3

Solve in the set of real numbers the equation :$[x]\cdot\{x\}=2007x$, where $[x]$ is the whole part of x and $\{x\}$ is the fractional part of x

First thing to mention is that $\{x\}\in(0,1)$, we can simply check if the equation when verifies when $x=0$

I tried writing $x=[x]+\{x\}$ and replacing it in the equation. But I got to nothing useful. Hope one of you can help me!Thank you!

emacs drives me nuts
  • 10,390
  • 2
  • 12
  • 31
IONELA BUCIU
  • 1,157
  • 1
  • 17
  • Although the symbol $$ is often used by computer languages for multiplication, in mathematics is is more common to use $$ for convolution. – GEdgar Mar 17 '24 at 17:32

2 Answers2

2

Let $\lfloor x \rfloor = m$ and $\{x\} = t$. Then your equation says $m t = 2007 (m + t)$. Rewrite it as $$ (m - 2007)(t - 2007) = 2007^2$$

Now $-2006 > t - 2007 \ge -2007$, so $m - 2007$ (which is an integer) is between $2007^2/(-2006) = -2008.00049\ldots$ and $2007^2/(-2007) = -2007$. The possibilities are $m - 2007 = -2008$ and $m - 2007 = -2007$, i.e. $m = -1$ (and then $t = 2007/2008$ so $x = -1/2008$) or $m = 0$ (and $t = 0$ so $x=0$).

Note: this is using the definition $\{x\} = x - \lfloor x \rfloor$. Some people define the fractional part of a negative number $x$ as $-\{-x\}$.

D S
  • 4,961
Robert Israel
  • 448,999
0

L.H.S $$f(x)=[x]*\{x\} \le y=x << 2007x$$ so there is no root but $x=0$ to make it clear $$0\le x <1 \to 0*\{x\} =2007x \to x=0\\ 1\le x <2 \to 1*\{x\}=2007x \\ 0\le LHS <1 ,2007 \le RHS=2007x <4014 \to \text{no-solution}$$ and do it for $x<0$ $$-1\ \le x <0 \to \ [x]*\{x\}=2007x \\(-1)*\{x\}=2007x \\(-1)*(x-[x])=2007x\\-1(x+1)=2007x\\x=\frac {-1}{2008} \ \checkmark$$ if you check for $x<-1$ there is no solution . for example $$-2\le x<-1 \to [x]*\{x\}=2007x \\(-2)*\{x\}=2007x\\ -2< LHS \le 0 \\-4007\le RHS <-2007 \\no-solution $$so there is two solution $x=0,\frac {-1}{2008}$

Khosrotash
  • 24,922