I understand the method used to prove injectivity or surjectivity, however, I am confused as to how to handle a function that presents itself as a set $(m,n)$.

In the example, $f:Z \rightarrow Z \times Z$, $f(x) = (x^2, x + 1)$

We know that $x^2$ is not injective as we can find for example $f(-1) = f (1) = 1$.

However, we also know that the second part of the equation is injective as we can deduct that:

$x + 1 = y + 1$

$x = y$

The form $(m, n)$ of the function is what is really confusing me.

  1. Would it be correct to conclude that the function is injective since it will produce a set of unique $(m,n)$ values for each $x$?
  2. If so, how would we define such a function to be surjective since there will exists coordinates for which there is no corresponding $x$? Would I be correct to conclude that the function is not surjective?

  3. Also, if a function was in the form $f(x) = z + 5$, would it be correct to conclude that the function is not well defined, and thus we can't determine if it's surjective\injective due to the fact that it's based on a ambiguous variable $z$?

  • 1
    $\begingroup$ Are you sure this is a function from $\mathbb{Z} \to \mathbb{Z}$ and not to $\mathbb{Z}^2$? $\endgroup$ – ImHereSometimes Oct 17 '17 at 21:08
  • $\begingroup$ You are absolutely correct. My apologies, I will edit it! $\endgroup$ – hisoka Oct 17 '17 at 21:08
  • $\begingroup$ All your answers are correct! $\endgroup$ – Bram28 Oct 17 '17 at 21:19

Answering your questions:

  1. You are correct to assume $f$ is injective. A function $f(x)$ is injective iff $f(x) = f(y) \implies x = y$. It is true that $f$ returns two values, but the criterion is the same. If $f(x) = f(y)$, in particular the two second coordinates are the same, but like you said, $x + 1 = y + 1 \iff x = y$;

  2. You are correct to assert $f$ is not surjective, as there is no $x$ such that $f(x) = (-1, 1)$, for example. An example of a surjective function could be a function that goes around in a spiral (for non-negative $x$) covering all points in $\mathbb{Z}^2$. That would be $f(0) = (0, 0), f(1) = (1,0), f(2) = (1,1), f(3) = (0,1), f(4) = (-1, 1), f(5) = (-1, 0), \cdots$, but I don't know how to write that in a clean way. For $x < 0$ we could just take anything you like.

  3. Also right! Defining $f(x) = z+5$ is nonsensical, unless you previously stated that $z$ is some constant. In that case $f$ would be a constant function.

  • 1
    $\begingroup$ How is your $g$ surjective? For what value of $x$ does $g(x)=(0,0)$? $\endgroup$ – G Tony Jacobs Oct 17 '17 at 22:14
  • $\begingroup$ @GTonyJacobs For none, I don't know what I was thinking D: Please check my new example. $\endgroup$ – RGS Oct 17 '17 at 22:15

$$f:\mathbb{Z} \to \mathbb{Z}^2$$

If $f(x)=f(y)$, then we have $(x^2, x+1) =(y^2, y+1)$, hence $x+1=y+1$ of which we can conclude that $x=y$, hence it is injective.

It is not surjective, for example, $(2,3)$ is not in the image set.

As for your last question, what do you mean by $z$? I don't think it is well defined.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.