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 1)  Assuming f finite everywhere, (*) <code></code>> x<>y <code></code>> f(x)<>f(y)

 2)  Exchanging x and y in (*) we see that f(-x) = -f(x).

 3)  a <> 0 <code></code>> f((a-a)/(a+a)) <code> (f(a)-f(a))/(f(a)+f(a)) </code><code>> f(0) </code> 0.

 4)  a <> 0 <code></code>> f((a+0)/(a-0)) <code> f(a)/f(a) </code><code>> f(1) </code> 1.

 5)  x<>y, y<>0 <code></code>> f(x/y) <code> f( ((x+y)/(x-y) + (x-y)/(x-y)) / ((x+y)/(x-y) - (x-y)/(x-y)) </code> f(x)/f(y) <code></code>> f(xy) <code> f(x)f(y) by replacing x with xy and by noting that f(x'''1) </code> f(x)'''1 and f(x'''0) = f(x)'''f(0).

 6)  f(x'''x) <code> f(x)'''f(x) </code>=> f(x) > 0 if x>0.

 7)  Let a<code>1+/2, b</code>1-/2; a,b satisfy (x+1)/(x-1) <code> x </code><code>> f(x) </code> (f(x)+1)/(f(x)-1) <code></code>> f(a)<code>a, f(b)</code>b.  f(1//2) <code> f((a+b)/(a-b))</code> (a+b)/(a-b) <code> 1//2 </code><code>> f(2) </code> 2.

 8)  By induction and the relation f((n+1)/(n-1)) = (f(n)+1)/(f(n)-1)
we get that f(n)=n for all integer n.  #5 now implies that f fixes
the rationals.

 9)  If x>y>0 (*) <code></code>> f(x) - f(y) = f(x+y)/f((x+y)/(x-y)) > 0 by #6.
Thus f is order-preserving.

Since f fixes the rationals '''and''' f is order-preserving, f must be the
identity function.

This was E2176 in "The American Mathematical Monthly" (the proposer was
R. S. Luthar).
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