Is $a \mapsto 2a$ bijective?

Question

Why $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x) = 2x$ is not always bijective? I selected as always bijective in a quiz, it said wrong, always injective!

Answer 1

What you selected is correct since you have $A=B=\mathbb R.$

Remember that $f:A\rightarrow B$ is a bijection $\iff$

• injective: $f(x)=f(y)$ implies $x=y$
• surjective: for all $y\in B$ there is some $x\in A$ such that $f(x)=y.$

Clearly your function is injective since $f(x)=f(y) \iff 2x=2y\iff x=y$ and it is also surjective since taking $x=\frac{y}{2}\in A$ we obtain $f(x)=2x=2\frac{y}{2}=y.$ Thus your function is a bijection.

If instead you had $A=B=\mathbb N$, then the function will still be injective, but it will NOT be surjective. This is because in this case there is no $x\in\mathbb N$ such that $f(x)=2n+1$ for some $n\in\mathbb Z_{\geq0}$ for example.

Moreover if you had $\forall x\in \mathbb R$ $f(x)=b$ for some constant $b$ (and $y\in\mathbb R$), then $f(x)$ is neither injective nor surjective.