Quadhelix 0 Posted December 12, 2009 Up to this point, you are absolutely correct. Just as 100 divided by 3 equals 33 and one third. 10 divided by 3 equals 3 and one third. 1 divided by 3 equals .3 and one third. In place of “and one third” you can say, “remainder one”. When you have an infinite string such as .333… , you technically drop the “and one third” part of the problem. If you try to multiply your infinite string of .333… by 3, (if it were possible to multiply an infinite string), you would get an infinite string of nines after the decimal. Which is the entire point of this argument: an infinite string of nines after the decimal, written as 0.999..., is another way of denoting 1. If you disagree, then please tell me what I would subtract from 1 in order to get 0.999... Nmenth suggested 0.000...R1. However, 1-0.000...R1=1.000...R-1. That is, the remainder is negative one. Thus, you are saying that 0.999...=1.000...R-1. What does a negative remainder mean, anyway. Then you contradict yourself. In order to equal 1.0, you have to start your problem by acknowledging the “and one third”. Therefore, .333…and one third, times 3, equals .999…and three thirds. Tah dah, you now have something that does indeed equal 1.0 just as Nmenth pointed out in his proofs way back on page one of this topic, and just as you yourself admitted to in the first part of your tirade. .999… does not equal 1.0, but .999…and three thirds does equal 1.0 plain and simple. Take a=0.3+0.03+0.003+...: you say that 0.333... is "infinity" and thus cannot be multiplied. However, you will agree that I can multiply, for example, 0.03 by any number that I wish - while the sum itself is is infinite, the individual elements are not. Let us then take 3*a: it is obvious that 3*a=0.9+0.09+0.009+... Additionally, 30*a=10*(3*a)=10*(0.9+0.09+0.009+...)=9+0.9+0.09+0.009+... Then, of course, you can take 30*a-3*a=(9+0.9+0.09+0.009+...)-(0.9+0.09+0.009+...)=9+(0.9+0.09+0.009+...)-(0.9+0.09+0.009+...)=9+(3*a-3*a)=9+3*(a-a); a-a=0, therefore 30*a - 3*a=9. However, 30*a - 3*a is, of course, the same as 27*a. By transitive property (if A=B and B=C, then A=C), this means that 27*a=9, which means that 3*a=1. Therefore, a=1/3 However, we started out by saying that a=0.3+0.03+0.003+...; this sum is apparently equal to 1/3. In other words, you are asserting that 0.3+0.03+0.003+...=/=0.333... Wtf? .999... does not equal 1, but .999... + 3/3 (or 1) equals 1? You sir are bat**** insane. I'm sorry, do you actually have an argument, or do you just expect me to cave to your royal fiat? Share this post Link to post
Remomeister 0 Posted December 13, 2009 (edited) Which is the entire point of this argument: an infinite string of nines after the decimal, written as 0.999..., is another way of denoting 1. Take a=0.3+0.03+0.003+...: you say that 0.333... is "infinity" and thus cannot be multiplied. However, you will agree that I can multiply, for example, 0.03 by any number that I wish - while the sum itself is is infinite, the individual elements are not. Don’t misquote us. We say 0.333… is infinite, not infinity. Now, if you think you can multiply an infinite number by any number, just because you can do so with finite numbers, I’ll ask you what I asked Dr. D. What is 0.333… times 4? It can’t be multiplied because in our system, multiplication begins with the furthest digit to the right. When there is no furthest digit to the right, multiplication is not possible. You have a syntax error, an illegal operation. Let us then take 3*a: it is obvious that 3*a=0.9+0.09+0.009+... Additionally, 30*a=10*(3*a)=10*(0.9+0.09+0.009+...)=9+0.9+0.09+0.009+... Then, of course, you can take 30*a-3*a=(9+0.9+0.09+0.009+...)-(0.9+0.09+0.009+...)=9+(0.9+0.09+0.009+...)-(0.9+0.09+0.009+...)