Please any help would be awesome...

[twin dew]

Kindly dm

[rapid rampart]

Hint: consider the pairs (a, b) and (c, d) as complex numbers a + bi and c + di.
If you don't know how to work with them:
The main thing is that i has the property i^2 = -1.
Addition: (a + bi) + (c + di) = (a + c) + (b + d)i.
Multiplication: (a + bi)(c + di) = ac + (ad + bc)i + bdi^2 = (ac - bd) + (ad + bc)i. Seems familiar?

[wise atlas]

Sort of, it's just really hard for me because I took a long break from math

Thanks for helping but I don't know how to expand off of what you told me. I researched about your suggestion earlier but I'm still stumped

[rapid rampart]

You expand it as usual multiplication, only considering that i^2 = -1.

Thus, (a + bi)(c + di) = (ac - bd) + (ad + bc)i.

You can see that this completely coincides with our ⊡ operation.

So, now you just need to verify that this multiplication is still commutative and associative, like regular multiplication.

[wise atlas]

I can't seem to figure it out...

[rapid rampart]

Well, let's see.
First, commutativity.
(a, b) ⊡ (c, d) = (ac - bd, ad + bc)
(c, d) ⊡ (a, b) = (ca - db, cb + da)
They are equal. So, ⊡ is commutative.

Try doing associativity. The expressions will be long, so just do it slowly and carefully.

[wise atlas]

I don't know...

How do you make it so easy

I tried introducing a new variable but I hit a block again

this is actually miserable

[rapid rampart]

Well, let's see. We just need to carefully multiply.
((a, b) ⊡ (c, d)) ⊡ (e, f) = (ac - bd, ad + bc) ⊡ (e, f) = (e(ac - bd) - f(ad + bc), f(ac - bd) + e(ad + bc))
(a, b) ⊡ ((c, d) ⊡ (e, f)) = (a, b) ⊡ (ce - df, cf + de) = (a(ce - df) - b(cf + de), a(cf + de) + b(ce - df))

Now just verify that those are the same.

[wise atlas]

Can you explain the bold part?

((a, b) ⊡ (c, d)) ⊡ (e, f) = (ac - bd, ad + bc) ⊡ (e, f) = (e(ac - bd) - f(ad + bc), f(ac - bd) + e(ad + bc))

Thank you so much btw...

[rapid rampart]

The first term is first*third - second*fourth. The second is first*fourth + second*third.

[wise atlas]

And that's just a rule?

Is there a name for it

I've noticed that pattern but I didn't know what to google

[rapid rampart]

Well' that's the definition of our operation.

[wise atlas]

Oh

wow

you are godsent btw... forrest gump

ty so so much