[seqfan] Re: Please name this function.

[franktaw at netscape.net]

I thought this stuff was better known than it apparently is.

Firstly, Ed did not specify that f must be continuous.

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For the real numbers, a function satisfying these identities is either 
f(x) = 0 or the identity. (The case f(x) = 0 is skipped over by Neil; 
the step f(x*1) = f(x)f(1) => f(1) = 1 should be ... => f(1) = 1 or 
f(x) = 0.) Otherwise follow Neil's argument up to the line starting "f 
is continuous".

Now for x > 0, f(x) = f(sqrt(x))^2 >= 0. Now if a/b < x < c/d, the 
differences x-a/b and c/d-x are positive, so f(a/b) <= f(x) <= f(c/d), 
forcing f(x) = x as we bound x more tightly.

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Charles asserted that f(x+y) = f(x) + f(y) implies f(x) = k*x. This is 
only true assuming f is continuous; otherwise the axiom of choice 
implies that there are (uncountably) many f satisfying this identity.

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For the complex numbers, besides the zero function and the identity, 
the complex conjugate is another homomorphism. It is the only other 
continuous homomorphism on the complex numbers, but - again assuming 
the axiom of choice - there are (uncountably) many discontinuous ones.

Franklin T. Adams-Watters

-----Original Message-----
From: Neil Sloane <njasloane at gmail.com>

Ed, let me expand on my answer. You said x,y need not be integers,
so I assumed your f is a map from the reals to the reals.
Now watch:
f(x+0)=f(x)+f(0) -> f(0)=0
f(x*1)=f(x)f(1) -> f(1)=1
f(x+x)=2f(x)=f(2x)=f(2)f(x)->f(2)=2
Simly f(k)=k for k integer
Simly f(1/k)=1/k
Simly f(j/k)=j/k
f is continuous, so f(x)=x for x real
So f=identity

On Sat, Sep 8, 2012 at 11:57 PM, Ed Jeffery <lejeffery7 at gmail.com> 
wrote:

> Thanks Neil, but the identity map would just be f(x+y) = x+y and 
f(x*y) =
> x*y, which is not what I notated.
>
> > The identity map!
>
> >>* What is the name of a function that is both additive and
> multiplicative but*>>* not arithmetic (in the sense that x and y are
> usually not integers), such*>*> that*>>**>*> f(x+y) = 
f(x)+f(y)*>>**>*>
> and*>>**>*> f(x*y) = f(x)*f(y)?*
>
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>



--
Dear Friends, I have now retired from AT&T. New coordinates:

Neil J. A. Sloane, President, OEIS Foundation
11 South Adelaide Avenue, Highland Park, NJ 08904, USA
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com

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