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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 $\mathbb{Q}$?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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