Science Memes @mander.xyz fossilesque @mander.xyz 8 mo. ago

teachings

67 comments

-3 feels like cheating.
- 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
  
  But for the Joker, that's the real solution.
Doesn’t x also equal -3?
- You found the other
  
  Now kiss
- 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)
  
  Fund the sqrter!
  
  Sqrt
  
  hehe
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 Gⁿ which includes the R and C roots...
  
  For example the grade 4 blade (3e1e2e3e4)^2 = 9 in G⁴
  
  Complex roots are covered because the grade 2 blade (e1e2)^2 = -1 making it identical to i so Gⁿ (n>=2) includes C.
  
  Gⁿ also includes all the scalars (grade 0 blades) so all the real roots are included.
  
  Gⁿ 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 Gⁿ 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 norm² 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.
  
  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.
  
  Then you can extend to arbitrary algebra
- 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'm guessing that you were one of those "I won't ever use all this math" kind of students?
  
  Or you were just shit at maths and don't have any idea how useful it is because you avoid it like the plague.
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!
- What, no. It's... Eh close enough.
- TAU IS BETTER
  
  /obligatory
- Help how do i take factorial of pi
-3 = 3
- 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.
  
  I gotta say second half of that goes over my head, but I raise my hat to you
  
  it starts out okay but 6/0 is not inf.
- Absolutely.
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
  
  Oh okay. Don't mind me trying to over analyze things.
- (-3)^2 = 9 as well
This only ever got handed down to us as gospel. Is there a compelling reason why we should accept that (-3) × (-3) = 9?
- You can look at multiplication as a shorthand for repeated addition, so, for example:
  
  3x3=0 + 3 + 3 + 3 = 9
  
  In other words we have three lots of three. The zero will be handy later...
  
  Next consider:
  
  -3x3 = 0 + -3 + -3 + -3 = -9
  
  Here we have three lots of minus three. So what happens if we instead have minus three lots of three? Instead of adding the threes, we subtract them:
  
  3x-3 = 0 - 3 - 3 - 3 = -9
  
  Finally, what if we want minus three lots of minus three? Subtracting a negative number is the equivalent of adding the positive value:
  
  -3x-3 = 0 - -3 - -3 - -3 = 0 + 3 + 3 + 3 = 9
  
  Do let me know if some of that isn't clear.
  
  This was very clear. Now that I see it, I realize it’s the same reasoning why x^(-3) is 1/(x^3):
  
  2 × -3 = -6 1 × -3 = -3 0 × -3 = 0 -1 × -3 = 3
  
  Thank you!
  
  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:
  
  0 = 0 * (-3) = (3 + -3) * (-3) = 3 * (-3) + (-3) * (-3) = -9 + (-3) * (-3).
  
  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.
  
  so, by adding 9 to both sides, we get:
  
  9 = 9 - 9 + (-3) * (-3).
  
  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.
- Here's another example:
  
  A) -3 × (-3 + 3) = ?
  
  You can solve this by figuring out the brackets first. -3 × 0 = 0
  
  You can also solve this using the distributive property of multiplication, rewriting the equation as
  
  A) -3 × (-3 + 3) = 0
  (-3 × -3) + (-3 × 3) = 0
  (-3 × -3) - 9 = 0
  (-3 × -3) = 9
  
  If (-3 × -3) didn't equal 9 then you'd get different answers to equation A depending on what method you used to solve it.
- Same reason that a double negative makes a positive.

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