Krdytkyu

We have this function: $f(x)=2x+1$ where $f:mathbb{R} to mathbb{R}$.

How can I prove that it is bijective?

I still visit school so an explanation for each step would be appreciated.

I know what injective and surjective means and I know that I have to prove both in separate steps, but I don't really understand how this should be done.

As you rightly state, we need to show that $f$ is both injective and surjective.

$textbf{$f$ is injective}$

Recall the definition of an injective function: A function is injective if, whenever $f(x)=f(y)$, it must follow that $x=y$.

Now, suppose $f(x)=2x+1$ is equal to $f(y)=2y+1$, then $$2x+1=2y+1 Rightarrow x=y,$$ so $f$ is injective.

$textbf{$f$ is surjective}$

For $f$ to be surjective, it must follow that every $z in Bbb R$ can be written as $z=f(m)$ for some $m$. We observe that $m=frac{z-1}{2}$ is indeed an element of the real numbers so we can and so $f$ is surjective.

Since $f$ is both injective and surjective, by extension we have that it's bijective as desired.

I want to present an alternative interpretation of what thesmallprint introduced in his answer, because - particularly the injective one - is very helpful to keep in mind.

Recall our function is $f : mathbb{R} to mathbb{R}$ defined by $f(x) = 2x + 1$.

Prove $f$ is injective

By definition, a function $f$ is injective if, whenever $f(x) = f(y)$, we have $x = y$. We can also look at this in the contrapositive since - $f$ is also injective if, whenever $x neq y$, $f(x) neq f(y)$. Sometimes, this angle of approach is easier to look at, since it is basically "unequal inputs have unequal outputs."

This also establishes the notion of injectivity as $f$ passing a horizontal line test: graph the function, and, if at any point you can place a horizontal line through the function and it crosses at $2$ or more points, then it is not injective. You should be able to easily visualize the graph of our given $f$ and confirm it, at least heuristically, in this sense.

Formally, notice that

$$x neq y ;;; Rightarrow ;;; 2x neq 2y ;;; Rightarrow ;;; 2x + 1 neq 2y + 1 ;;; Rightarrow ;;; f(x) neq f(y)$$

Thus, $f$ is injective.

Prove $f$ is surjective

Consider elements $y$ in the codomain $mathbb{R}$. What would the corresponding input $x$ under $f$? Obviously,

$$x = frac{y-1}{2}$$

To each real number $y$, then, you can associate a corresponding $x$ as the input to $f$. Thus, $f$ is surjective.

Prove $f$ is bijective

This follows by definition of bijectivity: as $f$ is injective and surjective, it is bijective.

Another notable tenet: since bijectivity implies the existence of a two-sided inverse, you could also show bijectivity by showing the existence of such an inverse. (Injectivity gives the left-sided inverse, and surjectivity the right. Given the time since the posting of this question, it has probably already been introduced to you in your coursework.) For example, let us write $f$ as

$$y = 2x + 1$$

and then solve for $x$, which yields, as before,

$$x = frac{y-1}{2}$$

Thus, if $f^{-1}(y) = frac{y-1}{2}$, where $f^{-1} : mathbb{R} to mathbb{R}$, if you show

$$(f circ f^{-1})(x) = (f^{-1} circ f)(x) = x$$

then $f$ is bijective. You can probably do this in your head.