Re: 4=3 A maths joke

Antonio Perez wrote :

On May 16, 12:44 am, amy666 <tommy1...@xxxxxxxxxxx>
wrote:
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

d a*b = da * b + db * a thus n * (1 times) + 1 (n
times)
= 2n and not n.

thus 2n = 2n

divide by n => 2 = 2.

****************************************************

I don't have the faintest idea what you did above,
but most probably
is another nonsense...

you weak pathetic fool !!!

i just took up your challenge and found the mistake in your paradox.

not the faintest idea ? i solved your problem !!

i used the product rule for derivatives, but apparently you dont know that (a*b)' = a'*b + b'*a.

you dont understand your own paradox nor the concept of derivative.

indeed you dont have the faintest idea about your own paradox , you dont even recognize the solution towards it and you dont know about derivatives of products.

so if you post things you dont understand your the one selling nonsense.

you weak pathetic fool !!!



By the way, what about those papers you said you
wrote on L-functions?

Regards
Tonio

why do you want them ? you cant even understand these posts here in this thread.

since you dont know about derivatives , i dont think you know what L-functions are.


tommy1729
.

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