.999... is equal to 1

My opponent plans on enlightening me, you, and the entire mathematical community. Sweet! I accept the definitions. I will prove that the decimal representation 0.999... is exactly equal to 1 in the real number system. I will do this claim and response style. Claim: 0.999... is equal to 0.999..., not 1 Response: It is equal to both. One of the main misconceptions that leads people to come to my opponent's conclusion is that each number should have its own unique decimal expansion. This is not a property of the real numbers, so we cannot make a conclusion based on it. Claim: In the CON R1 proof, 9x=8.999...1 or 8.99forever with a one at the end. Response: Like my opponent said, it is impossible to have something AFTER an infinite series of numbers. To counter this, he claims that it works backwards, going from the 1 at the end to the 8 at the beginning. Note that besides being completely invalid mathematically, this does not solve any problems, since it means the 8 at the beginning is actually 'after' an infinite series of numbers, an erroneous concept. Also, if I asked what the impact of that 1 is on the value of the number, the only valid answer in the real numbers would be 'zero,' assuming a one could exist 'after' an infinite number of decimal places. Claim: 0.999... is the number closest to one, but not equal to it. Response: One method of establishing equality between X and Y is "X - Y = 0". This can be written as "X = Y - 0", or in our specific case "0.999... = 1 - 0" Now obviously, my opponent will contend that the difference is not zero, and therefore this equation is invalid. However, let us consider something. 0.999... can be written as a convergent infinite sequence as such: (0.9, 0.99, 0.999, 0.9999, …) Now my opponent may not be ready to admit that the sequence is equal to the limit of the sequence [which is 1], but we can see one important thing from this. The difference between this number and zero shrinks every time we go to the next value in the sequence. Since the difference between this number and one is constantly getting smaller in this sequence, it does not have a fixed finite value, since going to the next iteration will always make the difference smaller. Indeed, this difference would not even be able to be written as a finite number. The only conclusion then to make, besides the conclusion that the difference equals zero, is that the difference is some positive, nonzero infinitesimal. However, by way of the Archimedean property of the real numbers, nonzero infinitesimals do not exist. Here is a link to an article explaining the Archimedean property of the real numbers: http://en.wikipedia.org... Thus, the only valid conclusion is that the difference between 0.999... and 1 is zero, making the numbers equal. Claim: This problem arises because of the base 10 number system. Response: Actually, this phenomenon exists within every number system. In base 8, 0.777... equals 1. In base 2, 0.111... equals 1. It is also true that in all of these number systems 1 = 1, but that does not preclude these other decimal representations from also equaling one. Now, I will present another proof for which "there's a one _at the end_ of the _non-terminating_ decimal" will not become an issue. 1] 1/3 = 0.333... This is an exact equivalence. 2] 3 * 1/3 = 3 * 0.333... A basic tenet of algebra is that doing the same thing to both sides of a valid equation leaves us with a valid equation. 3] 1 = 0.999.... Using the multiplication methods of fractions and decimals respectively, we get this equation. This avoids my opponent's (completely invalid) claims about a number "at the end" of a non-terminating decimal. I challenge my opponent to show me the supposed mathematical fallacy used in this proof. Remember, unless there is a fallacy or bad premise used in a proof, its conclusion must be true. I will now give the floor back to my opponent. Thank you.