0.999... = 1 | Prove Me Wrong

Each of these 4 arguments masquerading as proofs reflect a fundamental misunderstanding of numbers, algebra, and mathematics in general. I will speak to them one at a time:1.a) x = 0.999...b) 10x = 9.999...c) 10x - x = 9.999... - 0.999...d) 9x = 9e) x = 1This proof is a knock-off of that given by William Byers (also cited on Wikipedia) in 'How Mathematicians Think: Using Ambiguity, Contradiction, and Paradox to Create Mathematics (2007)'. What Pro perhaps fails to realize is that Byers did not intend this as a bona fide proof. Rather, he intended it as a trick to illustrate ambiguity and contradiction. "Students who did not accept the first argument sometimes accept the second argument, but, in Byers' opinion, still have not resolved the ambiguity, and therefore do not understand the representation for infinite decimals." It is in this category that I place Pro. There is a gross fallacy being committed here in the movement from step c to step d. One cannot simply subtract an irrational number from another and expect to get a rational number. In fact, mathematical laws tell us that given an irrational number x and another irrational number y, if x + y = a and x - y = b, it is not possible for both a and b to be rational. You can find the proof here, if you'd like: https://goo.gl...Suppose for a second that steps a through d are all correct and that 10x - x = 9x = 9. Here, to use the previous illustration, we will say that x = 10x and y = x. As has been proved by mathematicians far smarter than I, it is not possible for both x + y (10x + x) and x - y (10x - x) to be rational. Simplified, that means it is not possible for both 11x and 9x to be rational. Since Pro claims that 9x = 9 and thus x = 1, it must be 9x that is rational. Therefore, the only choice, based on a combination of mathematical proof and Pro's own allegations, is for 11x to be irrational. However, Pro used this "proof" to show that x = 1. If x = 1, then 11x = 11, which is rational. Since this is simply not mathematically possible, we have a proof by contradiction that x does not equal 1 (at least by this logic).2. 1/3 = 0.333...2/3 = 0.666...3/3 = 0.999...This is, again, faulty logic. One cannot say that "1/3 = 0.333..." and then multiply each side by 2 to get "2/3 = 0.666". This is totally disregarding the nature of irrational, repeating decimals. In order to use this actual logic, most likely by induction, Pro would have to lay out steps like the following, placing ends to the unending decimals and showing that the same result follows.2a.1/3 = 0.32/3 = 0.73/3 = 1.0Since 0.3 * 2 != 0.7, and 0.3 * 3 != 1.0, this does not work.2b.1/3 = 0.332/3 = 0.673/3 = 1.00Since 0.33 * 2 != 0.67 and 0.33 * 3 != 1.00, this does not work.2c.1/3 = 0.3332/3 = 0.6673/3 = 1.000Since 0.333 * 2 != 0.667 and 0.333 * 3 != 1.000, this does not work.I think you get the point...3.What fraction/decimal number comes in between 0.999... and 1. Any two numbers that are different must have a difference of something. That something will be somewhere in between the two numbersEx. 1.5 is in between 1 and 2With all due respect, this is just silly. As an example -After 0.999999 there could be a 8. Then, clearly, there would be a number between it and 1 (0.9999998 < 0.9999999 < 1). To this, you, as an astute mathematician may say, "wait a second, you can't just remove the '...' at some point and replace it with a value!" To that, I would say that this point only holds if you can show that there is a fraction that leads to the number 0.999....Surely, 1/3 = 0.333... but I would ask, what fraction equals 0.999... Of course, you cannot say that 1/1 = 0.999... because that would be using the conclusion to prove the conclusion, otherwise known as the logical fallacy, "begging the question."4.Put in 0. with 20 (it must be 20 or it may not work, 9's after it in a calculator and press =. It will say it equals 1.The response to this should be evident to anyone who has ever taken basic math and used a calculator. This is simply a result of the limitations of whatever computation device/calculator you're referencing. Apparently, with this calculator, 20 digits beyond the decimal place is the point at which it chooses to round. On an old calculator, that might have been at the 3rd decimal place, or the 5th, or the 19th. Using a sufficiently old calculator, typing in 0.6 = would result in a 1. I don't think Pro wishes to argue that 0.6 = 1.I believe I have sufficiently taken each of Pro's 4 arguments/"proofs" and refuted each one using sound, established, proven mathematical and computational theory. I welcome any rebuttal or comments from the voting public.Cheers!