wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> 2=1
        
(Message started by: Burl V. Minnis on Oct 31^st, 2004, 9:54am)

Title: 2=1
Post by Burl V. Minnis on Oct 31^st, 2004, 9:54am

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.

Title: Re: 2=1
Post by Obob on Oct 31^st, 2004, 10:17am

There was almost certainly a division by zero somewhere in the argument, making it fallacious.

Title: Re: 2=1
Post by Grimbal on Nov 2^nd, 2004, 3:46am

Or a square root of -1.

Title: Re: 2=1
Post by sok on Nov 3^rd, 2004, 5:46am

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

Title: Re: 2=1
Post by Grimbal on Nov 3^rd, 2004, 6:37am

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.

Title: Re: 2=1
Post by sok on Nov 6^th, 2004, 3:21am

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

Title: Re: 2=1
Post by towr on Nov 6^th, 2004, 4:07am

on 11/06/04 at 03:21:15, 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

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.

Title: Re: 2=1
Post by JocK on Nov 6^th, 2004, 4:26am

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.

Title: Re: 2=1
Post by srn347 on Sep 2^nd, 2007, 9:14pm

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.