Talk:Mathematical fallacy/Archive 1

message: .9999bar = 1
.9999bar = .3333bar + .3333bar + .3333bar
.3333bar = 1/3
1/3 + 1/3 + 1/3 = 1
therefore...
.9999bar = 1
HAHA!
:Seems to be a valid proof. Mathematically, they are the same number expressed differently (if you take the sum of 9 x 10^-n as n tends to infinity the difference between it and 1 goes as 10^-n, which tends to zero.) So they are equal.WolfKeeper
:: you're wrong though. .3333bar does not = 1/3 - but it is infinitely close. I'd like someone to prove that 1/3 = .3333... because they are just as much the same thing as pi = 22/7 (ie they're not equal) 68.6.112.70 03:05, 11 January 2006 (UTC)
replies: text: The bar means there is no last 9. Its been pretty well established it is in fact 1, not approaching 1. This is something people tend to resist, but it follows from the definition of limits and recuring decimals. See the .999... article Brentt 21:20, 17 March 2007 (UTC)
thread_title:  .9999bar = 1 

message: anyone know the proof of woman=evil? I know it involves time=money and money=\sqrt{evil} --Taejo | Talk 09:21, 24 October 2005 (UTC)
thread_title:  woman = evil 

message: :On the other hand it is possible to construct useful mathematical systems where 1 is technically equal to 2. Mathematics in domains modulo 1 are one example. In such domains 0.5 + 0.5 = 0 = 1 = 2...
Umm... does this even have anything to do with division by zero? I think this doesn't really belong in the article. - furrykef (Talk at me) 08:53, 9 December 2005 (UTC)
thread_title:  1 = 2 

message: I added an invald proof involving integration by parts but I haven't been able to identify the invalid step yet. If anyone can find it, please add that. The proof for women = evil goes as follows: 
*Women = Time * Money
**Time = Money
*Women = Money^2
**Money = Sqrt(Evil)
*Women = Evil
Splat 02:06, 24 December 2005 (UTC)
thread_title: Integration by parts

message: How about :
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Proof that the largest real number is 1 :
Let x be the largest real number. Therefore x^2 = x, which rearranges to x(x-1) = 0. This has solutions x=0 and x=1. We know 1>0, therefore 1 is the largest real number.
The fallacy is that we've assumed that there exists a largest real number, which is not true. There is no largest real number, because you can always "add one" to any suggested upper bound.
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Someone might want to explain it a bit more elegantly than me, but you get the basic premise of it.
:: You have the reason wrong. A "real number" as currently definined has no largest real number. However, if you took as a *given* that x is the largest real number, then you would assume *not* that x^2 = x, but that the set of all real numbers is not closed over addition or multiplication. This is to say that x^2 might not be a member of the set. For example, say you have a set {1,2,3,4} and you postulate that 4 is the largest of the set, 4^2 is obviously not equal to 4. 4^2 is still 16, it just isn't part of that set. Fresheneesz 20:23, 17 March 2006 (UTC)
replies: text: Also, even if there were a largest real number, would there be any reason to assume it has the property that x^2==x? (I know it seems obvious, but...you know)Brentt 22:17, 16 March 2007 (UTC)
thread_title: 1 is the greatest real number

message: I wanted to edit this section. Please let me know if this is accurate. Editing the part after QED:
The error here is that the associative law cannot be applied to an infinite sum unless the sum would converge without any parentheses. In this particular argument, the second line gives the series of partial sums 0 + 0 + 0 + ... (which converges to 0). However, without parenthesis, the series would be 1 - 1 + 1 - 1 + ..., which is divergent. Therefore, the third line employs an illegal usage of the associative law.
:It seems to me that this argument is circular.  The "invalid proof" is in fact a valid proof of the statement that the associative law does not hold in general for infinite sums. -- Dominus 16:07, 18 March 2006 (UTC)
replies: text: And I just realized that the "Conclusion" section of the article makes exactly this point.  -- Dominus 17:40, 18 March 2006 (UTC)
thread_title:  Editing "Proof that 0 equals 1" 

message: There's a section of the article stating that i=the sqrt of -1. The only problem is, there is no sqrt of -1. When it gave an invalid proof, it stated that i squared = 1. If i = the sqrt of -1, then wouldn't i2 be -1? If so, wouldn't this just be another proof that -1 = 1? AstroHurricane001 12:38, 18 October 2006 (UTC)
replies: text: i is the imaginary square root of -1.  See imaginary number for a better explanation than I can ever hope to give. QWERTY | DVORAK 21:36, 24 October 2006 (UTC)
thread_title:  i=? 

message: Power = Work / Time
Knowledge = Power
Time = Money
Thus:
Knowledge = Work / Money
Knowledge × Money = Work
Money = Work / Knowledge
As Knowledge → 0, Money → infinity. Damn it. The less you know, the more you make. Pity really. Paidgenius 16:37, 17 December 2006 (UTC)
thread_title:  Look at this... 

message: allow x to equal 1.
:x-1 = xx0
:cancel out the x's
:-1 = 0
...thus we prove that ALGEBRA is broken, as real numbers disproves this:
x = 1
: (1-1) = (1x0)
:which leaves us with 0 = 0 —The preceding unsigned comment was added by Tar7arus (talk • contribs) 14:03, 10 February 2007 (UTC)
replies: text: Yours is an equation and not an identity - you are substituting a value of x.  x42bn6 Talk 17:21, 10 February 2007 (UTC)
thread_title:  'Nother one 

message: :I'm not really sure where the error is in this proof (I think the last step) but if someone knows I think it is a good one to add.
:e ^ (i * pi) + 1 = 0
:e ^ (i * pi) = -1
:(e ^ (i * pi)) ^ 2 = (-1) ^ 2
:e ^ (2 * i * pi) = 1
:e ^ 0 = 1
:e ^ (2 * i * pi) = e ^ 0
:ln(e ^ (2 * i * pi)) = ln(e ^ 0)
:2 * i * pi = 0
:Therefore, either 2, i, or pi must equal 0.
:Any ideas on where the error is? 67.167.99.102 04:04, 18 April 2007 (UTC)
replies: text: The error occurs when you take logs. Logs of complex numbers are multivalued. If you use the principal branch, then \log(e^{2i\pi}) = 0 = \log(e^0). --Zundark 06:56, 18 April 2007 (UTC)
thread_title:  i, pi or two equals 0 

message: How fares this proof that 2 = 0?
2 = 1 + 1
2 = 1 + 11/2
2 = 1 + ((−1) × (−1))1/2
2 = 1 + ((−1)1/2 × (−1)1/2)
2 = 1 + ((−1)1/2)2
2 = 1 + (−1)1/2×2
2 = 1 + (−1)1
2 = 1 − 1 = 0
—Preceding unsigned comment added by 84.48.198.46 (talk • contribs) 21:11, 23 October 2007
:We already have this one in the article, stated as a proof that 1 = −1. --Zundark 21:22, 23 October 2007 (UTC)
thread_title:  Proof that 2 = 0