## Can you be injective but not surjective?

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Injective, but not surjective; there is no n for which f(n)=3/4, for example. (4) In each part, find a function f : N → N that has the desired properties. (a) Surjective, but not injective One possible answer is f(n) = L n + 1 2 C, where LxC is the floor or “round down” function.

### Can a function be both not injective and not surjective?

An example of a function which is neither injective, nor surjective, is the constant function f : N → N where f(x) = 1. An example of a function which is both injective and surjective is the iden- tity function f : N → N where f(x) = x.

**Is 2x 1 injective or surjective?**

The answer is “It depends.” If f:R→R then the function is both surjective and injective. For every x∈R we have f(12(x−1))=2(12(x−1))+1=(x−1)+1=x. Thus f is surjective.

**Is injective always surjective?**

An injective function need not be surjective (not all elements of the codomain may be associated with arguments), and a surjective function need not be injective (some images may be associated with more than one argument).

## How do you show not surjective?

not surjective. To show a function is not surjective we must show f(A) = B. Since a well-defined function must have f(A) ⊆ B, we should show B ⊆ f(A). Thus to show a function is not surjective it is enough to find an element in the codomain that is not the image of any element of the domain.

### How do you know if a function is injective or surjective?

If the codomain of a function is also its range, then the function is onto or surjective. If a function does not map two different elements in the domain to the same element in the range, it is one-to-one or injective.

**Is 2x injective?**

For instance, f(x)=2x from Z to Z is injective.

**Is 2x 1 a Bijective function?**

Originally Answered: Is f(x) =2x+1 is injective or surjective? Assuming that the domain of x is R, the function is Bijective. i.e it is both injective and surjective.

## How do you prove a function is not surjective?

To show a function is not surjective we must show f(A) = B. Since a well-defined function must have f(A) ⊆ B, we should show B ⊆ f(A). Thus to show a function is not surjective it is enough to find an element in the codomain that is not the image of any element of the domain.