This proof also shows that the original proposition implies Lior is the pope (which is harder to prove by other means).

]]>Define f(x) over the real numbers such that f(0) = 0, and f(x) = 1 for x != 0.

Further, let g(x) = x.f(x).

By considering the cases x=0 and x!=0, it is clear that g(x) = x for all x.

Supposing that your assertion is correct, we divide both sides by x to get

g(x)/x = 1 for all x

i.e. f(x) = 1 for all x

Now since this equation must hold for all x as you stated, we set x = 0 to get f(0) = 1. i.e. 0 = 1.

Since you have proven that 0 = 1, I deduct 10 further points, putting you at -20. (you can still have some points back for interest though.)

]]>Meanwhile, x turned into n?

I award you a score of -10 points for that section. (but you can have some points back for an interesting read.)

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