u/Batman_AoD

New SPP terms, "limbostic" and "limbosic".

SPP, it's true that I disagree with nearly every comment you write in this sub, and am vanishingly unlikely to ever find agreement. But I want you to know that I greatly enjoy and appreciate when you make up new words, even though I generally consider them nonsense. I think others who disagree with you enjoy it as well, and that's probably why many of them participate in the sub.

This post isn't intended to prove that 0.999... = 1, but it is intended to show why one of SPP's favored arguments, the "lack of a limbo kicker", isn't actually convincing to anyone who disagrees with him.

In the quoted post, SPP phrases the argument as "you need to add just the right touch of a limbo 1 to generate a 1 from the summation 0.999...9 + 0.000...1."

This is obviously true if the "..." in 0.999...9 indicates a large-but-finite number of 9s. That is, 1 - 0.999...9 = 0.000...1, where the number of 0's between "." and "1" is one less than the number of 9's.

But, [Morpheus voice] what if I told you that the "right touch of a limbo 1" is this number: 0.000...0999...?

There's obviously an ambiguity of notation here, because the first "..." is finite but arbitrarily large, whereas the second is actually infinite. So let's adopt the notation proposed by u/Inevitable_Garage706 in this post: (0_n) means n 0's for some large n, and (9_∞) means infinite 9's.

I am not going to prove here that 0.(0_n)(9_∞) is equal to 0.(0_n-1)1. But I will point out that they are equal if and only if 0.999... = 1, and for the same reasons. So, for anyone who accepts that standard equality, then there is indeed a unique "limbo kicker" for every 0.(9_n) that is already part of 0.(9_∞). Because for any n, 1 = 0.(9_n) + 0.(0_n-1)1 (this is SPP's "limbo kicker"); and it is also the case that 0.(9_∞) = 0.(9_n) + 0.(0_n)(9_∞). (Note that here SPP may object that it's actually 0.(0_n)(9_∞-n), but those are only different if ∞ is some kind of finite-but-increasing value, or a transfinite ordinal.) So, for someone who believes (i.e. sees why) 0.(9_∞)=1, it's easy to see that:

0.(9_∞) = 0.(9_n) + 0.(0_n)(9_∞) = 0.(9_n) + 0.(0_n-1)1 = 1

...because the limbo kicker was always inherent in 0.(9_∞) itself.

u/SouthPark_Piano, I agree with you that the "time element" is not part of how infinity "really" works.

But the problem is that you keep using time words, and your statements don't make sense without them.

You wrote:

> You will always be able to find a zero to the right hand side of the wavefront 9. And if you reckon that is where it ends, then think again brud, because it is certainly no longer a zero already, even though it was a zero a moment ago.

The only possible meaning for this is that the value is changing over time. Without a "time element", you can't have a "wavefront" that changes 0's into 9's.

• What are you quoting from? I can't find it with google: https://www.google.com/search?q=%22meaning+the+number+continues+to+grow+in+length+limitlessly%22&udm=14
• Why do you say "google is your friend" when it has so many results saying 0.999...=1?
• Why did you lock your comment immediately?

>You cannot unbelieve your own eyes brud.

Maybe that's why I'll never believe SPP, because I've already seen proofs in more consistent systems "with my own eyes."

>As long as you sign the contract, then you can generate a 0.999... of your own too.

...and is that supposed to sound like something I should want to do? I'm no fan of the real numbers, but they're way more interesting than SPP's system where the only important attribute is "a leading 0. guarantees magnitude less than 1".

Terence Tao has a lot more reputation than most people do riding on "standard math" being correct. So, surely, saying he'll eventually become a convert is a bit presumptuous, u/SouthPark_Piano.