)
would be correct. That way you get both 3 and -3 for x.
That's the way your math teacher would do it. So the correct version of the statement in the picture is: "if x^2 = 9 then abs(x) = 3"
Cool! Makes sense to me. Honestly, I've never done it this way, but it's so clean. Love it. Thanks.
Fund the sqrter!
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Also math teacher...
"Show your work"
Middle school math memes
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.
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.
You lost me at "arbitrarily complicated," sorry.
Lost me at "I think". I don't, apparently.
I don't think therefore I never was
I wish it was that easy! I stopped thinking and I'm still stuck here... Stupid Descartes
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.
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Me, a statistician: "if chi-square equals 9 then chi equals 3... What??"
My teacher explained as sqrt(poop^2) = abs(poop). Yes, he wrote poop on the blackboard.
He should have drawn a pile of poop instead 💩 (preferably without a face)
Oh, I know this one! It's pi!
-3 = 3
Absolutely.
Adding 3 on both sides
3-3=3+3
0 = 6
1•0 = 6
1 = 6/0
1 = inf
Multiplying e^(iπ) on both sides,
e^(iπ) = - inf
iπ = ln|-inf|
π = ln|-inf| ÷ i
1/0 isn't infinity though.
Like if everything else is true.
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)
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I know the math but I still feel like I'm out of the loop somehow?
There's nothing more to this than linking the star wars quote to the -3. That's it lol
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(-3)^2 = 9 as well
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