Find all functions $f:R\rightarrow R$ such that for every two real numbers $x\neq y$ the equality
$f(\dfrac{x+y}{x-y})=\dfrac{f(x)+f(y)}{f(x)-f(y)}$
is satisfied.
FE from real to real
- asif e elahi
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- Joined:Mon Aug 05, 2013 12:36 pm
- Location:Sylhet,Bangladesh
- Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
Re: FE from real to real
I don't have access to a pc currently, so just a sketch. Let $P(x,y)$ be the given equation. $P(x,0)$ implies $f(0)=0,f(1)=1$. $P(xy,y)-P(x,1)$ gives $f(xy)=f(x)f(y)$. So write $P(x,y)$ as $\frac {f(x+y)}{f(x-y)}=\frac {x+y}{x-y}$. Now let this be $Q(x,y)$. Use $P(2,1)P(3,1),P(4,1)P(5,1),f(6)=f(2)f(3)$ to prove $f(2)=2$. Now use induction to prove $f(n)=n$ for all naturals. Extend it to rationals. Use the multiplicative property and $Q(x,y)$ to prove $f$ is increasing. So $f(x)=x$ for all reals.
বড় ভালবাসি তোমায়,মা
- asif e elahi
- Posts:185
- Joined:Mon Aug 05, 2013 12:36 pm
- Location:Sylhet,Bangladesh
Re: FE from real to real
How you got (a) $f(xy)=f(x)f(y)$ and (b) $\frac {f(x+y)}{f(x-y)}=\frac {x+y}{x-y}$ ?Tahmid Hasan wrote:I don't have access to a pc currently, so just a sketch. Let $P(x,y)$ be the given equation. $P(x,0)$ implies $f(0)=0,f(1)=1$. $P(xy,y)-P(x,1)$ gives $f(xy)=f(x)f(y)$. So write $P(x,y)$ as $\frac {f(x+y)}{f(x-y)}=\frac {x+y}{x-y}$. Now let this be $Q(x,y)$. Use $P(2,1)P(3,1),P(4,1)P(5,1),f(6)=f(2)f(3)$ to prove $f(2)=2$. Now use induction to prove $f(n)=n$ for all naturals. Extend it to rationals. Use the multiplicative property and $Q(x,y)$ to prove $f$ is increasing. So $f(x)=x$ for all reals.