Roberto
Joined: 03 Jun 2010, 15:18 Posts: 136

Exercise [16.08] b
I attach an alternative solution, in which equation N=1/2((a+b)^2+3a+b) is explicitly solved finding a(N) and b(N).

robin
Joined: 26 Mar 2010, 04:39 Posts: 109

Re: Exercise [16.08] b
That's quite an interesting way to tackle it. I didn't even try to explicitly find the inverse.
But don't you also need to show that the inverse mapping is surjective? Otherwise the inverse you found could be mapping to a subset of NxN and then f would not necessarily have to be 11.
For example f: R > R+ given by f(x) = x^2 is not 11 even though it has an inverse g: R+ > R given by g(y) = sqrt(y) (which is not surjective).

Roberto
Joined: 03 Jun 2010, 15:18 Posts: 136

Re: Exercise [16.08] b
The idea was to demonstrate bijectivity of f by finding its inverse g and showing that the compositions fg and gf are the identity; this ensures that both f and g are surjective and injective. In your example gf is not an identity: gf=sqrt(x^2)=abs(x)!=x , and this reveals that g is not surjective (and f not injective). But I understand I was not so much explicit about that, and the free use of auxiliary functions made that even less obvious; so I attach a more explicit version of my solution.
