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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.
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Solution
f : R → R, 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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