If one is given the following:

$$A = \{(x, y)\mid x \in \mathbb{R}, y \in \mathbb{Z}, y = \lceil x \rceil\},$$ a relation from $\mathbb{R}$ to $\mathbb{Z}$.

How would I be able to tell whether or not it is injective or surjective?

I checked if it was a function, which i think it is. However I do not know how to proceed from here.

  • $\begingroup$ Suggestion for injective: Do you know the definition? Try some values. When $x = 0.5$ what is $y$? When $x = 0.75$ what is $y$? $\endgroup$ Oct 30, 2014 at 14:41
  • $\begingroup$ I thought injective since it is just line but I just needed verfication $\endgroup$
    – dreamin
    Oct 30, 2014 at 14:43
  • $\begingroup$ Think a little bit more about injective. What is the definition of injective? For surejective, can you find something mapping to $n \in \mathbb{Z}$? $\endgroup$
    – Jacob Bond
    Oct 30, 2014 at 14:44

1 Answer 1


To see if it is surjective, simply check if every element $y\in\mathbb Z$ can appear in $A$. This means:

  1. Take an arbitrary $y\in \mathbb Z$
  2. Find such an $x\in \mathbb R$ that $(x,y)\in A$.

On the other hand, if you want to prove a function is not surjective, simply find one particular value of $y$ such that $(x,y)$ is not in $A$ for any value $x$.

For injectivity, if you want to prove injectivity, take two pairs $(x_1, y_1)$ and $(x_2, y_2)$ such that $y_1=y_2$. If you can conclude that $x_1=x_2$, then the function is injective.

If you want to prove that the function is not injective, simply find two values of $x_1,x_2$ and one value of $y$ such that $(x_1,y)$ and $(x_2,y)$ are both in $A$.


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