The following functions all have
the set with elements one, two, three and four,
as both their domain and codomain.
For each, determine whether it is only injective,
only surjective, bijective or neither injective
nor surjective.
For a quick review, a function is injective,
provided every element of the codomain is the image of,
at most one element from the domain.
Injective functions do not have repeated elements
from the codomain in the range
but may miss elements.
And a function is surjective provided every element
of the codomain is the image
of at least one element from the domain.
Surjective functions do not miss elements
from the codomain in the range, but may have repeats.
And a function is bijective
if it is both injective and surjective.
So looking at the first function we want to focus
on the second row, which gives us the outputs
and the codomain, which is the set
with elements one, two, three and four.
So why do I notice all the elements
in the codomain do not appear in the range?
And therefore the function is not surjective.
Surjective functions do not miss elements
from the codomain in the range, but may have repeats.
And the function is also not injective
because injective functions did not have repeated elements
from the codomain in the range, but may miss elements.
So because the element of two from the codomain is repeated,
the function is also not injective.
Number one is neither injective nor surjective.
Now looking at number two, notice all the elements
from the codomain do appear in the range
and therefore the function is surjective.
Surjective functions do not miss elements
from the codomain in the range, but may have repeats.
We don't have repeats, but every element
in the codomain is in the range.
And the given function is also injective
because notice how we do not have repeated elements
from the codomain in the range.
Since number two is both surjective and injective,
it is bijective.
Bijective functions are functions where every element
in the codomain is the image of exactly one element
from the domain, which is what we have here.
For number three and number four,
we'll go ahead to determine the two line notation
for the given functions.
It's much easier to determine the type
of function when we have the function in two line notation.
When x is equal to one we have the opposite value
of negative four, which is four.
When x is equal to two we have the opposite value
of negative three, which is three.
When x is equal to three we have the opposite value
of negative two, which is two.
And when x is equal to four we have the opposite value
of negative one, which is one.
So once again, notice how each element
in the codomain does appear in the range.
And therefore the function is surjective.
Surjective functions do not miss elements
from the codomain in the range.
We can also say every element in the codomain is an image
of at least one element from the domain.
None of the elements from the codomain
are repeated in the range
and therefore the function is also injective.
Or we can say it is inductive
because every element of the codomain is an image
of that most one element from the domain.
And once again, every element in the codomain is the image
of exactly one element from the domain.
And therefore number three is also a bijective.
And now let's write the two line notation for number four.
Where f of x equals x divided by two.
If x is even and f of x equals
the quantity x plus one divided by two, if x is odd.
So when x is equal to two,
we have two divided by two or one.
When x is equal to four,
we have four divided by two, which is two.
That takes care of the even values from the domain.
And now for the odd values from the domain,
when x is equal to one,
notice how we have two divided by two, which is one.
And when x is equal to three
we have three plus 1 which is four,
four divided by two is two.
In this case, notice all the elements
from the codomain are not in the range.
And therefore the function is not surjective,
again surjective functions do not miss elements
from the codomain in the range.
Or we can also say it's not surjective
because every element of the codomain is not an image
of at least one element from the domain.
And the function is also not injective
because notice how the elements
of one and two from the codomain are repeated.
Injective functions do not have repeated elements
from the codomain in the range
or we can say it's not injective
because every element of the codomain is not an image
of at most one element from the domain.
So because the function is not injective and not surjective,
we say it's neither injective nor surjective.
I hope you found this helpful.