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# .9999999 bar does not = 1

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 Voting Style: Open Point System: 7 Point Started: 8/19/2011 Category: Science Updated: 6 years ago Status: Voting Period Viewed: 3,472 times Debate No: 17959
Debate Rounds (4)

 Pro 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.Report this Argument Con 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... = 1x = 0.999...Multiply both sides by 10...10x = 9.999...Subtract 0.999..., or "x" from both sides...9x = 9Divide both sides by 9...x = 1 With that having been said, I rest and await my opponent's reply. Report this Argument Pro 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" 10x-x= 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.Report this Argument Con I thank my opponent for the kind welcome. Refutation1.) "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 = 8Despite 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 = ∞ - 1A different approach to the same problemTo 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 = 1I rest and await my opponent's responseReport this Argument Pro 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. =-1Report this Argument Con "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... = 98.888... - 0.888... = 89.999... - 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. Report this Argument Pro truthseeker613 forfeited this round. Con StrykNyne forfeited this round.
12 comments have been posted on this debate. Showing 1 through 10 records.
Posted by 000ike 6 years ago
You're debating a fact. Did you WANT to lose?
Posted by RoyLatham 6 years ago
The proof using 10x = 9.999... is correct. Another way is to sum the series 9/10 + 9/100 + ...

This is a solved problem that is not debatable.
Posted by truthseeker613 6 years ago
Despite that there was one fewer decimal point, the result remained the same, because there are an infinite number of eights."

how does the fact that there are infinite 8's make it that the results will be the same?
Posted by CD-Host 6 years ago
Wow is this one math error after another after another. Drop me a line when this is all over.
Posted by StrykNyne 6 years ago
Feel free to post an argument as a comment if you'd like.
Posted by truthseeker613 6 years ago
My bad that was meant to be a comment not an argument. O well.
Posted by StrykNyne 6 years ago
Apparently, unicode is broken in plain text. (Still forgetful, it would seem.)
Infinity raised to the zero is undefined.
Posted by StrykNyne 6 years ago
Disregard that, &#8734;^0 = 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.
Posted by StrykNyne 6 years ago
I apologize for any and all mathematical errors on my part, both by terminology, and miscalculation.
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.
Posted by gizmo1650 6 years ago
I think this is the first time I have seen the con side of this debate make more mathematical mistakes than the pro.
3 votes have been placed for this debate. Showing 1 through 3 records.