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Question
Show that the function f : R* → R* defined by f(x) = `1/x` is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true if the domain R* is replaced by N, with the co-domain being the same as R?
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Solution
It is given that f : R* → R* is defined by f(x) = `1/x`
For one-one:
Let x, y ∈ R* such that f(x) = f(y)
⇒ `1/x = 1/y`
⇒ x = y
∴ f is one-one.
For onto:
It is clear that for y ∈ R*, there exists x = `1/y ∈ R_\ast` (as y ≠ 0) such that
f(x) = `1/((1/y))` = y
∴ f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function g: N → R* defined by g(x) = `1/x`
We have,
g(x1) = g(x2)
⇒ `1/x_1 = 1/x_2`
⇒ x1 = x2
∴ g is one-one.
Further, it is clear that g is not onto, as for 1.2 ∈ R*, there does not exist any x in N such that g(x) = `1/1.2`.
Hence, function g is one-one but not onto.
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