# Prove that $f:\mathbb{R}\times\mathbb{Z}\rightarrow\mathbb{R}$, $f(x,y)=\frac{2}{\pi}\arctan x+2y$ is injective

Recently in my discrete mathematics class we defined an injective function $$f:A\hookrightarrow B$$ in the following way:

A function $$f:A\rightarrow B$$ is injective if for all $$a,b\in A$$, $$a\neq b\rightarrow f(a)\neq f(b)$$.

or equivalently, $$f(a)=f(b)\rightarrow a=b$$. I attempted to extend this definition to multivariable functions:

A function $$f:A_1\times A_2\times\dots\times A_m\rightarrow B_1\times B_2\times\dots\times B_n$$ is injective if for all tuples $$(x_1,x_2,\dots,x_m),(y_1,y_2,\dots,y_m)\in A_1\times A_2\times\dots\times A_m$$, $$f(x_1,x_2,\dots,x_m)=f(y_1,y_2,\dots,y_m)\rightarrow (x_1,x_2,\dots,x_m)=(y_1,y_2,\dots,y_m).$$

Using this definition, I attempted to construct some examples of injective multivariable functions. I haven't yet come up with any functions $$f:\mathbb{R}^2\hookrightarrow\mathbb{R}$$, but I believe that $$f:\mathbb{R}\times\mathbb{Z}\hookrightarrow\mathbb{R}$$, $$f(x,y)=\frac{2}{\pi}\arctan x+2y$$ is an injective function based on the above definition. My reasoning is that since $$y=\arctan(x)$$ is injective and bounded below by $$y=-\frac{\pi}{2}$$ and above by $$y=\frac{\pi}{2}$$, the $$\frac{2}{\pi}\arctan(x)$$ part of $$f$$ only can change the value of $$f$$ by any amount between $$-1$$ and $$1$$. The $$2y$$ part, then, changes the "level" of the function, so when $$y=0$$, $$f$$ takes on values between $$-1$$ and $$1$$; when $$y=1$$, $$f$$ takes on values between $$1$$ and $$3$$; etc.

My difficulty is in proving that such a function is actually injective. The method we've used in class, i.e. manipulating the equation $$f(a)=f(b)$$ to show that $$a=b$$, doesn't easily work in this case, since it requires going from single values to tuples of values. I'm looking for a method to show that this function is injective (or for a counterexample if my example doesn't work! But preferably for a method to solve these kinds of problems). My goal is eventually to extend this simpler problem to construct a function $$f:\mathbb{R}^2\hookrightarrow\mathbb{R}$$, but I wanted to start with one that was easier to visualize first. With this in mind, is there a good method to prove that this function is injective?

Assume $$\frac{2}{\pi}\arctan(x) + 2y = \frac{2}{\pi}\arctan(x')+2y'$$.

Then $$2(y-y') = \frac{2}{\pi}(\arctan(x')-\arctan(x))$$.

What do you know about the left hand side? How many options are there for the right hand side then?

• I see how it follows that $x=x'$ if the left-hand side is zero, i.e. $y=y'$. How do we show that $y=y'$ though? Mar 4, 2019 at 22:35
• Can the left hand side be anything other than zero? Think about the bounds on the right hand side you could write down based on stuff you've already discussed in your question. Mar 4, 2019 at 22:36
• (+ what are y and y')? Mar 4, 2019 at 22:39
• So $-1<\frac{1}{\pi}\big(\arctan(x')-\arctan(x)\big)<1$. Then $-1<y-y'<1$, and since $y,y'\in\mathbb{Z}$, $y-y'$ must be zero. Is that correct? Mar 4, 2019 at 22:41
• Exactly. This probably isn't as general of a method as you were hoping for, but the same kinds of "play around with the equality" tricks can work. Mar 4, 2019 at 22:47

Let $$g(x,y)=f(x,y)/2$$. Proving $$f$$ is injective is the same as proving $$g$$ is injective.

You can note that $$-\frac{\pi}{2}<\arctan x<\frac{\pi}{2}$$ so that $$-\frac{1}{2}<\frac{1}{\pi}\arctan x<\frac{1}{2}$$ and therefore $$y-\frac{1}{2} Hence, $$y_1 implies $$y_1-\frac{1}{2} and, in particular, $$g(x_1,y_1)\ne g(x_2,y_2)$$, for every $$x_1$$ and $$x_2$$.

Thus $$g(x_1,y_2)=g(x_2,y_2)$$ implies $$y_1=y_2$$ and you can finish up.