.9999999 bar does not = 1
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Voting Style:  Open  Point System:  7 Point  
Started:  8/19/2011  Category:  Science  
Updated:  5 years ago  Status:  Voting Period  
Viewed:  3,363 times  Debate No:  17959 
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In response to the thread that there is a lack of debates I'm trying to liven things up here.
Resolved: .99999ect. (aka bar) does not equal 1. (this is not as obvious as it might seem.) BoP shared. No semantics. No you tube video's. no new points in final argument. I thank my opponent for presenting this debate, and I hope that what follows proves to be informative, and enriching. The initial idea of .9999999 bar (henceforth referred to as "0.999...") not being equal to one comes from the false assumption that 0.999... is a "number", when 0.999... is in fact, a geometric convergence which is equal to 1. There are many explanations which express the exact same, but few go far as to explaining why. Why 0.999... is equal to one is what I propose to explain throughout this debate. Why 0.999... is not a "number" In this section, I intend to prove that 0.999... is neither a rational, nor an irrational number. A rational number is defined as "any real number of the form a / b , where a and b are integers and b is not zero." Source: http://dictionary.reference.com... An irrational number is defined as "a number that cannot be expressed as a ratio between two integers and is not an imaginary number. If written in decimal notation, an irrational number would have an infinite number of digits to the right of the decimal point, without repetition." Source: http://dictionary.reference.com...; Thus, based on the knowledge that 0.999... cannot be expressed in the form of "a / b, when a, and b are integers, and b is not zero", and the knowledge that 0.999... is a repeating decimal, we can safely say that it fits into neither of the above definitions. With that knowledge in hand, what precisely is 0.999...? Well, to put it blankly, 0.999... is a concept, not at all unlike √(1); however, much unlike √(1), 0.999... can be resolved into a real, rational number using simple algebra. Which brings us to our... Proof that 0.999... = 1 x = 0.999... Multiply both sides by 10... 10x = 9.999... Subtract 0.999..., or "x" from both sides... 9x = 9 Divide both sides by 9... x = 1 With that having been said, I rest and await my opponent's reply. 

I thank my opponent for accepting this debate. And I would like to welcome him to ddo.
Seeing that my opponent is familiar with the subject I will get strait to the point. I take issue with the Prof at the following point: "...Subtract 0.999..., or "x" from both sides... 9x = 9" 10xx= 9x However, the other side of the equation is not true. when x was multiplied by 10, ".999..." became 9.9999.... what is left out is the fact that there is now one less 9 at the end. Now, one may be tempted to say that since there are infinite 9's this shouldn't matter. However contrary to popular belief, in the world of infinity, not all infinities are created equal. inf. 1 = inf. However, inf.  1  inf. = 1. So, when x is multiplied by 10 there are "inf. 1". 9's following the decimal point. As opposed to x, which has inf. 9's following it. Thus when subtracted, we do not have "9" but rather a value approaching 9. Thus while .999... is obviously approaching 1, it is not equivalent to 1. I thank my opponent for the kind welcome. Refutation 1.) "when x was multiplied by 10, ".999..." became 9.9999.... what is left out is the fact that there is now one less 9 at the end." Another example to demonstrate the concept used in this argument. x = 0.888... 10x = 8.888... 8.888...  0.888... = 8 9x = 8 Despite that there was one fewer decimal point, the result remained the same, because there are an infinite number of eights. The same can be said of any and all infinitely repeating decimals, including 0.999... 2.) "inf. 1 = inf. However, inf.  1  inf. = 1." The issue with this statement is that infinity, as a concept, can be applicable in some situations, and not applicable in others. Infinity cannot be balanced with any real number. When you start to use infinity to balance infinity, you are stepping outside of the boundaries of real numbers, thus defeating the concept. In addition, I would like to point out that the second equation used in this argument is essentially ∞  1 = ∞  1 A different approach to the same problem To the point, an infinitesimal difference ie., (a  (1/∞)) is not a difference within the real, nor complex number system, because it cannot be quantified. (∞ + a ) = ∞ {a = R} That is to say that you can add any real number to infinity, and it will still equal infinity. (My apologies for the reiteration, but I feel that it is best to hold a mathematical debate with consistency in mind.) Provided that we state that the above is true of infinity, we can make some assumptions about the concept of 0.999... Seeing as 0.999... is an infinitesimal amount less than 1, we can define 0.999... by using a quotient. That having been said, 0.999... = 1  (1 / ∞) = (∞  1) / ∞ Algebra tells us that (∞  1) / ∞ = (∞ + (1)) / ∞ Now, armed with the above knowledge that "you can add any real number to infinity, and it will still equal infinity," and knowing that 1 fits within the real number system, we can state that "∞  1 = ∞" Thus, we can also state: (∞ + (1)) / ∞ = ∞ / ∞ = ∞^0 = 1 I rest and await my opponent's response 

Despite that there was one fewer decimal point, the result remained the same, because there are an infinite number of eights.
I showed that the fact that there are inf. 8's doesn't help since: inf.  1  inf. =1 "how does the fact that there are infinite 8's make it that the results will be the same?" The results are the same because the lack of one decimal point is irrelevant when dealing with an infinite number of decimal points, regardless of whether or not "inf.  1  inf. = 1" Simply put, 9.999...  0.999... = 9 8.888...  0.888... = 8 9.999...  0.999... ≠ 8.999... Multiplying 0.888..., or any decimal by 10 moves the decimal point one place to the right, exactly the same as 0.999... but, as we can see with the previous example, the movement of the decimal point has no impact on the result. In short, having "one fewer decimal place" does not change the result. 

truthseeker613 forfeited this round.
StrykNyne forfeited this round. 
3 votes have been placed for this debate. Showing 1 through 3 records.
Vote Placed by LaL36 4 years ago
truthseeker613  StrykNyne  Tied  

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Total points awarded:  3  0 
Reasons for voting decision: One thing left me thinking that the pro said. 10x=9.99999....
There is one less nine.
Vote Placed by QT 5 years ago
truthseeker613  StrykNyne  Tied  

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Reasons for voting decision: Pro did not adequately refute Con's case.
Vote Placed by RoyLatham 5 years ago
truthseeker613  StrykNyne  Tied  

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Reasons for voting decision: This really should not be a debate topic, as any standard math text shows Con's proof is correct.
This is a solved problem that is not debatable.
how does the fact that there are infinite 8's make it that the results will be the same?
Infinity raised to the zero is undefined.
I am apparently rather forgetful.
My sincere apologies for butchering mathematics.
You can disregard the second half of my second round argument.
Although, my refutation still stands.
I am, regrettably, not as well learned within the field of mathematics as I would like.
I do, however, think that my point stands, regardless of semantical errors.
I'd be happy to clarify, at the request of the reader.