I have been asked to determine whether-
$h : \mathbb{N} \rightarrow \mathbb{N}$ defined by $h(x) = x^3$ is injective, surjective or bijective. And why?
3 Answers
$\begingroup$The function here is injective but not surjective and therefore not bijective.
To see that it is injective, note that $h(x)$ is strictly increasing on $\mathbb{N}$. So, if $n,m \in \mathbb{N}$ and $n \neq m$ then either $n>m$ or $m>n$. Without loss of generality, assume that $n>m$. Then we have $n^3 > m^3.$ So, if $n \neq m$ then $h(n) \neq h(m).$
To see that it is not surjective, consider $2 \in \mathbb{N}$ and note that there does not exist a natural number $x$ such that $x^3 = 2$.
$\endgroup$ 3 $\begingroup$Make sure you know what the definition of injection, surjection, and bijection are before answering these questions. Note that there are several equivalent definitions of what it means for a function to be invertible, one of which is that it is one of the above three definitions, another is that
$f(a) = f(b)$ implies $a = b$
You can use the given function to directly prove that.
Now as for surjectivity. The $h:\mathbb{N} \rightarrow \mathbb{N}$ restriction is important. Can you think of a natural number that, when cubed, equals 2?
$\endgroup$ $\begingroup$To show a function is injective, you want to show that If $f(x) = f(y)$ then $x = y$
So let $h(x) = h(y)$ Then $x^3 = y^3$ and when we cube root each side we get $x = y$. Therefore it is injective
To show a function is surjective, for any element in the codomain we have to show their is an element in the domain that maps to it. Is there an natural number that when we cube it we get 10? No. So clearly this function is not surective over the natural numbers.
To be bijective it has to be injective and surjective, so we know its not a bijection. Thus it is just an injection.
$\endgroup$ 2
