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More options Oct 24 2012, 10:39 pm
Newsgroups: sci.math, rec.puzzles, alt.math.recreational
From: William Elliot <ma...@panix.com>
Date: Wed, 24 Oct 2012 19:39:32 -0700
Local: Wed, Oct 24 2012 10:39 pm
Subject: Re: today's puzzle

On Wed, 24 Oct 2012, Herman Rubin wrote:
> On 2012-10-24, William Elliot <ma...@panix.com> wrote:
> > On Tue, 23 Oct 2012, PT wrote:

> >> g(n) is a function of any integer n, positive or negative, which
> >> produces an integer value, with conditions:

> >> a) g(g(n)) = n
> >> b) g(g(n + 2) + 2) = n
> >> c) g(0) = 1

> > Let f(n) = 1 - n.

> > ff(n) = f(1 - n) = 1 - (1 - n) = n
> > f(f(n + 2) + 2) = f(1 - (n + 2) + 2) = f(1 - n) = n
> > f(0) = 1

> >> 1. Determine g(n)
> > g = f.

> >> 2. Prove your solution is unique.

> From (a), g(g(g(n))) = g(n), so g is 1-1.

Only if g is surjective.

> Then from (b), g(n+2) = g(n) - 2.

> From (a) and (c), g(1) = 0. so we have g(0) and g(1), and increasing
> n by 2 decreases g(n) by 2, so g is uniquely defined.

Ok, provided g is surjective, you've determined g(n) for all n >= 0.
What about g(n) for n < 0?