Nikita: the latest news

Nikita expands his internet presence by attacking the Russian functional programming community: http://ru-declarative.livejournal.com/108182.html

Cute!

Part 5: If n is a natural number, then we have that either ∀n f(n) = 0, or we have that ∀n f(n) = n.

It was shown in part 2 that if f(1) is not zero, then f(1) = 1.

Suppose that f(1) = 0, and n is a natural number. Then we have that:

f(n)
= f(1 + 1 + ... (n times) ... + 1)
= f(1) + f(1) + ... (n times) ... + f(1)
= 0 + 0 + ... (n times) ... + 0
= 0 XD

Now suppose it is the case that f(1) = 1. Then:

f(n)
= f(1 + 1 + ... (n times) ... + 1)
= f(1) + f(1) + ... (n times) ... + f(1)
= 1 + 1 + ... (n times) ... + 1
= n XDD

Part 6: If n is an integer, then either ∀n f(n) = 0, or ∀n f(n) = n.

In part 4 it was shown that f(-x) = -f(x). If f is zero on the naturals, we have, for all negative integers m:

f(m) = -f(-m) = -0 = 0 XD

If f is id on the naturals, we have for all such m:

f(m) = -f(-m) = -(-m) = m XDD