Mathematical ambiguity when accepting 0.99...=1

First of all, I was saddened, seriously, with the fact that I failed to close parentheses, making my equation illegible. Very well, secondly I will address your response to the proof that all numbers are equal. You specifically said that "one gives you the original equation, and the other gives you something impossible". I do realise the "flaw" (there is no flaw, the asker simply ommitted the fact that there are and - values for square roots, but the conclusion that a=b is still a possible solution) is discussed, so I ask you: How do you know, that one answer is impossible? What I am trying to show here is simply that as there are ways to show that 1=0.99..., there are was to show that they are not the same as well. Homogenous digital roots : Homogenous in a physical sense. Where did the definiton of dr(a) come from? I made my own definition of dr(a) , I kind of said that. No I have no link of the derivation or the equation itself found on the web, but I will link it, because nobody uses that form, most probably, unless someone also happened to have derived dr(a) as I did, which is highly unlikely but possible. I derived it using Graham et al's definition of fract(x). This time I will link the equation directly [1], and may I say it works for real whole numbers if you use dr(a) and for real numbers if you use drF(x) where x=a/(3b-1), a and b being what I mentioned before. However for clarity's sake I will link WolframAlpha's neat presentation below with the correct equation. (Note that WolframAlpha does not even plot drF(x), at least on my smartphone, and sometimes it does not even give the right output, ...at least on my smartphone, so I highly suggest using a calculator, my computation times out :/) Now, when you ask why I plugged in tan(pi/2) as countable infinity, I certainly did not mean it to stand for countable infinity. From what I understood from what you said, "Since infinity is infinitely long, no dividing or separating into separate chunks to add together later will lead to anything less than infinity, so dsum(infinity) is undefined.", you were talking about "normal" infinity. Since drF(x) is only intended to work where x has a real domain, tan(pi/2) can be said to tend to "normal" infinity."Complex" infinity, per se, is only applicable if x is a complex variable. I simply did that to go along with what you were saying. And the reason I did that was this : I wanted to show you that even if you let 9n tend to normal infinity, the digital root would still be 9. I did not have to do this at all. You asked in the comments for me to elaborate what I specifically did when I let x tend to infinity, so I will explain it all here with the rest of what I was saying. Please refer to the equation I linked for this. I will also use limits this time. As previously stated: drF(x)=dr(dr(dr(a)/dr(3b-1)))) Now you can agree, instead of writing all of it and risk making a syntax error somewhere, we can simply work it out bit by bit. If one lets b=1 and a tend to infinity; drF("infinity")=dr(dr(dr("infinity)/dr1)=dr(dr(dr("infinity")/dr(2))) As a approaches infinity, we can see that dr(a) will tend to nothing but 5.4. [2] (Link below. Of course we cannot plug in a=infinity, but we can check where it is going as a gets closer, as we do in limits) You were actually very right when you pointed out my mistake with the tangent function, hopefully it is cleared now and for that I apologize. Below you will find a link of drF(x) in its full form, (which again gives me a computational time out) and a link to an online limit calculator with the equation dr(a) already plugged in: (The limit was taken for only dr(a) as only it has a in it, by using the formal method of getting closer to a target, maybe you will have more luck with setting the limit directly to infinity, but when I tried it the computation timed out) [1] http://m.wolframalpha.com...(9%2Fpi)(arctan(tan(pi(((5.4%2B(9%2Fpi)arctan(tan(pi((%5B5.4%2B(9%2Fpi)arctan(tan(pi((x-0.9)%2F9-0.5)))%5D%2F%5B5.4%2B(9%2Fpi)arctan(tan(pi((3b-1-0.9)%2F9-0.5)))%5D-0.9)%2F9-0.5))))-0.9)%2F9-0.5))))&incCompTime=true[/1] [2] http://m.wolframalpha.com... [/2] P.S. As I have a feeling the first link might not work: drF(x) =5.4+(9/pi)(arctan(tan(pi(((5.4+(9/pi)arctan(tan(pi(([5.4+(9/pi)arctan(tan(pi((x-0.9)/9-0.5)))]/[5.4+(9/pi)arctan(tan(pi((3b-1-0.9)/9-0.5)))]-0.9)/9-0.5))))-0.9)/9-0.5))))