One plus One is equal to Two ( 1 + 1 = 2 )
Post Voting Period
The voting period for this debate has ended.
after 0 votes the winner is...
It's a Tie!
Voting Style:  Open  Point System:  7 Point  
Started:  8/1/2013  Category:  Miscellaneous  
Updated:  4 years ago  Status:  Post Voting Period  
Viewed:  1,460 times  Debate No:  36261 
Debate Rounds (3)
Comments (14)
Votes (0)
The purpose of this debate is to either solidify or challenge the renowned mathematical and philosophical debate regarding whether or not 1 + 1 = 2 I NEGATE the topic... As a courtesy, I am allowing my opponent to administer the first argument.
v 

Unfortunately, it has become apparent that my opponent made an accidental typo; to conserve debating space, I will begin the argument. 1 + 1 = 2 is commonly percieved as an axiom of arithmetic and mathematics; proving to be a foundation of mathematical advancement and progression. However, mathematics itself is an incomplete theory. Fallacy and argumentum verbosium have dictated some of the unjust paths of mathematics, and are subject to reasonable doubt. I propose that this equation, disregarding its apparent simplicity, is subject to reasonable doubt. 1 + 1 = 2 relies on the numerical value of '1' to be stagnant and unquestionably fundamental. Yet, 1 can have two equivelants: one of which being 0.999... First, one must address 0.999... as an infinite sum, like such: 0.999... = Σ[+∞, n=1] (9/10^n) = (9/10) + (9/10^2) + (9/10^3) +... To find the limit of this geometric sum: Let, x = (1/10) + (1/10^2) + (1/10^3)+... Given that 9/10 = 0.9, 9x = 0.999... Given that 10/10 = 1, 10x = 1 + (1/10) + (1/10^2) +..., thus, 10x  x = 1 This implies x = 1/9 Hence, 0.999... = 9 · 1/9 = 1 Q.E.D. This proof can be reiterated in numerous ways, involving concepts ranging from modular fractions to abstract summation. With 0.999... = 1 established, one can formally conclude that: 1 + 1 = 0.999... + 0.999... = 1.999...8
These numbers don't add up! "Given that 9/10 = 0.9, 9x = 0.999... Given that 10/10 = 1, 10x = 1 + (1/10) + (1/10^2) +..., thus, 10x  x = 1" "9x = 0.999..." "thus, 10x  x = 1" "9x = 0.999..." "thus, 10x  x = 1" "9x = 0.999..." "thus, 10x  x = 1" "9x = 0.999..." "thus, 10x  x = 1" "9x = 0.999..." "thus, 10x  x = 1" 

Perhaps my proof was slightly unclear. Your rebuttal suggests that my numbers do not "add up," yet you have misinterpretated the foundations for each implication. As 0.999... is an infinite geometric sum, it can be expressed by its sum expansion: (9/10) + (9/10^2) + (9/10^3) +. . . In order to prove that 0.999... = 1, you must solve a limit by substituting this value within an established equation. If x = (1/10) + (1/10^2) + (1/10^3) +... Thus, 9x = 9 · [(1/10) + (1/10^2) + (1/10^3) +...] = (9/10) + (9/10^2) + (9/10^3) +... = 0.999... If one were to multiply x by 10, 10x = 10 · [(1/10) + (1/10^2) + (1/10^3) +...] = 1 + (1/10) + (1/10^2) + (1/10^3) +... Thus, by subtracting [(1/10) + (1/10^2) + (1/10^3) +...], you are subtracting x, resulting in: 10x  x = 1. Granted that 9/9 = 1, this equation implies that x = 1/9. By substituting 1/9 as x in the original equation, 9x = 0.999..., you get: 0.999... = 9 · 1/9 = 9/9 = 1. Thus, 0.999... = 1. Substituting 0.999... as 1 in the equation 1 + 1 = 2 does not result with 2, it equals 1.999...8. However, this may unclarified and misleading, due to paradoxal conundrum when expressed as an infinite geometric sum. So, to provide a different mathematical perspective: 0.999... = lim[n→∞] (0.99...9) = lim[n→∞] (Σ[n, k=1] (9/10^k) ) = lim[n→∞] (1  (1/10^n)) = 1  lim[n→∞] (1/10^n) = 1  n  This simplification of limits is justified by the Archimedean property of the real numbers. Yet, this does not explain how the sum of an infinite geometric series, 0.999..., can equal two things at once. It is a calculus technicality that 0.999... = 1, as proved, yet it is numerically less than 1. Thus, the calculus says that 0.999... + 0.999... = 1 + 1 = 2, yet numerically it cannot be equal to 2 because 0.999... is not signified as 1. The actual number, 1, has two simulataneous values, which when applied to 1 + 1 can both equal 2 and not equal 2. This disproves the validity of 1 + 1 solely equalling 2 because reasonable doubt has been offered to subject it (this may be the area in which my opponent misunderstood my proposition). guiltygear forfeited this round. 
