is x-[x] defined from all reals to reals ,(where [x] is the greatest integer function ) one-to-one and onto?

Question

dan815 · Answer

@ganeshie8

shadowlegendx · Answer

dan you fanboy c;

dan815 · Answer

explain this  gane

dan815 · Answer

u gota be a fan of gan

ganeshie8 · Answer

http://www.mathwords.com/h/horizontal_line_test.htm

ganeshie8 · Answer

if we restrict the domain to  (n, n+1]
n = natural number 

then, it will be one-to-one

ganeshie8 · Answer

n = integer*

parthkohli · Answer

Recall that$$x - \lfloor x \rfloor = \{x\}$$Where $\{x\}$ is the fractional part of $x$.  Do you think that no two different real numbers have the same fractional part?  Also, do you think that all real numbers can be there in the fractional part of a number?

anonymous · Answer

@parth i think the ans is no for both the question.am i right??

parthkohli · Answer

Yes, if you say that the function is $\mathbb R \to \mathbb R$, then the answer for both should be no.

parthkohli · Answer

The range of your function is only $[0,1)$ which doesn't cover all real numbers.  And $f(2.5) = f(1.5)$ for example.

anonymous · Answer

ohk i get it now thanks a lot to all those who helped me...this is really a very good site