Is the function injective and surjective?

Question

I want to check if the function $f:\mathbb{N}\times\mathbb{R}\rightarrow \mathbb{R}, \ (x,y)\mapsto \frac{3}{2}y-x$ is injective, surjective, bijective.

I have done the following:

• Let $(x_1, y_1), (x_2,y_2)\in \mathbb{N}\times\mathbb{R}$ with $(x_1, y_1)\neq (x_2,y_2)$. Then we have that $x_1\neq x_2$ or/and $y_1\neq y_2$.

  Therefore we get \begin{equation*}f(x_1, y_1)=\frac{3}{2}y_1-x_1\neq \frac{3}{2}y_2-x_2=f(x_2, y_2)\end{equation*} This means that $f$ is injective. Is this correct?

• Let $z\in \mathbb{R}$. Let $z=f(x,y)$, now we want to calculate $(x,y)$. \begin{equation*}z=f(x,y)\Rightarrow z=\frac{3}{2}y-x\end{equation*}

  How do we continue?

Answer 1

The function is not injective, so it cannot be bijective either. However, the function is surjective.

As J.W. Tanner mentioned, $f(1, 2/3) = f(4, 8/3)$ implies that the function cannot be injective. Your proof is incorrect since you assume injectivity to start with. In order to show injectivity, you need to start with $f(x_1, y_1) = f(x_2, y_2)$ and show that $x_1 = x_2$ and $y_1 = y_2$ hold.

Now, I claim the function $f$ is surjective. In order to show this, we need to show that for every real number $z$, there exists a natural number $x$ and a real number $y$ so that $f(x, y) = z$. There are many ways to do this. I will show you one way.

For any real number $z$, set $x = 1$ and $y = 2(z + 1)/3$ so that we have $x \in \mathbb{N}, y \in \mathbb{R}$. It follows that

$$f(x, y) = \frac{3}{2} \cdot \frac{2}{3}\left(z + 1\right) - 1 = z.$$

Since this holds for any real number $z$, we conclude the function is surjective.

Answer 2

My two cents (mostly to see what other people think of that point of view):

sure the definition of injective is

for every $x_1$ and $x_2$ in the domain of $f$, if $x_1\neq x_2$, then $f(x_1)\neq f(x_2)$

but I would personally recommend using the contraposition

for every $x_1$ and $x_2$ in the domain of $f$, if $f(x_1)=f(x_2)$, then $x_1=x_2$

They are logically equivalent, but equalities are easier to handle than non-equalities (not sure how to call the $\neq$ symbol). Indeed, if $a=b$, then if you perform any operation (that is, a function) on both sides, you obtain a new equality. For example, if $a=b$, then $2a=2b$, $\frac{a}{3}=\frac{b}{3}$, $\ln(a)=\ln(b)$, $\sin(a)=\sin(b)$, $a^2=b^2$,... and more generally $g(a)=g(b)$ for any function $g$. On the contrary, functions don't preserve non-inequalities in general. For example, if $a\neq b$, then you are not guaranteed that $a^2\neq b^2$ or $g(a)\neq g(b)$ in general. This happens precisely when $g$ is injective.

And I believe that any proof that tries to use the first form of the definition will resort to proof by contradiction at some point ("if $x_1\neq x_2$, then (...) so $f(x_1)\neq f(x_2)$. Indeed, if $f(x_1)=f(x_2)$, we would have...(...), so $x_1=x_2$"), so it better to start the proof "the right way" from the start.

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Back to your problem, $f(x_1,y_1)=f(x_2,y_1)$ means that

\begin{equation*}\frac{3}{2}y_1-x_1 = \frac{3}{2}y_2-x_2\end{equation*}

Written that way (a linear equation), it is intuitive that there are solutions with $(x_1,y_1)\neq(x_2,y_2)$, for example $(x_1,y_2)=(2,1)$ and $(x_1,y_2)=(2,\frac{8}{3})$.

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As for the subjectivity, the function is a linear function (*) from a space of dimension two into a space of dimension 1, so intuitively the function is probably subjective. To see it, you can consider partial function, for example, $g(y)=f(1,y)=\frac{3}{2}y-1$.

Since $g$ is a linear function with nonzero slope, its range is $\mathbb R$ so it is surjective. This also hints that $f$ is not injective: by using all possibilities for the first variable $x$, we are bound to have repetitions of the output.

(*) Not exactly since the first variable is an integer.