.9 repeated is equal to 1.

Ok, since Pro doesn't think that I can bring a provable piece of logic to the discussion, lets get this debate more heated, shall we? _________________________________________________________________________________________________ Your first argument in Round 3: ".3 repeated is a third. Please prove that it is not. Without any logical fallacies. " .3 repeated is a decimal correct? And decimals are a product of long division and/or by typing an equation into a calculator correct? You can "solve" 1/3 by either using a "standard calculator" on your computer, or by doing it by hand (long division), or perhaps using a different calculator as well. (If you use the "standard calculator" on your computer, and by using this definition of "standard calculator", I'm referring to the one that automatically comes pre-built into a computer and you have to do the following directions on a computer to open: Start-All Programs-Accessories-Calculator) The point of me showing the "standard calculator" definition is to show a tool that we can both use and have knowledge of, and therefore can eliminate any confusion with regards to entering the information into the calculator. If you "solve" the equation 1/3 (One divided by three) on your calculator or on the "standard calculator" that I showed above, you should get an answer that will look like this, 0.33333333333333333333333333333333. Or if you prefer to "solve" the equation by using long division, you will get the same answer. You can go on and on trying to solve the problem, but will never reach the answer. Nine subtracted from ten equals one which in turn causes the problem to constantly go on forever. Therefore since the problem can never be solved, it never actually equals 1/3. The calculator just shows the number that is closest to one third (1/3), because that is what it is programmed to do and because what it is limited to, due to the information that I have previously stated. Furthermore, if you get .3 repeated and add it together three times or multiply it three times, it equals .9 repeated. If you decide to add it three times on either a calculator or by writing it down, it will equal .9 repeated. Here I will add .3 repeated by adding it 3 times: 0.33333333333333333333333333333333 + 0.33333333333333333333333333333333 = 0.66666666666666666666666666666666 0.66666666666666666666666666666666 + 0.66666666666666666666666666666666 = 0.99999999999999999999999999999999 Here I will multiply .3 repeated by 3: 0.33333333333333333333333333333333 x 3.00000000000000000000000000000000 = 0.99999999999999999999999999999999 As you can see, you get .9 repeated either way. However, if you choose the multiply it on the "standard calculator", you will notice that you do indeed get 1. This happens because the calculator rounds and the answer in return is 1. _________________________________________________________________________________________________ Your second argument in Round 3: "1-.9 is not .01. That would equal a number larger than 1. " (I would like to make the point that Pro has only shown two pieces of evidence to support his argument that .9 subtracted from 1 doesn't equal .01 and that it would equal a number larger than 1. I would also like to note that last round; I identified 0.01 as 0.01 repeated. I am unsure if Pro noticed this, but I assume he has because in the first round Pro said, ".9 repeated (which for the sake of the argument we will call .9)" so that would mean .1 or .01 repeated for the sake of the argument would mean .1 or .01. ) First off, I would have liked it if Pro would have clarified his argument. Does the word "that" refer to the solution of .9 subtracted from 1, or does "that" refer to the possible different solution that seems it may be numerically different than .01? I would like Pro to clarify the message he was getting across, as one could interpret "that" many ways. Since I have identified two possible clarifications for what "that" could actually mean, (with reference to this sentence) I will respond to both with counter arguments below: First, if you were referring to the solution of 1 minus .9, the solution does indeed equal .1 or .01 repeated. For however many 9's you put after the nine in .9, that many zeros is put in front of the one in .1. For example, if you were to perform the following equation on your calculator; 1 minus .9 or .99 or 0.999999999999999999999999999999, the answer would be .1 or .01 or 0.000000000000000000000000000001. Since you clarified that .9 was to be meant as .9 repeated, that means that .1 or .01 is .1 repeated or .01 repeated. Another reason is that since you chose to label .9 with infinite 9's after .9, I can select to represent the .9 as .99, or .999999 or however many 9's is needed to represent that. Therefore, that means that .9 in your argument could simply have meant .99 which subtracted from 1 equals .01. Second, if you were referring to the possible different solution that seems it may be numerically different than .01; this response pertains to that notion. Now, you seem to say that the solution of 1 minus .9 would equal a number larger than 1 (this point of view is only coming from my second interpretation). I hope we can agree on a couple things here. .9 is a positive number correct? So how can the solution of 1 minus .9 equal a number greater than 1? Even if you follow the assertion that .9 is equal to 1, 1 minus .9 would equal 0. Is 0 equal to 1? No, it definitely isn't. So that being said, how would the solution equal a number larger than 1? _________________________________________________________________________________________________ Your third argument in Round 3: "I don't have to round, this reasoning is only true once you have proved that .9 is not 1. Otherwise, this argument is completely null due to the fact that it is based upon a false piece of logic. " I'm assuming what you mean is that you don't have to round .9 repeated in order for it to be equal to 1. If that is the case then that means they are two entirely different numbers. You then go on to mention that the reasoning is only true once when I have proved that .9 isn't 1, yet that is evidence right there that .9 isn't equal to 1 since it has to be rounded in order to equal it. I believe I have provided sufficient amount of evidence to prove that .9 isn't 1 so by your logic that means I can use rounding as a reason to prove that .9 isn't equal to 1. So that would in effect counter your argument asserting that my argument is null due to a "false piece of logic". _________________________________________________________________________________________________ Your fourth "statement and/or argument" in round 3: "I urge the voters to vote Pro because Con cannot bring a provable piece of logic to the discussion. " I would have to say that this is somewhat bad conduct by Pro. You really shouldn't urge and/or persuade the voters to vote for you until it is the last round. You also make the assertion that I cannot bring any provable piece of logic to the discussion. I would like to point out an interesting point you bring there. Since cannot and can not mean the same thing, you are essentially saying that Con can not have the ability to bring a logical argument to this debate. So if I were to bring a logical argument to the debate than I will have successfully countered your assertion and immediately made it false. Well, I would encourage you to at least be more proffesional about conduct. Thank you.