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If (f:\mathbb{N}\to\mathbb{N}) is defined by (f(n)=n+1), where (\mathbb{N}={1,2,3,\ldots}), what type of function is (f)?

Answer and explanation

Correct answer: One-one but not onto

Step 1: If (n_1\neq n_2), then (n_1+1\neq n_2+1), so the function is one-one. Step 2: The codomain (\mathbb{N}) contains (1), but there is no (n\in\mathbb{N}) such that (n+1=1). Step 3: Therefore, the function is one-one but not onto.

Related tags

Natural-Number-FunctionOne-OneOntoFunctionsRelations And FunctionsMathematicsClass 12 Mcq

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