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Topic: 2=1 (Read 1037 times) |
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Burl V. Minnis
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I have a question. I once saw an algebraic equation that solved as 2=1, (or 1=2). It was an entire equation and the procedures used to reach the solution. I think it started out as two polynomials, separated by an equal sign, (or equal to each other) and after a few mathematical procedures were applied to the problem, it derived out as 2=1, (or 1=2). The riddle was to find out what mathematical procedure was applied to the equation incorrectly to make that happen, (like multiplying both sides of the equal sign by ½, or something, to get X by itself). Are you at all familiar with this riddle? Please email me back.
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Obob
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There was almost certainly a division by zero somewhere in the argument, making it fallacious.
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Grimbal
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Re: 2=1
« Reply #2 on: Nov 2nd, 2004, 3:46am » |
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Or a square root of -1.
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sok
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i know the riddle you are referring to
a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b)
a + b = b
b + b = b
2b = b
2=1
note that the ^ indicates power of
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Grimbal
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Re: 2=1
« Reply #4 on: Nov 3rd, 2004, 6:37am » |
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And of course, you can not say
(a+b)(a-b) = b(a-b) => a + b = b
Consider for example a=b=1, (a+b) = 2, (a-b) = 0
The claims translates to
2*0 = 1*0 => 2 = 1
which is wrong, because in effect you divide both sides by 0.
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sok
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Re: 2=1
« Reply #5 on: Nov 6th, 2004, 3:21am » |
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actually i disagree
there is nothing mathmatically wrong with saying
(a+b)(a-b) = b(a-b) => a+b = b
on both sides,
you are dividing my (a-b)
which is correct
the mathematically example you have provided did not wrong due to another reason
this is because the hypothesis is used in the proof
the first line claims that a=b
thus this should be the proof
but in line 6, a substitution was done which used this again
if this is your hypothesis,
what you are trying to prove,
you cannot use it again in your proof
eg HYPOTHESIS ~ a cat is a dog
.. proof ..
.. proof ..
.. sub cat for dog ..
.. proof ..
thus a cat is a dog
it is nonsense
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towr
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Some people are average, some are just mean.
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Re: 2=1
« Reply #6 on: Nov 6th, 2004, 4:07am » |
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on Nov 6th, 2004, 3:21am, sok wrote:actually i disagree
there is nothing mathmatically wrong with saying
(a+b)(a-b) = b(a-b) => a+b = b
on both sides,
you are dividing my (a-b)
which is correct |
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No it, isn't
a+b = b => (a+b)(a-b) = b(a-b)
is correct, but the other way should be
(a+b)(a-b) = b(a-b) => a+b = b OR a=b
Otherwise it implies a=0, but (a+b)(a-b) = b(a-b) is also true for every case where a=b and a [ne] 0.
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Wikipedia, Google, Mathworld, Integer sequence DB
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JocK
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Re: 2=1
« Reply #7 on: Nov 6th, 2004, 4:26am » |
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Why would you clasify 'a=b' in above "derivation" as an hypothesis?
if it helps, just add the word 'given' at the start:
Given: a = b
a*a = a*b
a*a - b*b = a*b - b*b
etc.
It should be clear that the above "derivation" is equivalent to the two-step "proof":
For any x: x[sup2] - x[sup2] = x(x - x)
Rearranging: (x+x)(x-x) = x(x-x)
Dividing by (x-x) gives: x+x = x
which leads to the contradiction. Clearly, the zero-division causes this contradiction.
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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srn437
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the dark lord rises again....
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Re: 2=1
« Reply #8 on: Sep 2nd, 2007, 9:14pm » |
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Since it divides by zero, it is an invalid proof which prooves nothing. There are many more, like that 1>1 or 0=1. Check wikipedia or the 1>1 post.
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