Let 𝑓: ℕ ⟶ ℕ by defined by 𝑓(𝑥) - 2𝑥 for all 𝑥 ∈ ℕ where ℕ is the set of natural numbers. Showthat 𝑓 is one - one but not onto function.
Question
Let 𝑓: ℕ ⟶ ℕ be defined by
𝑓(𝑥) = 2𝑥 for all 𝑥 ∈ ℕ where ℕ is the set of natural numbers.
Show that 𝑓 is one-one but not onto function.
Solution
Sure, let's break this down:
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One-to-One (Injective) Function: A function is said to be one-to-one (or injective), if the pre-image of every element in the range is unique. In other words, if different elements in the domain have different images in the co-domain.
To prove that 𝑓 is one-to-one, let's assume 𝑓(𝑎) = 𝑓(𝑏) for some 𝑎, 𝑏 ∈ ℕ. This implies 2𝑎 = 2𝑏. Dividing both sides by 2, we get 𝑎 = 𝑏. Hence, 𝑎 and 𝑏 are not different, which means 𝑓 is one-to-one.
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Onto (Surjective) Function: A function is said to be onto (or surjective), if every element in the co-domain has a pre-image in the domain.
To prove that 𝑓 is not onto, we need to find an element in the co-domain ℕ that does not have a pre-image in the domain ℕ.
Let's consider the number 1. There is no natural number 𝑥 such that 2𝑥 = 1, because 2𝑥 is always an even number for any natural number 𝑥. Therefore, 1 does not have a pre-image in the domain, which means 𝑓 is not onto.
So, the function 𝑓: ℕ ⟶ ℕ defined by 𝑓(𝑥) = 2𝑥 is one-to-one but not onto.
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