Surjectivity $f:\mathbb Z \to \mathbb Z$, $f(x) = 5x$. Is the function $f(x)=5x$ surjective if $f:\mathbb{Z}\to \mathbb{Z}$?
I believe it is not as $f\left(\dfrac{x}{5}\right) = x$ can be rational, not an integer. Could someone confirm this?
Thank you.
 A: A function $f: \mathbb{Z} \to \mathbb{Z}$ is surjective if for any $n\in \mathbb{Z}$, you can find a $m\in \mathbb{Z}$ such that $f(m) = n$. That is, you want to be able to get any given integer as output of your function.
Now, your function $f(x) = 5x$, could that ever be equal to $7$? That is, can you find an integer $m$ such that $f(m) = 5m = 7$?
A: A function $h: X \to Y$ is surjective if and only if for any $y \in Y$, there exists an $x \in X$ such that $h(x) = y$. 
In your case, $X = Y = \mathbb Z$, and $n \mapsto 5n$. Is there some integer $m \in \mathbb Z$ such that there is no  $n$ such that $5n = m$? Yes: For example, $3$ is an integer, let's call it $m$. Is there any integer $n$ that gives us $5n = 3$?
No, there isn't any such integer $n$: so the function $f:\mathbb Z \to \mathbb Z,\;f(n) = 5n$ is not surjective.
Alternatively, the image of $f:\mathbb Z \to \mathbb Z$, $f(x) = 5x\;$  is given by $\;\{5x\mid x\in \mathbb Z\} \subset \mathbb Z$. 
$$\text{Is}\;\;\{5x\mid x\in \mathbb Z\} = \{\ldots, -10, -5, 0, 5, 10, \ldots\} = \{\ldots, -2, -1, 0, 1, 2, \ldots\} = \mathbb Z\quad ?$$
If not, then $f$ cannot be surjective.
A: You seem to be rather confused by the answers you are getting.
You have a function $f:\mathbb Z \to \mathbb Z$.
Your suggestion is that this cannot be surjective, because $f(\frac n5)=n$ and for some $n$, and $1, 6, 7$ are examples, $\frac n5$ is a rational number, not an integer.
The problem with this as a proof is that you can't apply $f$ to a non-integral rational number, because it is only defined on the integers. It can be extended, of course, to a function on the rationals, reals and complex numbers - but that would strictly be a different function from the one you have originally been given.
Some respondents have made a technical repair to the proof, by identifying integers $n$ not divisible by $5$ and showing (or suggesting) that they are not in the image of $f$ ie that $f^{-1}(n)$ does not exist. This is a more accurate approach than saying (the equivalent to what you have put) that $f^{-1}(n)=\frac n5$ is not an integer.
A: The function $f:\mathbb{Z}\rightarrow \mathbb{Z}$ with $f(x)=5$ is not surjective because there is no $a \in \mathbb{Z}$ such that $f(a) = 6$. 
A: Hint: Is there an integer $x$ satisfying $5x=1$?
A: So here is an answer to your question. For a correct result to your exercise there are already enough answers.
since the domain of $f$ is $\mathbb{Z}$, for $x/5$ to be an integer, $x$ has to be a  an integer multiple of $5$ so the value of $f(x/5)$ will still be an integer and not a rational number. So your assumption is not correct.
A: Surjection is possible only when Range = Codomain
So only focus on the output(Range) . . 
$f(x)=5x$ I should get ALL integers as output
 If however i happen to say that there is atleast one integer in the codomain which does not come in the set of my range then it is enough contradiction for me to say that function is not surjective .
Now take 2 for example , it can never occur in range of $ f(x)=5x$ for integer inputs .
This is enough evidence for me to say that the function is NOT SURJECTIVE .
