Proving the bijection of a function

Question

I am having some trouble understanding why the function $f : R \rightarrow R $ defined by $f(x) =2x$ would be bijective. This is because I cannot understand how can the function be surjective? Wouldn't the function only give us even numbers as outputs? So if odd numbers like $5,7$ are part of the codomain but not of the range then how is the function surjective? Thank you for your help

Answer 1

Note that both your domain and codomain are the real numbers. So for any $y \in \mathbb{R}$, ask yourself for what $x\in\mathbb{R}$ it is the case, that $f(x) = y$.
That is $2x=y$ or $x=\frac{y}{2}$

Answer 2

If the function is mapped from R to R, then for any real X (integer or otherwise), $f^{-1}(x/2)=x$. However, if we map from Z to Z, then you’re right; for odd x, the value x/2 would not be in our domain, so our function could not be surjective.

Answer 3

If $f:\mathbb Z\to\mathbb Z$ is prescribed by $n\mapsto2n$ where $\mathbb Z$ stands for the set of integers then $f$ is not surjective since there is no integer that satisfies $2n=3$.

If $f:\mathbb R\to\mathbb R$ is prescribed by $x\mapsto2x$ where $\mathbb R$ stands for the set of real numbers then $f$ is surjective, since $f(0.5x)=x$ for every $x\in\mathbb R$.