find if possible, examples of functions f: N -> N having following proporties.?

Question

a) m f(m) b) m f(m)>f(n)
c) m f(m)=f(n)
d) if m,n are odd, then m f(m) Try as many as you can.

--> or -> is arrow pointing rigth

Answer 1

a: f(x) = x

b: No such function exists. Proof: ∀n∈(N \ 1), 1< n, so f(1) > f(n) ∀n∈(N \ 1). However, since f(1)∈N, there are only finitely many numbers less than f(1), and infinitely many numbers which must be assigned to a number less than f(1). Therefore, by the pigeonhole principle, f cannot be injective. However, suppose p≠q. Then either pf(q) or f(q)>f(p), respectively, and in either case f(p)≠f(q). However, p≠q→f(p)≠f(q) is the definition of an injective function, so f must be both injective and not injective, which is impossible. Therefore, f does not exist. Q.E.D.

c: f(x) = 1

d: f(x) = {x if x is odd,
1 if x is even}

Note to previous poster: he said functions from N to N, not from N to Z. f(x)=-x is not a valid function, because ∀n∈N, ¬(-n∈N)

Answer 2

a) f(x) = x
b) f(x) = -x
c) f(x) = 0
d) f(x) = if x is odd then x else 0
3 < 8 and f(3) = 3, f(8) = 0, so f(3) > f(8)
4 < 7 and f(4) = 0, f(7) = 7, so f(4) < f(7)

EDIT: Oops!, I used Z instead of N

a) is correct: there are many examples of strictly increasing functions

b) impossible, as formally proved by Pascal
An easier explanation is that on N you cannot have a strictly decreasing function. Starting with f(1), must have f(2) < f(1), and f(3) < f(2), .... but this cannot continue because you are always limited by 0.