I have been tasked with checking if the following functions are injective, surjective or bijective:

(a) $f:\mathbb {N} \to \mathbb {N} ,n \to n^4$

(b) $f:\mathbb {R} \to [−1,1], x \to \sin(x)$

Now, I have no idea how to approach these functions since there are no numbers included. I know that for a function to be injective, for $f(x_1)=f(x_2)$ has to apply $x_1=x_2$. However, I have no idea what to plug here since there is only $n^4$.

My guess for (a) is that it is injective, while on (b) I am completely stumped.

  • $\begingroup$ To show the first function is injective, you must show that if $m^4 = n^4$, then $m = n$, where $m, n \in \mathbb{N}$. $\endgroup$ – N. F. Taussig May 8 at 16:32

For injectivity in part (a), if $f(x_1)=f(x_2)$, then by the definition of $f$, we have $x_1^4=x_2^4$. Does this imply that $x_1=x_2$ (keeping in mind that $x_1$ and $x_2$ are nonnegative integers)? For surjectivity, is it true that any $m \in \mathbb{N}$ satisfies $m=n^4$ for some $n \in \mathbb{N}$?

For injectivity in part (b), if $f(x_1)=f(x_2)$, then $\sin(x_1)=\sin(x_2)$. Does this imply that $x_1=x_2$ (where now $x_1$ and $x_2$ can be any real numbers)? For surjectivity, is it true that for any $y \in [-1,1]$ there is $x \in \mathbb{R}$ such that $\sin(x)=y$?

  • $\begingroup$ I understand what needs to be satisfied, but I do not know how can I check if it's satisfied. Do I plug in any random number from ℕ? When I plug a number in does it have to be the same on the right side as well? $\endgroup$ – Ario May 8 at 17:32
  • $\begingroup$ @Ario These questions are about infinitely many possible values of $x_1$ and $x_2$, so you can't prove them by choosing values of $x_1$ and $x_2$. However, you can disprove them by providing a single counterexample. For injectivity in part (a), do you think it's true that whenever $x_1^4=x_2^4$, then $x_1=x_2$? Could you manipulate the equation $x_1^4=x_2^4$ to show this? If not, can you come up with an example of $x_1$ and $x_2$ such that $x_1^4=x_2^4$ and yet $x_1 \neq x_2$? $\endgroup$ – kccu May 8 at 17:39
  • $\begingroup$ For surjectivity, if you are given an arbitrary $m \in \mathbb{N}$, can you find $n \in \mathbb{N}$ so that $n^4=m$? Could you, for instance, give a formula for $n$ in terms of $m$? If not, can you find an example of $m \in \mathbb{N}$ so that there is no $n \in \mathbb{N}$ that satisfies $n^4=m$? $\endgroup$ – kccu May 8 at 17:41
  • $\begingroup$ So would that make the first function surjective? Since there is always $m$ to satisfy $n^4=m$ as we are working with natural numbers. (Example - $1^4=1$,$2^4=16$,$3^4=81$ and so on..). And it is injective since for every number from the domain there is one in the co-domain. (It would be different if we were working with negative numbers as well since there would be $1^4=(-4)^4=1$) $\endgroup$ – Ario May 9 at 8:59
  • $\begingroup$ The second one is not injective nor bijective. There are multiple numbers from the domain that have the same image in co-domain. Making it non-injective. It is surjective since every output has a image in the domain. Am I correct? $\endgroup$ – Ario May 9 at 9:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.