Q.Show that the function $f : \mathbf{N} \to \mathbf{N}$ defined by $f(x) = 2x$ is one-one but not onto.

One-one: suppose $f(x_1)=f(x_2)$. Then $2x_1=2x_2 \Rightarrow x_1=x_2$. So $f$ is one-one.

Not onto: the range of $f$ is $\{2,4,6,\dots\}$, the even natural numbers. But the codomain is all of $\mathbf{N}=\{1,2,3,\dots\}$. Take $y=1$: solving $2x=1$ gives $x=0.5\notin\mathbf{N}$ …

Prove injectivity algebraically as usual, then exhibit one specific odd natural number and show solving $f(x)=y$ gives a non-natural ( …

• Forgetting the codomain is stated as $\mathbf{N}$, not the range itself, so onto must be checked against all of $\mathbf{N}$. …