Saved Bookmarks
| 1. |
Show that the Signumfunction `f: R->R`, given by`f(x)={1, if x >0 0, if x=0-1, if x |
|
Answer» `f: R to R and f(x) = " " f(x) = {{:(1","if x gt 0),(0","if x =0),(-1","if x lt 0","):}` `because " " f(1) = f(2) = 1 ` ` therefore f ` is not one-one. Again `2 in R` and there does not exist `x in R` for which `f(x) = 2`. `therefore f` is not onto. Therefore, `f` is neither one-one nor onto. |
|