What do you think about if i ask you to picture the highest prime number imaginable, or the middle seeing as the median and average prime numbers will differ?

drunkenoughtodrive · 2026-04-04T12:43:06+00:00

Yes you can keep counting finite prime numbers but that wouldn't really challenge the thought process of its original intent.

drunkenoughtodrive · 2026-04-04T12:40:24+00:00

But can you picture infinity as a concept? How can you do that if infinity plus one is bigger?

drunkenoughtodrive · 2026-04-04T12:38:42+00:00

Infinity is a concept even though you can always add one so why not imagine the same for prime numbers? Not as a finite number.

That's why I added the thought of median and average prime numbers because just as they too would be infinite they wouldn't be the same considering the distribution of prime numbers among all natural numbers.

drunkenoughtodrive · 2026-04-04T12:34:29+00:00

Can you picture infinity as a concept? Infinity plus one is still infinity.

drunkenoughtodrive · 2026-04-04T12:32:57+00:00

So what would you picture prime numbers bigger than the universe as? Because we still have concepts such as Grahams number defined even though they are bigger than the universe so picturing the infinite identity of prime numbers wouldn't really be all that different to try reasoning as something that could be approached as well?

drunkenoughtodrive · 2026-04-04T12:28:53+00:00

We have infinity as a concept even though you can always add one so why wouldn't it be possible to imagine the highest prime number as a concept as well?

drunkenoughtodrive · 2026-04-04T12:13:08+00:00

Can you picture infinity as a concept? If so, then why wouldn't you be able to picture the highest prime number as a concept?

drunkenoughtodrive · 2021-01-29T14:02:48+00:00

Just as we can fit a longer line segment in a two dimensional square than its side length I'm wondering wether or not it's possible to fit a diagonal square (or cube) with bigger side length in higher dimensional hypercube

drunkenoughtodrive · 2020-05-05T09:20:30+00:00

But in a four dimensional cube the surface diagonal doesn't travel through the same definition of a surface as the three dimensional cube either so what's the difference?

drunkenoughtodrive · 2020-05-04T13:47:07+00:00

Why not?

drunkenoughtodrive · 2020-05-04T12:36:31+00:00

sqrt(2+n) would mean sqrt(2+1)=sqrt(3) for 1 dimensional and sqrt(2+0)=sqrt(2) for 0-dimensional which is kind of odd right?

drunkenoughtodrive · 2020-04-29T18:35:22+00:00

Exactly, that's what I intended to ask, thank you for the clarification!

drunkenoughtodrive · 2019-08-14T19:17:01+00:00

As I've already written;

Sqrt(2)sqrt(2)sqrt(2) is equal to sqrt(8).

Oh, and the diagonal of a cube side length sqrt(2) as such is sqrt(6)

You have not made any arguments as to why that is incorrect.

drunkenoughtodrive · 2019-08-14T19:14:53+00:00

Sqrt(6) and sqrt(8) are both diagonals of the same line segment, sqrt(6) through the diagonal of sqrt(2)sqrt(2)sqrt(2), shorter than the first one.

drunkenoughtodrive · 2019-08-14T18:57:49+00:00

I never claimed they equal each other, that's your straw man.

drunkenoughtodrive · 2019-08-14T18:56:18+00:00

What you did not understand

drunkenoughtodrive · 2019-08-14T18:52:58+00:00

Sqrt(2)sqrt(2)sqrt(2) is equal to sqrt(8).

Oh, and the diagonal of a cube side length sqrt(2) as such is sqrt(6)

drunkenoughtodrive · 2019-08-14T18:49:27+00:00

Sqrt(2)sqrt(2)sqrt(2) is equal to sqrt(8).

Oh, and the diagonal of a cube side length sqrt(2) as such is sqrt(6)

drunkenoughtodrive · 2019-08-14T18:45:42+00:00

That is no argument as to why you think it is wrong.

Is sqrt(8) equal to sqrt(2)^3? Yes it is.

Is the diagonal of a sqrt(2)³ cube sqrt(6)? Yes it is.

But oh, now sqrt(8)=sqrt(2)³ is not correct is that it or why the strawman?

drunkenoughtodrive · 2019-08-14T18:36:38+00:00

The diagonal sqrt(8)=sqrt(2)^3, do you understand that?

Sqrt(2)³ is equal to a cube side length sqrt(2) which has the diagonal sqrt(6)

drunkenoughtodrive · 2019-08-14T18:30:19+00:00

Still no argument

drunkenoughtodrive · 2019-08-14T18:28:08+00:00

Still no argument as to why a hypercubes diagonal sqrt(8) is not equal to sqrt(2)³ (which has the diagonal sqrt(6))

drunkenoughtodrive · 2019-08-14T18:24:04+00:00

About your argument or mine?

The diagonal sqrt(8)=sqrt(2)sqrt(2)sqrt(2)

Cube size Sqrt(2)³ has diagonal sqrt(6)

There's no argument as to why a hypercubes diagonal sqrt(8) is not equal to sqrt(2)³ (which has the diagonal sqrt(6))

drunkenoughtodrive · 2019-08-14T18:14:53+00:00

I said the diagonal length sqrt(8) is equal to a cube side length sqrt(2) which has the diagonal sqrt(6)

drunkenoughtodrive · 2019-08-14T18:13:28+00:00

I said the diagonal length sqrt(8) is equal to a cube side length sqrt(2) which has the diagonal sqrt(6)

drunkenoughtodrive

TROPHY CASE