Prove that $f$ is onto

Question

Let $X=\{55n\mid n\in\Bbb N\}$ and let $f$ be a function $f: X \rightarrow X$ defined as $f(n)=55n$

a) Prove that $f$ is one-to-one

b) Prove that $f$ is onto

a) $55x = 55y$

$$x = y$$. Hence proven it is one-to-one

b) For all $y$, there exist at least an $n\in X$ so that: $f(n)=y$.

I know what an onto function is but I'm having hard time figuring out how to do it. I maybe overthinking this. If anyone can help me if would be greatly appreciated.

Answer 1

Assuming $X=\{55\cdot n\mid n\in\mathbb{N}\}$, it is true that $f:X\rightarrow X$ is injective, it is however not onto, since there is no $x\in X$ such that $f(x)=55$.

This is because if $x\in X$ then $x=55n$, for some $n\in\mathbb{N}$, and so we need to solve $$f(x)=55\cdot 55\cdot n=55$$ $$55n=1$$ which has no solutions.