=9+(3*a-3*a)=9+3*(a-a); a-a=0, therefore 30*a - 3*a=9. However, 30*a - 3*a is, of course, the same as 27*a. By transitive property (if A=B and B=C, then A=C), this means that 27*a=9, which means that 3*a=1. Therefore, a=1/3 Not quite. However, we started out by saying that a=0.3+0.03+0.003+...; this sum is apparently equal to 1/3. We said right from the start, that 1/3 does Not equal 0.333… and some of you agreed at one point that you must always “carry the one”. One divided by three then equals 1/3. (That little slash between the “1” and the “3” means “divided by”). Therefore, One divided by three equals 0.333...AND ONE THIRD (also known as “remainder one”). You people keep dropping off an essential mathematical element to get your phony 0.999… argument which drops the “remainder one”, three times. In other words, you are asserting that 0.3+0.03+0.003+...=/=0.333… That assertion is absolutely incorrect! 0.3+0.03+0.003+...does=0.333… But does NOT equal 1/3. Don’t misquote us! Edited December 13, 2009 by Remomeister Share this post Link to post
Quadhelix 0 Posted December 13, 2009 (edited) Don’t misquote us. We say 0.333… is infinite, not infinity. Now, if you think you can multiply an infinite number by any number, just because you can do so with finite numbers, I’ll ask you what I asked Dr. D. What is 0.333… times 4? It can’t be multiplied because in our system, multiplication begins with the furthest digit to the right. When there is no furthest digit to the right, multiplication is not possible. You have a syntax error, an illegal operation. Irrelevant: you might not think that I can multiply 0.333..., but you can multiply 0.3 and 0.03 and 0.003 and... So, you asked what 4*0.333... would be. That is easily solved. 4*0.333...=4*(0.3+0.03+0.003+...) 4*(0.3+0.03+0.003+...)=4*0.3+4*0.03+4*0.003+... 4*0.3+4*0.03+4*0.003+...=1.2+0.12+0.012+... I'll leave you to tell me what 1.2+0.12+0.012+... is (hint: it's 1.333...). Not quite. Refute my argument then. You have asserted that 0.3+0.03+0.003+... is not equal to 1/3, but you have failed to find any flaw in my argument. We said right from the start, that 1/3 does Not equal 0.333…Except that I just proved that 0.3+0.03+0.003+...=1/3, therefore, from the start, you have been wrong. Therefore, One divided by three equals 0.333...AND ONE THIRD (also known as “remainder one”). You people keep dropping off an essential mathematical element to get your phony 0.999… argument which drops the “remainder one”, three times. 0.333... is infinitely long. Where does the remainder go? {Edit: See Below.} That assertion is absolutely incorrect! 0.3+0.03+0.003+...does=0.333… But does NOT equal 1/3. Don’t misquote us! I'm not misquoting you: you claim that 0.333... =/= 1/3. However, I proved that 0.3+0.003+0.003+...=1/3; a proof that you failed to refute. Therefore, if 0.333... is not 1/3, then 0.333... cannot be 0.3+0.03+0.003+... either. Edit: I just realized something today: when dividing 1 by 3, the remainder of 1 doesn't disappear, but the remainder's effect on the answer does disappear. To clarify, let us begin our long division. We divide 1 by three and get 0.3 with a remainder of 1 (which works, because we haven't finished dividing). We know that 0.3 is not 1/3, but we can ask how close we are. Thus, we take (1/3)-0.3; 1/3 can be rewritten as 10/30, and 0.3 can be rewritten as 3/10 or 9/30, meaning that (1/3)-0.3=(10/30)-(9/30)=(1/30). We can rearrange this to find the equation 1/3=0.3+(1/30). Then, we can ask what we get for (1/3)-0.33. We can rewrite this as ((1/3)-0.3)-0.03: we know that the first part is 1/30, so 1/3-0.33 can also be written as (1/30)-0.03. A little arithmetic shows then that (1/3)-0.33=1/300, which is to say that (1/3)=0.33+(1/300). Therefore, as 0.33...33 reaches n decimal places, we find that (1/3)=0.33...33 + 1/(3*10^n); this 1/(3*10^n) is your mythical "and one-third" remainder. However, as goes to infinity, 0.33...33 goes to 0.333... while 1/(3*10^n) goes to 0. Edited December 14, 2009 by Quadhelix Share this post Link to post
