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Question
Let A = {0, 1} and N be the set of natural numbers. Then the mapping f: N → A defined by f(2n – 1) = 0, f(2n) = 1, ∀ n ∈ N, is onto.
Options
True
False
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Solution
This statement is True.
Explanation:
Given, A = {0, 1}
f(2n – 1) = 0, f(2n) = 1, ∀ n ∈ N
Thus range of f is {0, 1}
So, the mapping f: N → A is onto.
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