-3 id the hidden character of the solution, like evil ryu or devil jin.

# teachings

Submitted 4 weeks ago by fossilesque@mander.xyz to science_memes@mander.xyz

https://mander.xyz/pictrs/image/7a77a0d9-bf97-4b35-aa7a-8ab42497c005.jpeg

## Comments

## namelivia@lemmy.world 4 weeks ago

## lseif@sopuli.xyz 4 weeks ago

just a glimpse into my dark and twisted mind

## ZILtoid1991@lemmy.world 4 weeks ago

But for the Joker, that’s the real solution.

## Gemini24601@lemmy.world 4 weeks ago

Doesn’t x also -3?

## dankestnug420@lemmy.ml 4 weeks ago

Sqrt(9)

## UnRelatedBurner@sh.itjust.works 4 weeks ago

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)## Evil_Shrubbery@lemm.ee 4 weeks ago

Fund the sqrter!

## Kusimulkku@lemm.ee 4 weeks ago

Sqrt

hehe

## Crashumbc@lemmy.world 4 weeks ago

Also math teacher…

“Show your work”

## merari42@lemmy.world 4 weeks ago

Middle school math memes

## xkforce@lemmy.world 4 weeks ago

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.

## GnomeKat@lemmy.blahaj.zone 4 weeks ago

I think you can make arbitrarily complicated roots if you move over to

**G^n^**which includes the**R**and**C**roots…For example

`(3e1e2e3e4)^2 = 9`

in**G^4^**, complex roots are covered because`e1e2^2 = -1`

making it identical to`i`

so**G^n^***(n>=2)*includes**C**.**G^n^**also includes all the vectors so any vector with length 3 will square to 9 because`u^2 = u dot u = |u|^2`

## ouRKaoS@lemmy.today 4 weeks ago

You lost me at “arbitrarily complicated,” sorry.

## MBM@lemmings.world 4 weeks ago

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.

## someacnt_@lemmy.world 4 weeks ago

Then you can extend to arbitrary algebra

## Beetschnapps@lemmy.world 4 weeks ago

To translate: As a child learning math this means “ignore math, the explanations don’t explain anything real, they only explain math. So instead focus on language and the arts.”

## overcast5348@lemmy.world 4 weeks ago

I’m guessing that you were one of those “I won’t ever use all this math” kind of students?

## Maalus@lemmy.world 4 weeks ago

Or you were just shit at maths and don’t have any idea how useful it is because you avoid it like the plague.

## Patrizsche@lemmy.ca 4 weeks ago

Me, a statistician: “if chi-square equals 9 then chi equals 3… What??”

## space@lemmy.dbzer0.com 4 weeks ago

My teacher explained as sqrt(poop^2) = abs(poop). Yes, he wrote poop on the blackboard.

## NeatNit@discuss.tchncs.de 4 weeks ago

He should have drawn a pile of poop instead 💩 (preferably without a face)

## Sweetpeaches69@lemmy.world 4 weeks ago

Oh, I know this one! It’s pi!

## jol@discuss.tchncs.de 4 weeks ago

What, no. It’s… Eh close enough.

## Reddfugee42@lemmy.world 4 weeks ago

TAU IS BETTER

/obligatory

## mexicancartel@lemmy.dbzer0.com 4 weeks ago

Help how do i take factorial of pi

## hemko@lemmy.dbzer0.com 4 weeks ago

-3 = 3

## mexicancartel@lemmy.dbzer0.com 4 weeks ago

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

## hemko@lemmy.dbzer0.com 4 weeks ago

I gotta say half of that goes over my head, but I raise my hat to you

## ech@lemm.ee 4 weeks ago

Absolutely.

## bleistift2@feddit.de 4 weeks ago

This only every got handed down to us as gospel. Is there a compelling reason why we should accept that (-3) × (-3) = 9?

## notabot@lemm.ee 4 weeks ago

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.

## bleistift2@feddit.de 4 weeks ago

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

## affiliate@lemmy.world 4 weeks ago

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.

## Davel23@fedia.io 4 weeks ago

Same reason that a double negative makes a positive.

## ImplyingImplications@lemmy.ca 4 weeks ago

Here’s a other 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.

## Maggoty@lemmy.world 4 weeks ago

I know the math but I still feel like I’m out of the loop somehow?

## kszeslaw@szmer.info 4 weeks ago

(-3)^2 = 9 as well

## JackbyDev@programming.dev 4 weeks ago

There’s nothing more to this than linking the star wars quote to the -3. That’s it lol

## Maggoty@lemmy.world 4 weeks ago

Oh okay. Don’t mind me trying to over analyze things.

## Neato@ttrpg.network 4 weeks ago

-3 feels like cheating.

## ech@lemm.ee 4 weeks ago

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.

## knorke3@lemm.ee 4 weeks ago

just wait for roots of imaginary numbers :)