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| Did you figure it out? |
| Yes |
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89% |
[ 25 ] |
| No |
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[ 3 ] |
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| Total Votes : 28 |
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 Posted: Tue Nov 07, 2006 7:04 pm
see if you can find out what's wrong with this proof:
we are given that x + y = z
so, assuming this, we can make two equations:
4x + 4y = 4z , and
5z = 5x + 5y
by combining (adding) these equations, we obtain another true equation:
4x + 4y + 5z = 5x + 5y + 4z
4x + 4y - 4z = 5x + 5y - 5z
4(x + y - z) = 5(x + y - z)
4=5
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Posted: Wed Nov 08, 2006 5:49 am
Just want to share a similar "proof", but about 1=2:
Assume a = b
a² = ab
a² + a² - 2ab = ab + a² - 2ab
2(a² - ab) = a² - ab
2 = 1
This also proves that any consecutive number is equal.
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Posted: Wed Nov 08, 2006 11:41 am
It also divides by 0, which is bad. I believe that this is somewhat relevant:  On a more serious note, your proof goes as follows: a² = ab a*0 = a(2a - 2b) = a² + a² - 2ab = ab + a² - 2ab = a(b + a - 2b) = 0 2*0 = 0 2 = 1 And since anything times 0 is 0 in any ring/field, etc., we get that anything is equal to 1, by your reasoning. Fortunately, in almost all of the number systems that matter, you can't divide by 0.
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Posted: Wed Nov 08, 2006 3:53 pm
Aww...now the secret answer is out. Oh well, anyone have any other cool divide by zero proofs like that?
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Posted: Wed Nov 08, 2006 4:34 pm
Consider int (1/x) dx, where int comes from me being too lazy to find an integral sign.
Integration by parts, letting u = 1/x and dv = dx, gives that du = (-1/x^2)dx and v = x, so therefore
int (1/x) dx = uv - int v du
= x/x - int x(-1/x^2) dx
= 1 + int (1/x) dx
Thus, subtracting int (1/x)dx from both sides gives 0 = 1.
Find the problem.
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Posted: Wed Nov 08, 2006 7:14 pm
Yes, I figured it out back in my high school days. The error in the 2=1 proof is the same error in the 4=5 proof. As for the 0=1 proof, a constant makes the difference. It is often neglected to add a constant variable in getting antiderivatives or indefinite integrals. (It reminds me of someone whose only deduction in a long test is forgetting to add a c in his final answer. Now, I know that adding a constant is crucial.) Also, it reminds mo of a joke: Two mathematicians are in a bar. The first one says to the second that the average person knows very little about basic mathematics. The second one disagrees. As the first mathematician goes off to the washroom, the second calls over the waitress and tells her that after his friend has returned, he will call her over and ask her a question. All she has to do is answer 'one third x cubed'. The first guy returns and the second proposes a bet to prove his point, that most people do know something about basic math. He says he will ask the blonde waitress an integral, and the first laughingly agrees. The second man calls over the waitress and asks 'what is the integral of x squared?'. The waitress says 'one third x cubed' and while walking away, turns back and says over her shoulder 'plus a constant'! There is more invalid proofs in wikipedia: http://en.wikipedia.org/wiki/False_proofCheck it out!
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Posted: Thu Nov 09, 2006 12:51 am
The integration by parts of 1/x makes it quite clear that something's fishy, even discounting the obviously false result. It would be much better to obfuscate it a bit in order to catch the unwary student, a la Int[ cot x dx ] = Int[ (sin x)'(1/sin x) dx ] = sin(x)(1/sin(x)) - Int[ (sin x) (1/sin x)' dx ] = 1 - Int[ (sin x)(-cos x/sin^2 x) dx ] = 1 + Int[ cot x dx ], therefore 0 = 1. I've shown this version to many people; it seems the more natural impulse it to check if it there is something wrong with the application of integration by parts using trigonometry.
