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Show that the Signum function f: R->R...

Show that the Signum function `f: R->R` , given by `f(x)={1,\ if\ x >0 0,\ if\ x=0-1,\ if\ x<0` is neither one-one nor onto.

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`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.
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