Define $h: \Bbb Q\to\Bbb Q$ by $h(x) = 2x + 1$. Is $h$ one-to-one? Why?

Question

Every rational number is different. x2 of every rational num will also be different. Add one to all sorts of different numbers, will yield a different number. Hence, I think it is one-to-one. Where am I wrong?

Answer 1

Your argument is correct.

Note that to show a function is one-to-one, normally we assume $f(x)=f(y)$ and show that $x=y.$

In this case$$2x+1=2y+1 \implies x=y$$ therefore it is one-to-one.

Intuitively you can see that your function is strictly increasing so it is one-to-one.

Or you can look at the graph which is a linear dust and it is one-to-one.

Answer 2

Hint Recall the definition of one to one function: Let $x,y\in\Bbb Q$ such that $h(x)=h(y)$. Can you prove that $x=y$ using simple algebra?

Answer 3

Let $y \in \mathbb{Q}$. Then:
$$ y=2x+1 \implies x=\frac{y-1}{2} \in\mathbb{Q} $$ Hence, $Im(f)=\mathbb{Q}$ and $f$ is surjective. Let $x_1,x_2 \in \mathbb{Q}, x_1 \neq x_2$. Then: $$ f(x_1)=f(x_2) \implies \\ 2x_1+1=2x_2+1 \implies \\ 2(x_1-x_2)=0 \implies \\ x_1=x_2 $$ There exists no different $x_1, x_2 \in \mathbb{Q}$ such that $f(x_1)=f(x_2)$. Therefore, the function is injective.
Since it is surjective and injective, it is bijective (also called one-to-one).