# Checking if a function is injective, surjective or bijective

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.

• To show the first function is injective, you must show that if $m^4 = n^4$, then $m = n$, where $m, n \in \mathbb{N}$. – 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 $$\sin(x)=y$$?
• @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 $x_1 \neq x_2$? – kccu May 8 at 17:39
• 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 $n^4=m$? – kccu May 8 at 17:41
• 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 $1^4=(-4)^4=1$) – Ario May 9 at 8:59