.999... is equal to 1

My opponent's first argument, one that concedes the entire point, is that 0.999... only equals 1 in the base 10 number system, and therefore you should vote CON because base 10 is not specified in the resolution. However, note that the base 10 number system is the default number system for all of humanity. It does not need to be specified. Under my opponent's flimsy reasoning, every mathematical formula must have an explicit specification of number system. This is ludicrous, considering that if no other number system is specified, base 10 is always assumed. My opponent also misrepresented my argument, claiming that I said that 0.888... equals 1 in base 8. This is not true, and not what I said. I claimed that 0.777... equals 1 in base 8. Thus, now that I have covered that the base 10 system is the default one, that first argument against me is invalid. Next, my opponent attempts to tell us that decimal representations are not always perfect. He attempts to do this by showing that 1/3 has a different decimal representation in base 9 than in base 10. In base 9, the 'decimal' is 0.3, whereas it is 0.333... in base 10. He then says that since 0.3 is not equal to 0.333..., decimals must be flawed. False. 0.3 in base 9 = 0.333... in base 10 That is true. Numbers have different representations in various number systems. For instance, 255 in hexidecimal [base 16] is FF, whereas it is 255 in decimal [base 10]. Indeed, the very word 'decimal' implies that we are talking about base 10, since base 10 is CALLED decimal. My opponent then goes to show us some more 'invalidness' in using decimal representations [by that I mean representations that use a decimal point]. On one side, he has the function 3*1/3 proof I used earlier. Using valid math, this side comes to the conclusion that 0.999... = 1. On the other side, he has the same function in base 9. 1/3 = 0.3 base 9 - this is true 3*1/3 =3*0.3 base 9 - this is also true 1 = 0.9 base 9 - this is half true You see, for that third step, 0.9 is not a valid number in base 9. The 9 numbers in base 9 are 0, 1, 2, 3, 4, 5, 6, 7, and 8. When one digit gets to the theoretical '9' it becomes a 1 in the digit to the left. 0.9 in base 9 does not exist, it is instead written as '1'. We can verify this when we use 'A', the single-digit representation of 10, in base 10. 1/2 = 0.5 2*1/2 = 2*0.5 1 = 0.A That is invalid, since A does not exist in base 10. It is instead written as a 1 in the column to the left of it, or 1.0 So as you see, math with decimals is not broken. My opponent has just made a mistake in his execution of them, pretending that the digit 9 exists in base 9. All fractions have exact decimal representations, whether they have recurring decimal places or not. For example, 1/7 = .142857... , where the portion that repeats is the entire 142857. It would have a different decimal representation in any other number system, but you would have to specify the number system if you wanted to set them equal to each other. 0.3 base 9 = 0.333... That is a true statement, as is any statement you will come to doing valid math on decimals, fractions, or whole numbers. Next, my opponent points out something that's truly laughable. I'll simply copy and paste what he said. "3. My opponent attempts to disprove my claims with Archimedean property, however there is such a thing as non-Archimedian property such as p-adic numbers which support my rational of ending or begining at decimal point." First off, the Archimedean property applies to the real numbers, which is what we are discussing. This is because, like base 10, real numbers rather than p-adic numbers are default. The 'non-Archimedean property' applies to the p-adic numbers [I was unable to find an actual property called called non-Archimedean, I assume it simply means that a system does not have the Archimedean property]. Now I probably know about as much as my opponent does about the p-adic system, which is net to nothing. However, let me show you something that really highlights how different the p-adic system is. "Whereas two decimal expansions are close to one another if they differ by a large negative power of 10, two 10-adic expansions are close if they differ by a large positive power of 10. Thus 3333 and 4333 are close in the 10-adic metric, and 33333333 and 43333333 are even closer." That's so far from traditional mathematics that it's almost ridiculous how obvious it is we aren't talking about p-adic numbers. We're talking about real numbers, since they are the standard in mathematics. And for real numbers, the Archimedean property applies, which proves that the difference between 0.999... and 1 is 0. A difference of 0 one of the basic ways of establishing equality. 1 - 0 = 0.999... 1 = 0.999... Finally, to my opponent's last objection: "4.Am I wrong or is it true that no calculator with base 10 will express a .999... . This sort of disproves the final step of the 1/3 proof via calculator, because when you times .333... by 3 it simply equals 1 not .999..., on a calc. ? Why won't the calculator express this step?" This is for the same reason that a calculator will give you 0.66666666667 for 2/3. Calculators cannot display an infinite number of decimals. Therefore, it will round at the last digit it can display. Rounding 0.99 to the nearest 0.1 is 1.0 Rounding 0.999 to the nearest 0.01 is 1.00 Rounding 0.9999 to the nearest 0.001 is 1.000 Rounding 0.99999 to the nearest 0.0001 is 1.0000 Get the point? It's simply a rounding process, since it can only display it to a certain finite number of decimals. And with that, it is now time for my opponent to make his closing arguments. Thank you.