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Question
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
Solution
f: R → R is given by,
`f(x) = {(1, if x > 0), (0, if x = 0), (-1, if x < 0):}`
It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
∴ f is not one-one.
Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R, there does not exist any x in domain R such that f(x) = −2.
∴ f is not onto.
Hence, the signum function is neither one-one nor onto.
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