Question Video: Describing an Injective Function | Nagwa Question Video: Describing an Injective Function | Nagwa

Question Video: Describing an Injective Function Mathematics • Second Year of Secondary School

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Choose the statement that describes an injective function.

07:02

Video Transcript

Choose the statement that describes an injective function. Option (A) for every element in the domain, there must be at least two corresponding elements in the range. Option (B) for every element in the codomain, there must be a corresponding element in the domain. Option (C) no element in the range can have more than one element in the domain corresponding to it. Option (D) for every element in the domain, there is exactly one element corresponding to it in the range. Or is it option (E) for every element in the range, there is more than one element corresponding to it in the domain and vice versa?

In this question, we’re asked to determine which of five given statements correctly describes an injective function. And the easiest way to do this is to start by recalling the definition of an injective function. We can recall that we call a function an injective function if every element in the range of that function corresponds to exactly one element of the domain. And if we check the five given options, we can see none of the five given options match this word for word. So let’s instead go through each option individually.

Let’s start with option (A). Option (A) tells us that for every element in the domain, there must be at least two corresponding elements in the range. But this presents a problem. This means we can input a value, and we must get two different outputs. In other words, the statement in option (A) does not represent a function. So in particular, it cannot represent an injective function. All functions must have one output per input. In other words, we can’t have multiple elements in the range representing single elements in the domain.

Let’s now move on to option (B). This tells us for every element in the codomain, there must be a corresponding element in the domain. And at first, it might seem that this is the correct answer. However, we need to be careful. We’re not told any information about the number of elements. And the easiest way to see why this does not necessarily represent an injective function is to see an example.

For example, let’s consider the following function 𝑓 defined by this arrow diagram. The domain of this function is the set containing one and two. And we have defined the codomain and the range of this function just to be equal to zero since we’re going to have this function only output zero. And it is worth noting the codomain and range of a function do not need to be the same. The range is the set of all possible outputs, but the codomain is the set containing the range. However, we can choose them to be the same since we’re choosing our function 𝑓.

We can see that our function 𝑓 satisfies statement (B). To do this, we need to check that every element of the codomain of our function has a corresponding element in the domain. And we can just check this from the given diagram. We have one element in the codomain we need to check has a corresponding element in the domain. In fact, it has two: 𝑓 evaluated at one is zero and 𝑓 evaluated at two is zero. However, using the same logic, we can show that this function is not an injective function since injective functions must have each element in the range corresponding to exactly one element of the domain. So because there are two elements which are mapped to zero in our function 𝑓, we know that it’s not an injective function. So the statement in option (B) can’t represent injective functions.

In fact, this represents a different type of function called a surjective function. These are functions with the same codomain and range. In other words, every element in the codomain of the function must have a corresponding element in the domain. So every element in the codomain is mapped onto. But this doesn’t help us answer the question. So let’s now move on to option (C).

Option (C) tells us that no element in the range can have more than one element in the domain corresponding to it. Let’s once again sketch a function diagram using arrows to try and determine what this property is telling us. Let’s start with the domain which is the set of elements one, two, and three and a codomain which is the set one, two, and three.

In statement (C), we’re told that no element in the range can have more than one element in the domain corresponding to it. And we can start by recalling the range of a function is the set of all outputs of the function. In other words, it will be all of the elements in the codomain which have an arrow pointing to them. So let’s say that we want one to be in the range of our function. This means there must be an element in the domain which corresponds to one.

We can choose any element in the domain to be this value. So let’s say that 𝑓 evaluated at one is one. This property now tells us that we cannot have more than one element corresponding to one. So statement (C) tells us we now cannot have 𝑓 evaluated at two is one. And now we can notice the similarity with the definition of an injective function. Every element in the range corresponds to exactly one element of the domain. This is just a slightly different wording of the same statement.

This time, we’re being told for any element in the range of our function, that means there is an arrow pointing to it in the codomain, there is exactly one element in the domain pointing to it. We can’t have two arrows pointing to the same element in the codomain. And so we’ve shown (C) is the correct answer. However, for due diligence, let’s also check options (D) and (E).

Option (D) tells us for every element in the domain, there is exactly one element corresponding to it in the range. We can once again consider this by using an arrow diagram. This time, we’re told that every element in the domain of this function corresponds to exactly one element in the codomain. But we’re not told any information about the uniqueness or number of arrows pointing to elements in the codomain.

For example, we could have that 𝑓 evaluated at one is one and 𝑓 evaluated at two is also one. We can see that every element in the domain of this function has exactly one element corresponding to it in the range. And of course, this is not an injective function since we can see there are two elements which output one: 𝑓 of one is one and 𝑓 of two is one. So option (D) does not represent an injective function.

However, we can recognize this definition as the definition of a function. This statement is just telling us for every input value, we have exactly one output value. This is a function. But it’s not an injective function, so let’s now move on to option (E).

Option (E) tells us for every element in the range, there is more than one element corresponding to it in the domain and vice versa. And although we could construct an arrow diagram representing this relation, we can already use the previous option. In this function, we can see every element in the range of this function corresponds to more than one element in the domain. And this is the first criteria of option (E). However, we’ve already explained that function 𝑓 cannot be an injective function. Just the first part of this statement tells us that this function cannot be injective, so option (E) is not correct.

Therefore, of the five given options, only option (C) describes an injective function. It’s one where no elements in the range can have more than one element in the domain corresponding to it.

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