Imo 2010-1

[Masum]

Find all functions $f:\mathbb R \to \mathbb R$ such that for all $x,y\in \mathbb R$ $f([x]y)=f(x)f([y])$ where $[a]$ denots the greatest integer less than or equal $a$

[Zzzz]

Ahh...a nice problem that gave me 5 imo points Here is my solution:

We are given $f(\lfloor x\rfloor y)=f(x)\lfloor f(y)\rfloor...(1)$
For $x=y=0$ in $(1),\ f(0)=f(0)\lfloor f(0)\rfloor$
$\Rightarrow f(0)=0$ or $\lfloor f(0)\rfloor=1$
Case 1: $\lfloor f(0)\rfloor=1$
For $y=0$ form $(1)$ we get $f(x)=f(0)$ for any $x$. So $f$ is a constant function.
Let $f(x)=c$. From $(1)$, $c=c\lfloor c\rfloor$
$\Rightarrow c(1-\lfloor c\rfloor)=0$
But, $c\ne 0$ as $\lfloor f(0)\rfloor =1$
So, $\lfloor c\rfloor=1 \Rightarrow c$ is any number between $1$(inclusive) and $2$(exclusive)
We can easily check, this solution satisfies equation $(1)$ (Checking is important for getting full marks )
So, from Case 1, we have got $f(x)=c$ where $1\le c<2$
Case 2: f(0)=0
Let $x$ be a number where $0<x<1$
From $(1)$, $f(0)=f(x)\lfloor f(y)\rfloor$
$\Rightarrow \lfloor f(y) \rfloor f(x)=0$
$\Rightarrow \lfloor f(y)\rfloor =0$ for all $y$ or $f(x)=0$ for $0<x<1$
Case 2.1 ( ) : $\lfloor f(y)\rfloor =0$ for all $y$
From $(1)$, $f(\lfloor x\rfloor y)=0$.
We can right any real number as a product of an integer and a real number. So, $f(z)=0$ for any real $z$.
Case 2.2 (last one ) : $f(x)=0$ for $0<x<1$
Let $y$ be a real number where $0<y<1$ So, $f(y)=0$.
From $(1)$, $f(\lfloor x\rfloor y)=0$
$\Rightarrow f(zy)=0\ [z=\lfloor x \rfloor ] ... (2)$
Remember, Here $z$ can be any integer and $y$ can be any real number between $0$ and $1$.
An interesting fact is, we can write any real number as a product of of an integer and a real number between $0$ and $1$. (Proof: Let $m$ be a positive real. $n$ is an integer greater than $m$. So $0<\frac{m}{n}{}<1$ and $n\cdot \frac {m}{n}=m$. If $m$ is a negative real, take $n$ less than $m$. For $m=0$, more than easy )
So, let $m$ be any real, then $m$ can be written as a product of a real number $r\ (0<r<1)$ and an integer $k$.
$\therefore m=kr$
$f(m)=f(kr)=0$ (from $\ (2)$).

So, the solutions are
$f(x)=0$ or $f(x)=c$ where $1\le c<2$.

[sourav das]

I have a question in Case 2.2 .
If $0<y<1$ then, how can i express the integers (not $0$) in form of \[\left \lfloor x \right \rfloor y\]
[I had this question when i saw the solution in newspaper ]

[*Mahi*]

I have a question in Case 2.2 .
If $0<y<1$ then, how can i express the integers (not $0$) in form of \[\left \lfloor x \right \rfloor y\]
[I had this question when i saw the solution in newspaper ]

Let $\frac x {\left \lfloor x \right \rfloor} = y$
Now,for all real $x$ , $0< y \leq 1$ (I think you can see why)
(Again a thing you should notice, $x,y$ are real numbers , not necessarily integers. )

@Zzzz Vai
2 mark kata gesilo keno?