.99999...(repeating) is equal to 1

A very well written response by my opponent. For this final speech, I want to take a look at all of my opponents proofs, and explain why they aren't valid, explaining why 1=/=.9(r) after each. Then I'll look to my opponents 3rd round and comment as necessary. [Division by 3] 1 = .9(r) --- --- = .3(r)=.3(r) 3 3 This proof says that the two must be equal because you can divide by 3 and get the same answer. This entire conflict is caused by the precision infinity allows. With each new '9' added to .9(r), it gets closer to 1. Just look at my proof by subtraction above. Whether you believe in the end result is irrelevant for now, just note that with each '9 after the decimal point, the number gets closer and closer to one. Now, what happens when the difference becomes so small that it's impossible to tell a difference between them? A great example of this is dividing by 0. It can't be done, but you can get really close. With 1/x, as x approaches 0, 1/x gets progressively bigger. Can x ever be 0? No, it'd give us an invalid answer. Same concept here. Eventually, the numbers become so close together that a difference is neigh impossible to tell, but exists nonetheless. [The Mysterious Ninths] The same fallacy is seen in the 1/9's proof. 9/9, we're taught, equals one. but 1/9 * 9 = .9(r). Surely they're the same thing? No. The numbers are just so close together that for all intensive purposes, they act the same. But a difference is there. I'll get to how we know this in a little (actually, I've already covered it, but I'll reaffirm it later) [Solving for 'x'] And to cover the final proof, found at the end of my opponents final speech. Infinity makes this proof an extremely hard one to grasp. With .9(r), when you multiply by 10, it becomes 9.9(r). But according to common math ideas, that would shift the decimal point. It'd be .9(r) with one less nine. But it's infinity! There is no end; there is no "one less 9. " Nevertheless, the two numbers are once again so close together that humans cannot tell a difference, but one exists. BUT HOW DO WE KNOW? Simple, really. Look to my Proof by Decimal from Rd.2. Both PRO and CON show examples of how .9(r)=1 and how .9(r)=/=1. We can't both be right! And we aren't. Prefer CON's argument because I've shown how it's easily mistaken that .9(r)=1 (Incredibly unnoticeable gap) yet proven that a gap exists. Just look at the ones digit of both numbers. 1.0 0.9(r) 1>0 thus 1>0.9(r) Don't discard all evidence used in this debate, instead adapt to fit the truth I've shown and the easily made misconception due to the accuracy of infinity. Proof By Subtraction ----------------------- We're dealing with infinity. .0(r)1 is legal, but hard to understand. In the real world, no it doesn't exist. It's simply a concept used to experiment with infinity, or representing an infinitely small number. .0(r)1 is as close to 0 as .9(r) is close to 1. Proof By Addition -------------------- You raise a good point on which is greater, .9(r) or .9(r)8. Infinity is certainly a tricky concept and I honestly don't know the answer to it. Even with this concession, however, the first argument in this speech still stands. It looks like this: Units digit is all you need to look at here. 0.9(r) 1.0 Because 1>0, 1.0>0.9(r). The misconception [that 1=.9(r)] occurs because infinity is so precise, so accurate, that the difference is so small, so insignificant, that the two numbers ACT as one, yet are different. You might say that .9(r) is 1's stunt double. (Hollywood would have a fit with this one, no? ) A miniscule difference exists, one that allows the two DIFFERENT numbers to ACT as the SAME.