No votes have been placed for this debate.
I can't believe its your understanding!!!
That's so substantial and relevant!!!
To a laymen it would seem if you apply different kinds of mathematics to the same sum then you can get different answers, therefore there's no point arguing one way or the other. It's like scoring a professional boxing match using the Olympic scoring system.  two different techniques to analyse the same problem while producing different outcomes.
P.S. I loved your comment  "Perhaps, before you start making accusations and questioning the mathematics in my proof, you should actually learn mathematics"  haha, talk about touchy!
To a laymen it would seem if you apply different kinds of mathematics to the same sum then you can get different answers, therefore there's no point arguing one way or the other. It's like scoring a professional boxing match using the Olympic scoring system.  two different techniques to analyse the same problem while producing different outcomes.
P.S. I loved your comment  "Perhaps, before you start making accusations and questioning the mathematics in my proof, you should actually learn mathematics"  haha, talk about touchy!
I understand your conception on the topic, and I am intrigued by your perception. However, to clarify my rebuttal's proposition, I am stating that the expression 1 + 1 can be solved in two different ways. Calculus questions the actual value of 1 by proposing its equivalence to an infinite series, where number theory defines 1 > 0.999... simply because their numerical value is different. When combining both premises, 1 + 1 becomes an immediate conundrum; no true answer can be properly distinguished, only accommodated for specific circumstance. Thus, 1 + 1 cannot equal just "2," there is room for reasonable doubt as to the meaning of the original expression. I am not disproving that it is equal to 2, I am also not proving it. The whole concept is unprovable.
However you say...
"The actual number, 1, has two simultaneous values, which when applied to 1 + 1 can both equal 2 and not equal 2. This disproves the validity of 1 + 1 solely equalling 2 because reasonable doubt has been offered to subject it"
It all seems a bit quantum to me (.....or at least fence sitting on your part!). You flicked a coin and called both sides just as it lands! lol.
It seems you're arguing that 1 + 1 doesn't equal 2 on the basis that it equals 2 and also equals something else (just because it doesn't 'solely' add up to 2 does not mean it does not add up to 2).. A horse has three legs as well as four. Pain exists as both pleasure and displeasure when experienced by different receivers. Stuff may 'exist' in two forms simultaneously.
Lacking any mathematical capacity I'll believe your proposition that one plus one DOES equal two  one plus one DOES NOT equal two.
....but with that in mind I can't see how can argue one way or the other.
As you previously stated, you are "no mathematician" and you obviously are not aware of the definition of an Infinite Sum nor a Geometric Series. If a series, 0.9 + 0.09 + 0.009 +... increases by the multiplication of a common ratio to infinity, then the sum of that series can be defined as:
]1;[W34;, n=1] a_n = a/(1  r)
Where,
a_n is a term within the series defined by the value of n
r is the common ratio in which each previous term is multiplied by to get the corresponding term
If you were to substitute 0.9 for a and 0.1 for r, you can easily find the sum of the Infinite Geometric Series that is 0.999... The sum is 0.9/0.9 = 1.
Perhaps, before you start making accusations and questioning the mathematics in my proof, you should actually learn mathematics.