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Question
Show that the modulus function f : R → R given by f(x) = |x| is neither one-one nor onto, where |x| is x if x is positive or 0 and |x| is − x if x is negative.
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Solution
f : R → R is given by,
f(x) = |x| = `{(x", if" x ≥ 0), (-x", if" x < 0):}`
It is seen that f(−1) = |−1| = 1, f(1) = |1| = 1
∴ f(−1) = f(1), but −1 ≠ 1.
∴ f is not one-one.
Now, consider −1 ∈ R.
It is known that f(x) = |x| is always non-negative. Hence, there is no element −1 in the codomain R that can get the value of f(x).
∴ f is not onto.
Hence, the modulus function is neither one-one nor onto.
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