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Show that f: R->R , given by f(x)=x-[x] ...

Show that `f: R->R` , given by `f(x)=x-[x]` , is neither one-one nor onto.

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We have given
`f(x)=x-[x]`
Now check one -one:
`f(x)=0 `for all `x in Z`
Hence, f is a many one function.
Now check onto:
Range`(f) =[0,1) ne R`
So, f is an into function.
...
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