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Let ๐‘“: โ„• โŸถ โ„• by defined by ๐‘“(๐‘ฅ) - 2๐‘ฅ for all ๐‘ฅ โˆˆ โ„• where โ„• is the set of natural numbers. Showthat ๐‘“ is one - one but not onto function.

Question

Let ๐‘“: โ„• โŸถ โ„• be defined by

๐‘“(๐‘ฅ) = 2๐‘ฅ for all ๐‘ฅ โˆˆ โ„• where โ„• is the set of natural numbers.

Show that ๐‘“ is one-one but not onto function.

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Solution

Sure, let's break this down:

  1. 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 ๐‘“ is one-to-one, let's assume ๐‘“(๐‘Ž) = ๐‘“(๐‘) for some ๐‘Ž, ๐‘ โˆˆ โ„•. This implies 2๐‘Ž = 2๐‘. Dividing both sides by 2, we get ๐‘Ž = ๐‘. Hence, ๐‘Ž and ๐‘ are not different, which means ๐‘“ is one-to-one.

  2. 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 ๐‘“ is not onto, we need to find an element in the co-domain โ„• that does not have a pre-image in the domain โ„•.

    Let's consider the number 1. There is no natural number ๐‘ฅ such that 2๐‘ฅ = 1, because 2๐‘ฅ is always an even number for any natural number ๐‘ฅ. Therefore, 1 does not have a pre-image in the domain, which means ๐‘“ is not onto.

So, the function ๐‘“: โ„• โŸถ โ„• defined by ๐‘“(๐‘ฅ) = 2๐‘ฅ is one-to-one but not onto.

This problem has been solved

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