Consider the identity function IN:N→N defined as IN(x)=x∀x∈N. Show that although IN is onto but IN+IN:N→N defined as (IN+IN)(x)=IN(x)+IN(x)=x+x=2x is not onto. $
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Step-by-Step Solution
Step 1: Understand the definition of an onto function
A function is considered 'onto' or surjective if every element in the codomain (the set where the function's outputs lie) is mapped to by at least one element from the domain (the set of inputs). In simpler terms, the range of the function must be equal to its codomain.
Step 2: Analyze the identity function IN(x)=x
The identity function IN(x)=x maps each natural number to itself. For any natural number y in the codomain, we can always find an x in the domain (which is y itself) such that IN(x)=y. Therefore, the range of IN is N, which is equal to its codomain, proving that IN is onto.
Step 3: Analyze the function (IN+IN)(x)=2x
Now, let's consider the function f(x)=(IN+IN)(x)=2x. The domain and codomain for this function are both the set of natural numbers, N={1,2,3,…}.
Step 4: Check if f(x)=2x is onto
To check if f(x)=2x is onto, we need to see if every element in the codomain N can be reached by the function. Let's pick an odd natural number from the codomain, for example, y=1. If f(x)=1, then 2x=1, which means x=21.
Step 5: Conclusion for f(x)=2x
Since x=21 is not a natural number, there is no element in the domain N that maps to 1 in the codomain. This means that odd natural numbers are not in the range of f(x)=2x. Since the range (even natural numbers) is not equal to the codomain (all natural numbers), the function f(x)=2x is not onto.