Author:  Roberto [ 23 Aug 2010, 13:38 ]
Post subject:  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).

 Attachments: Exercise_16_08.pdf [50.19 KiB] Downloaded 137 times

 Author: robin [ 25 Aug 2010, 06:02 ] Post subject: 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 1-1.For example f: R -> R+ given by f(x) = x^2 is not 1-1 even though it has an inverse g: R+ -> R given by g(y) = sqrt(y) (which is not surjective).

Author:  Roberto [ 30 Aug 2010, 10:27 ]
Post subject:  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.

 Attachments: Exercise_16_08 a.pdf [57.2 KiB] Downloaded 125 times

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