Mathematical ambiguity when accepting 0.99...=1

I thank you for accepting my challenge! First, we need to clarify what I mean by digital root. My counter-argument: You are right in a way, that is what the well known definition of dsum does, but not mine. I also looked up the digital root proof a few weeks back and to be honest I was not quite satisfied of its limitation, supposedly this limitation we are talking about right now, which is why I extended it a bit. In other words, the dsum(x) I used is a bit dfferent. Mine is an explicit input output function. Thus I shall post my algebraic definition of dsum(x), without going into lengthy derivation. First off we need to be clear on how we treat infinity here. You should agree that n, representing a countably infinite whole number of digits, is composed of whole numbers added together. Which means that, apart from the fact that dsum(infinity) is not dsum(9n), exactly: 1."Infinity" is not a real number and cannot be treated as such. 2. You were partially correct when saying that dsum(1/3) and dsum(1/6) and dsum(1/9)are all quite undefined but not the case when it is 0.99..., as for the digital root function, even if the number of 9s are -countably infinite-, as they already are, i.e. you can count them and tend to infinite digits in only that way, dsum(9n) is still 9, no matter how large n is. Infact if you insist anyway that 9n should be treated as simply "infinity", which is not so, my digital root function -also- returns 9 if one were to brazenly consider it a real number and use it algebraically, oddly enough. This is my definition of digital root of x, where x is real and digit-finite, or x is real and has a homogenous digital root no matter how many digits you consider from x, and from where: drF(x) represents the final digital root of x, x being any real number which is digit finite or homogenous in its digital root, x=a/(3b-1) dr(a) represents the digital root of a where a is a real whole number dr(3b-1) represents the digital root of 3b-1 where b is a real whole number drF(x) = dr[dr(dr(a)/dr(3b-1))] {Shorthand definition} Formal algebraic definition of dr(a) [which you can use to then formally define drF(x), as simply writing drF(x) as such is simply way too long and I will likely copy it wrong somewhere] dr(a) = 5.4+(9/pi)arctan(tan(pi({abs(a)-0.5-arctan(tan((pi*0.5)(2abs(a)-1))/pi-5.4}/9) , where abs(a) can be considered as sqrt(a^2). If you were to brazenly suggest to replace x by infinity and see what happens, you can do so by saying that x=infinity/(3(1)-1)=infinity, I will not oblige. Taking (infinity) as tan(pi/2), tan(infinity)=tan tan(pi/2), and as tan(pi/2) is undefined it is suitable to take tan(infinity)=tan(pi/2) That would actually result in 9 as an output, and I fully urge you to take my word with skepticism, write down drF(x) and do as I did. In other words, no matter how you view 9n to be, dr(9n) is still 9. You might ask, what makes tan(pi/2) special when there are other inputs of tan(c) which tend to infinity? Try them as well, they all give back drF(infinity)=9 Which I admit is mysterious, but also gives me freedom in saying, you were also wrong in another part of your argument: You asked me to give a reason why using normal algebra does not apply when "proving" that 0.99...=1. ...Well, this IS normal algebra, all of it. And by using normal algebra I have also shown how 0.99...=/=1. But, I will go along with what you have shown: x=0.99... 10x=0.99...*10=9.99... 10x-x=0.99...*9 x=0.99... Frankly what I am saying is that the algebraic proof is flawed in the sense that it shows not proves. Here is what I mean, and after you read it, you will appreciate the difference between showing and proving: In mathematics there are, literally countless examples of when some variable has only one value or two that make sense from all the possible outcomes. The others are contradictory or non sensical. Such an example, which is very commonly found, is one that claims to show that all numbers are equal, with actually no algebraic fallacy. Let me demonstrate. [1] Proof That All Numbers Are Equal? Date: 12/06/2005 at 17:00:34 From: RV Subject: All numbers are equal Theorem: All numbers are equal. Proof: Choose arbitrary a and b, and let t = a + b. Then a + b = t (a + b)(a - b) = t(a - b) a^2 - b^2 = ta - tb a^2 - ta = b^2 - tb a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4 (a - t/2)^2 = (b - t/2)^2 a - t/2 = b - t/2 a = b So all numbers are the same, and math is pointless. What is the error in this proof? I think that if a = b then a - b = 0 and we cannot multiply both sides of the equation by (a - b) because when a number is multiplied by zero the answer is always zero. [/1] In this demonstration, there is no logical fallacy unless a=b, but a=b is just one case. The rest are still valid. So I ask you, what makes the "proof" I have linked forward more or less valid than the "proof" you proposed that 1=0.99... If to safeguard all mathematical reason, the above "proof" that all numbers are equal needs to have one of the possible solutions disregarded, that a=b, then why does this not apply to your argument? (N.B. There is a response to the question put forward in the link I provided) [1] http://mathforum.org... [/1]