# Problem of the week 3

### Problem of the week 3

Find all functions - f: Q -> Q
such that
f(1) = 2 and f(xy) = f(x).f(y) - f(x+y) + 1
for all $x \in Q$, $y \in Q$

Last edited by Math Tutor on Sat Nov 28, 2009 2:09 pm, edited 1 time in total.
Math Tutor

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### Re: Problem of the week 3

teacher wrote:Find all functions - f: Q -> Q
such that
f(1) = 2 and f(xy) = f(x).f(y) - f(x+y) + 1
for all x ∈ Q, y ∈ Q

1) f(1)=f(1)f(1) - f(2) +1, so f(2) =3
Suppose f(n) = n+1. Induction on n: f(n) = f(n)f(1) - f(n+1)+1, f(n+1) = n+2. So f(n)=n+1 n natural
2) f(0)=f(0)n-f(n-1)+1, f(0)=1
3) f(-1)=f(-1)f(1)-f(0)+1, f(-1) = 0
Suppose f(-n) = -n+1, n natural. Induction on n shows f(-n) = -n+1 for all n nat
Up till now f(n) = n+1 for all n integers
4)f(1/p)=2f(1/p)-f(1+1/p)+1, so f(1+1/p)=1+f(1/p). Induction on n: f(n+1/p)=n+f(1/p), n natural
Secondly, for 1+pn and 1/p: f(1/p)+n=(2+pn)f(1/p)-f(1/p)-1-pn+1, f(1/p)=1+1/p
5)f(q/p) = (q+1)(1+1/p)-f(1/p)-q+1, f(q/p)=q/p

so n rational then f(n)=n+1
martin123456

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