1 + 1 = 2

[Olli]

SPACETIME

[Olli]

i² + j² + k² = ijk = -1

[jacdelad]

√(-shit²) Shit just got real

Erm...no. The root of a negative, real number is an imaginary number: https://en.wikipedia.org/wiki/Imaginary ... prov=sfla1

[Lord]

It's (-shit)² and not -(shit²)

[Michael Vogel]

Code: Select all

uno=1
eins=1
one=1

Debug "1 + 1 * (1-1) + 1 = ?"
Debug ""
Debug one + (Red(eins*(uno-Red(eins))) + uno)
Debug "...or..."
Debug one + (uno + Red(eins*(uno-Red(eins))))

PS: you can replace Red() by Int() or Bool() or whatever

[jacdelad]

It's (-shit)² and not -(shit²)

Ah sh..., you're right. Or not? https://www.wolframalpha.com/input?i2d= ... 5C%2841%29
How about √(|shit|) ?

[fsw]

6/2(1+2) = 9
or
6/2(1+2) = 1
?
The answer depends on your calculator...

[Olli]

From left to right :
6/2(1+2) = 3(1+2) = 3×3 = 9
6/[2(1+2)] = 6/[2×3] = 6/6 = 1

[Piero]

Heh... isn't that really great? I read a wonderful explanation of how he got the equation (deducting from sin/cos formulas) but cannot remember the book

[Piero]

6/2(1+2) = 9
or
6/2(1+2) = 1
?
The answer depends on your calculator...

Maybe I'm missing something, but if calculator respects parenthesis, isn't it always 9?

[Olli]

Heh... isn't that really great? I read a wonderful explanation of how he got the equation (deducting from sin/cos formulas) but cannot remember the book

+

6/2(1+2) = 9
or
6/2(1+2) = 1
?
The answer depends on your calculator...

Maybe I'm missing something, but if calculator respects parenthesis, isn't it always 9?

=



for piero : on 4th minute and 17th second, when the guy starts to shoot the e exponential symbol, cos and sin appear !!
Else you have this :
e^(ix) = cos(x) + iSin(x)

and if you want to see a simple man who walks, here is a link..

[Piero]

6/2(1+2) = 6*1/2*3

[Olli]

Note :

i is considered as a vector.

i is defined as i * i = (-1)

As (-1) * (-1) = 1,
i * i * i * i = 1

And i * i * i * i is also considered as a vector
This vector is as i / i.

If we represent the two vectors i and i / i on a right plane, we can consider that, if i^(4*k) = 1 (k being a relative integer number), then there are 4 directions for the vectors i^(4*k). i is constant. So the 4 angles between the 4 directions are the same ones. So, the vectors i^1 and i^0 are perpendicular on this right plane.

By extension, if we consider :
i * i = -1
j * j = -1
k * k = -1
In a space, where we forget i^0 as a vector, there is a mechanic to find several relative directions for i, j and k in this space (as i * i * i * i = 1) from this : i * k * j / 1 = 1.

[Piero]

from this : i * k * j / 1 = 1.

[Olli]

Nice tribute! Note that the notion of using three distinct imaginary numbers i, j and k to apply 3 directions in a space, had its first hours in Hamilton's mind.

What it is off topic, but it surprises me a lot, is the death we give to this brilliant mind :
<< He [Hamilton] died on 2 September 1865, following a severe attack of gout. >>

I am very surprised, because Jacques Psychriste discovered that colchicum was an antidote for gout attacks, 1300 years earlier !

Plus colchicum appears naturally in Europa every year on September, since the greek gold age !

I am subject to this disease, and never I will die by this disease, even if it would be 100 times more strong on me. Colchicum acts two hours after a taken, and we must calculate precisely and beforehand according to its weight (else it's a very powerful poison).

Hamilton was however a mathematician...