Fruitbox

Oct 4, 2006

Proof of … something

Let there be a number a such that b is a number such that there exists a number c such that a, b, and c are not numbers….
No, seriously:
Things are not always simple, we know that. Many problems can be simplified, but let’s do the opposite!

Let us start with a number… let’s call it x. Now this is equivalent to writing 2x-x, right? Now let’s substitute the second term with the derivative of x^2 divided by 2, and therefore we have:
2x-D(x^2/2), where D is the derivative function. Now if we add +1-1 it changes nothing:
2x-D(x^2/2)+1-1
one of the 1’s can be written as an integral, yielding:
2x-D(x^2/2)+1-int[0..1](dx)
Expanding the integral we have:
2x-D(x^2/2)+1-(x)[0..1]
But now remember x was equal to 2x-x, of course:
2x-D(x^2/2)+1-(2x-x)[0..1]
And the last x-term can be written as a derivative of x^2/2… … … All right. See the recursive pattern will never end, and therefore we might as well stop and conclude something nice, like: We start with a number, let’s call it x, and then we … never end with anything at all! Particularly nothing very interesting!

Don’t try this at home!

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