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Posted: Thu Nov 09, 2006 3:19 pm
Drake07 As for the 0=1 proof, a constant makes the difference. It is often neglected to add a constant variable in getting antiderivatives or indefinite integrals. (It reminds me of someone whose only deduction in a long test is forgetting to add a c in his final answer. Now, I know that adding a constant is crucial.) The funny thing is that you can tell people: "Okay, let's get rid of that constants problem by making them definite integrals:" int_1^2 (1/x)dx = uv - int_1^2 v du = x/x - int_1^2 x(-1/x^2)dx = 1 + int_1^2 (1/x)dx And we still have a problem... wink
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Posted: Thu Nov 09, 2006 5:47 pm
Quote: The funny thing is that you can tell people: "Okay, let's get rid of that constants problem by making them definite integrals:"
int_1^2 (1/x)dx = uv - int_1^2 v du = x/x - int_1^2 x(-1/x^2)dx = 1 + int_1^2 (1/x)dx Perhaps, you should also put bounds on x/x and will result to 2/2 - 1/1 = 0 and the rest should work. But I am not sure of this though. I just don't recommend such step because constants are very important. (Our math department is very strict in checking our solutions because math major students are expected to be logical and very clear in every step of the solution. Such manipulation of equation by making it definite integral would be unacceptable.) Anyway, this should also serve as a warning that integration by parts should be used as a last resort because you may never know when the constant makes the answer different.
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Posted: Thu Nov 09, 2006 6:53 pm
Let I = [a,b] be a closed interval that excludes zero. Since f(x) = x and g(x) = 1/x are Lipschitz on any such I, they are also absolutely continuous, so that the integration by parts formula for f,g is valid on I: Int_I[ g df ] = g(b)f(b) - g(a)f(a) - Int_I[ f dg ].
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Posted: Thu Nov 09, 2006 6:54 pm
Drake07 Quote: The funny thing is that you can tell people: "Okay, let's get rid of that constants problem by making them definite integrals:"
int_1^2 (1/x)dx = uv - int_1^2 v du = x/x - int_1^2 x(-1/x^2)dx = 1 + int_1^2 (1/x)dx Perhaps, you should also put bounds on x/x and will result to 2/2 - 1/1 = 0 and the rest should work. But I am not sure of this though. I just don't recommend such step because constants are very important. (Our math department is very strict in checking our solutions because math major students are expected to be logical and very clear in every step of the solution. Such manipulation of equation by making it definite integral would be unacceptable.) Anyway, this should also serve as a warning that integration by parts should be used as a last resort because you may never know when the constant makes the answer different. Yes, bounds on x/x are necessary of course. The trick is to not tell people that. And I find integration by parts to work fine as long as you put in the bounds; if you're dealing with indefinite integrals, well, they're indefinite so equality is always up to a constant anyway.
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Posted: Sat Mar 17, 2007 9:24 pm
Here is a good one...
5/0 = (5*0)/(0*0) = (5*0)/0
let X=0 so
(5*X)/X=5
like omg!! rofl
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Posted: Wed May 16, 2007 4:32 pm
`Fenix Here is a good one... 5/0 = (5*0)/(0*0) = (5*0)/0 let X=0 so (5*X)/X=5 like omg!! rofl But (5*0)/(0*0) doesn't equal 5/0, it's 0/0. Speaking of, is 0/0 = 1? xd
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Posted: Sun May 20, 2007 11:58 pm
Altik_0 `Fenix Here is a good one... 5/0 = (5*0)/(0*0) = (5*0)/0 let X=0 so (5*X)/X=5 like omg!! rofl But (5*0)/(0*0) doesn't equal 5/0, it's 0/0. Speaking of, is 0/0 = 1? xd 0/0 is everything and nothing at all. It's anything you want it to be, provided that you arrive at it as a limit and not via algebra. From an algebraic viewpoint it's a meaningless object, indeterminate, valueless (which is different from it being 0). Only in terms of an analytic viewpoint can you get it to mean anything, because on its own you can give it any value and no one would be able to confirm or deny.
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Posted: Wed Jun 27, 2007 9:44 pm
Don't drink and derive!
And do not miss use the number zero.
Cypher has been very importat in the advances in mathematics.
You should respect him instead of making fun of him.
Though it is funny 'proving' to un-smart ppl that 1=2. yes.
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