Define a function $f\colon\mathbb{R}\to\mathbb{R}$ which satisfies
$$f(xy)=f(x)f(y)-f(x+y)+1$$ for all $x,y\in\mathbb Q$. With a supp condition $f(1)=2$. (I didn't notice that.)

How to show that $f(x)=x+1$ for all $x$ that belong to $\mathbb{Q}$?

  • $\begingroup$ This is homework, right? Starting point: show that $f(0) = 1$. Try to continue with integers, then with rationals of the form $1\over q$, finish with $\mathbb Q$. $\endgroup$ – Elvis Jan 4 '12 at 10:56
  • 3
    $\begingroup$ This is not precisely true. The function $f$ which is identically equal to $1$ satisfies your functional equation. $\endgroup$ – André Nicolas Jan 4 '12 at 11:19
  • $\begingroup$ This question appeared in our Dhaka Regional MO yesterday.. :O Searching MSE give me this ... There was a special condition $f(2017) \not = f(2018)$ which excludes $f(x)=1$.. And asked to find $f(2017/2018)$ :3 How they use an old problem in MO :3 :| $\endgroup$ – Rezwan Arefin Jan 21 '17 at 18:15

Suppose $$\tag{1}f(xy)=f(x)f(y)-f(x+y)+1.$$ Put $x=y=0$ in $(1)$, we have $f(0)=f(0)^2-f(0)+1$, which implies that $f(0)^2-2f(0)+1=0$, or $(f(0)-1)^2=0$, i.e. $f(0)=1$.

Put $y=-1$ and $x=1$ in $(1)$ we have $$f(-1)=f(1)f(-1)=2f(-1),$$ which implies that $f(-1)=0$.

Now taking $y=1$ in $(1)$, we have $$f(x)=f(x)f(1)-f(x+1)+1=2f(x)-f(x+1)+1,$$ which gives $$\tag{2}f(x)=f(x+1)-1.$$ Since $f(1)=2$, by using $(2)$ and induction, $f(x)=1+x$ for all positive integers $x$. Since $f(-1)=0$, by using $(2)$ and induction again, $f(x)=1+x$ for all negative integers $x$.

Note that $(2)$ implies that $$\tag{3}f(x+q)=f(x)+q.$$ for any integer $q$ and for all $x$. Finally for any rational number $p/q$ where $p,q$ are integers, put $x=p/q$ and $y=q$ in $(1)$, we get $$\tag{4} f(p)=f(p/q)f(q)-f(p/q+q)+1=f(p/q)f(q)-[f(p/q)+q]+1$$ where we have used $(3)$ in the last equality. Since $f(x)=1+x$ for all integers $x$, it follows from $(4)$ that $$1+p=f(p/q)(q+1)-f(p/q)-q+1$$ which implies that $f(p/q)=1+p/q$, as required.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.