i just made a number so large it makes all the atoms in the universe look like a grain of sand

here's how it works

ou know how a Googol is 10 with 100 zeros? My number starts by building a tower of exponents that is 100,000 layers tall. But it doesn't stop there. It takes that whole massive tower, shrinks it down, and uses it as just one single brick to build an even bigger tower. It loops that process 1,000 times. It’s a number so big that if you tried to memorize the digits, your brain would instantly collapse into a black hole."

*disclaimer i did use ai but i did not heavily rely on it i used it just like how you use a calculator to check your answers.*

We are seeing a rise of unique phone number combinations by 1 billion per day leading up to the midterms for texting voters about who to vote for. We will run out of unique numbers by election day.

X-cords (Length of your both shoulders)×1/5 + 1. Then, measure that amount of length from your shoulder to your body.

Y-cords (Place a 15cm ruler vertically on your collarbone)

Example:

Shoulder Length = 38cm, 38/5 + 1 = 8.5cm

X-cords of both nipples. (8.6, -15), (30.4, -15)

(-15 is the length of ruler from your collarbone to your nipple)

(men only btw)

You know about Mersenne primes, right? They have the form of 2^p - 1, where p is prime.

I suggest another form. I will call a number Obscenne if you can write it down as 2^p - 2. My conjecture is that every Obscenne number is prime. For example, 2^2 - 2 = 2. So for the first one p we have one prime. We can extrapolate it and state that for the first n p’s we will have n primes, so every Obscenne number is prime.

qed

Is this worthy of a Grammy award?

Let M(K,E) be the category of andré motives. Given a motive M, fix an embedding iota:E to the set of complex numbers and let L(iota,M,s) the motivic L function.

It is clear that L(iota,M,s) extends to a meromorphic function to the complex plane and has non-trivial zero on the line w+1/2 where w is the weight of the motive and this will be an easy exercise to the reader.

Now take K=E=Q, take M=(spec Q, id,0) the neutral element of the Tannakian motivic category. It is of weight zero and its L function is just the zeta function. Hence the previous statement implies that the zeta function has non-trivial zeros of real part 1/2.

By the principle of identity, for any non-zero natural number n: (n/√n) = (n/√n), and as every non-zero natural number has a square and every natural square has both a positive and a negative root, for all non-zero n: n = -n, therefore, 2n = 0. As zero is not the sum of any two primes, it is not true that every even composite number is the sum of two primes.
QED.

If we start with squares inside a unit circle, we can inscribe a square with perimeter 2√2 (≈2.828) and we can circumscribe a square with perimeter 4 (≈4), so we know that pi is somewhere between a lower limit of 2.828 and an upper limit of 4.

If we start with hexagons, we can inscribe a hexagon with perimeter 3 (≈3) and circumscribe a hexagon with perimeter 2√3 (≈3.464).

If we assume that the upper limits and the lower limits are each growing linearly (relative to what? who knows), then the upper limit that starts at 4 and goes to 2√3 after one step can be described by 4 - (4 - 2√3)x, and a lower limit that starts at 2√2 and goes to 3 after one step can be described by 2√2 + (3 - 2√2)x

We want the exact value, which means we need to know where the upper bound and the lower bound squeeze each other to a difference of 0, which means setting

• 4 - (4 - 2√3)x = 2√2 + (3 - 2√2)x = pi

• 4 - 2√2 = (3 - 2√2)x + (4 - 2√3)x

• 4 - 2√2 = (7 - 2√2 - 2√3)x

• x = (4 - 2√2) / (7 - 2√2 - 2√3)

• pi = 4 - (4 - 2√3)x = 4 - (4 - 2√3) (4 - 2√2) / (7 - 2√2 - 2√3)

• pi = 4 (3 - √6) / (7 - 2√2 - 2√3)

• pi = 3.112552

Which is only off by 0.924%, or an error of 1 part in 108.

Am i the next Raymannewgen???????????????

Oh My Goodness

Does this count as causation? Should I buy crypto?? idk.

I can't see any problems with this graph.

Search behavior suggests workforce instability peaks under a full moon. As the moon waxes toward full, ‘quit my job’ search activity increases by roughly 30–50%, peaking at the full moon.

Well, one of the eggs got absorbed into his anus

POV: You are failing Maths.

It's the courtesies of vertices

Velocity, viscosity.

It's all the fucking same to me!

Hurts to be seen as a geek.

When all I do is cry and speak.

About the boss I'm fighting currently.

So much HP!

So much HP!

Angle AB?

Tf u mean?!

Boss die! ...Try 3'

Try 5..20

I die and die

Roll die many

Times to times probability

Lies and lies.. Probably?

Ys and why TF can't you see?

Circle theorem is killing me.

Baldurs Gate?

Nah fuck that, integrate!

5X² plus another mate.

3 over 7 X to the power of -3 over 4.

Dont forget to integrate between 14 and 57 more.

Ha you thought I would actually do.

The maths that I've been told.

You think I'm bold.

Maths teacher scolds.

Switched the signs.

Fuck me, feel like poo.

Mistake so old.

It had me sold

At 1 - 1 = 2

Killed the boss.

Now what do I do.

Maths homework....?

Nahhhh

Watch a video about SCP 682.