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 Exercise [16.08] b 
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Joined: 03 Jun 2010, 15:18
Posts: 136
Post 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).


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Exercise_16_08.pdf [50.19 KiB]
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23 Aug 2010, 13:38

Joined: 26 Mar 2010, 04:39
Posts: 109
Post 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).


25 Aug 2010, 06:02

Joined: 03 Jun 2010, 15:18
Posts: 136
Post 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.


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Exercise_16_08 a.pdf [57.2 KiB]
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30 Aug 2010, 10:27
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