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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.

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  • $\begingroup$ 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
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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 $\sin(x)=y$?

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  • $\begingroup$ 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

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