# Study the injectivity and surjectivity of the function f

Let $$f:\mathbb{R}\to\mathbb{R}$$, where $$f(x) = \begin{cases} 2x+1, & \text{if x is rational} \\ \sqrt2 x+3, & \text{if x is irrational} \end{cases}$$

The injectivity:

I worked out the injectivity of the function by finding a rational number $$x_1$$ and an irrational number $$x_2$$ such that $$f(x_1)=f(x_2)$$, but $$x_1!=x_2$$, which are $$x_1=2, f(2)=2\cdot 2+1=5$$ and $$x_2=\sqrt{2}$$, $$f(\sqrt{2})=\sqrt{2\cdot 2}+3=5,f(2)=f(\sqrt{2})$$ and $$2!=\sqrt{2}$$ which means that it's not injective.

The surjectivity:

If $$x$$ is rational $$\implies y=2x+1\implies x=\frac{y-1}{2}$$, and if $$x$$ is irrational $$x=\frac{y-3}{\sqrt{2}}$$;

I don't really know how to show that the function is surjective, but I know that when $$y=2x+1$$(the first equation), it will touch all the rational numbers, but I'm not sure about the second equation if there's an y it doesn't touch and if it doens't then it means that it's not surjective...

I was wondering if there is a general way to study the injectivity and surjectivity of these types of functions. Thanks for the help.

Your proof that $$f$$ is not injective is correct.
The given function $$f$$ is not surjective. Consider the irrational number $$\sqrt{2}+3$$. Since $$2x+1$$ is rational for any rational number $$x$$, we should find an irrational number $$x$$ such that $$\sqrt2 x+3=\sqrt{2}+3$$ which holds only if $$x=1$$ which is rational. Contradiction.
More generally, in the same way, we show that $$f(\mathbb{R})$$ does not include all irrational numbers of the form $$\sqrt{2}q+3$$ where $$q$$ is any rational number different from zero. As you already noted $$f(\mathbb{R})\supset \mathbb{Q}$$.
• Minor nitpick (very, very minor). You assume to solve $f(x) = \sqrt 2 +3$ that the solutions $x$ must be irrational. What if the solution $x$ is rational? (The answer is, of course ludicrously easy, that that could only occur if $x = \frac{\sqrt 2+2}2$ is rational, which it isn't). Dec 21, 2021 at 15:27