Funny

9/1 ≈ 8

16/2 is an almost exact replacement for 8. OP's name includes Fermat, and so he's probably smarter than me though.

9/1 is approximately 8, for extremely large values of 8.

Nah, the engineer probably designed it with a safety factor. You could probably even go 9/0 and be perfectly safe ;)

Have you considered running for Indiana governor? You have the right mindset.

https://en.wikipedia.org/wiki/Indiana_pi_bill

For that last one, how bad are we talking? I need to know soon, I have some important banking software I need to develop.

By the way guys, a very similar approximation for 8, which also starts at 9.000 for n=1, but quickly gets much closer to 8 for increasing n, is:

exp{-2(n-1)} + 8

It approaches 8 about as fast as the above method but this one has a simple formula that is usable in python etc.

gr8 m8, I r8 8/8

WOOOAH🤯

Witch! Begone foul demon, and take your dark sorcery with you!

I myself once learned 380 digits of π, when I was a crazy high-school kid. My never-attained ambition was to reach the spot, 762 digits out in the decimal expansion, where it goes "999999", so that I could recite it out loud, come to those six 9s, and then impishly say, "and so on!"

—Douglas Hofstadter

So, years ago in college in Linear Algebra our professor said to us to study about idempotent matrices. So I checked out that wiki page and saw the example for 2x2 matrix, that are composed by the numbers 3, -6, 1 and -2. And I was like wait a second, 3×-2=-6 there's no way they are not relationship there, so I started trying other numbers, and found and proved (using induction) that any n, -n(n-1), 1, -(n-1) is an idempotent matrix. At the test there were no questions about that, and I was short of 0.5 poits to pass the class without having to present a final exam and I told my professor that I spent a lot of time learning that and that even discovered something and proved he pass me the chart and asked me to proved it, after that he gave the missing points. Was really good.

Then you try to figure out why they do be like that

Base 3:

21 / 12 = 1.1012101210121012

gonna need this in every base

...all of them?

Well, simple. Jest substitute that 8 with the above approximation.

It contains the numbers 8x10^7 and 8x10^1, but not 8x10^0

This sequence approximates an integer to arbitrary precision, not 8 specifically though, and never perfectly.

I tried it out using other bases, and the rule seems to be that doing this in base n results in n-2 with remainder n-1. So it doesn't ever actually converge, but the remainder becomes small very fast.

(n * 8 + 1) / n