Re: 4=3 A maths joke
- From: Tonico <Tonicopm@xxxxxxxxx>
- Date: Thu, 15 May 2008 05:47:48 -0700 (PDT)
On May 15, 9:02 am, Albert <albert.xtheunkno...@xxxxxxxxx> wrote:
LOL This took me 5 minutes to work out what was wrong to my friend and
it was hilarious.
a+b=c where c > a and c > b
4a-3a+4b-3b = 4c-3c
4a+4b-4c=3a+3b-3c
4(a+b-c)=3(a+b-c)
4=3
Where's the flaw?
*****************************************************************
Want a slightly (VERY slightly) less boring and less known "hoax",
which at least uses some basic calculus? Here:
Let n be a natural number (non-zero, to be sure). As we all know, n^2
= n*n, and the right side can be put as n + n +....+n (n times), so:
** n^2 = n + n +...+ n (n times). Now evaluate the derivative in both
sides, using the well known rule for the sum:
** 2n = 1 + 1 +...+ 1 (n times) = n , and thus
** 2n = n ==> 2 = 1, dividing by n
Enjoy...and let's hope experienced mathematicians can refrain for a
while from postingn the solution.
Regards
Tonio
.
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