shown that f is even

Question

lets one function $f:\mathbb{R}-\{0\}\mapsto\mathbb{R}$ where $\mathbb{R}$ is the set of reals, lets $f$ such that $f\left(\frac{a}{b}\right)=f(a)-f(b)$ for every $a$ and $b$ in belonging to the domain of f, shown that $f$ is a even function.

attempt

for $(a,b)=(x,y)\Rightarrow f\left(\frac{x}{y}\right)=f(x)-f(y)$

for $(a,b)=(-x,-y)\Rightarrow f\left(\frac{x}{y}\right)=f(-x)-f(-y)$

for $(a,b)=(-x,y)\Rightarrow f\left(-\frac{x}{y}\right)=f(-x)-f(y)$

for $(a,b)=(x,-y)\Rightarrow f\left(-\frac{x}{y}\right)=f(x)-f(-y)$

so we want to proof that $f(a)=f(-a)$, we have

$$\begin{align} f\left(\frac{x}{y}\right)&=f(x)-f(y)=f(-x)-f(-y)\\ f\left(-\frac{x}{y}\right)&=f(-x)-f(y)=f(x)-f(-y) \end{align}$$

but i dont know how i can finish the proof, any hint?

Answer 1

Hint:

• First, prove that $f(1)=0$ (writing $1=\frac11$).
• Then, prove that $f(-1)=0$ (writing $-1=\frac{-1}1$ and $-1=\frac1{-1}$).
• Then, using this prove that $f(-x)=f(x)$ (writing $x-=\frac{x}{-1}$) for any $x\neq0$.

Answer 2

Subtract your equations to obtain $f(x/y)-f(-x/y)=f(x)-f(-x)=f(-x)-f(x)$