Determine the range of the function and whether it is one-to-one, onto, both or neither. Then, find all cases where $f(x) = x$ A. $f: \Bbb R\rightarrow \Bbb Z$, defined by the equation $f(x) =\lfloor x \rfloor$
B. $f: \Bbb N\rightarrow \Bbb N$, Where the least significant digit of $x$ is removed. For example, $f(13579) = 1357,\space f(619) = 61,\space f(4) = 0,\space f(0) = 0.$

For A, The function is onto as every number of the range will be matched with at least one number on the domain, but it isn't one-to-one because $f(2) = 2$ and $f(2.1) = 2$.

*

*The range of the function would be $ \Bbb Z$.


*As for the cases where $f(x) = x$, then $\forall x \in \Bbb Z$ such that $f(x) = x$
For B, Since every single digit number will equal $0$, the tens will equal $1$, twenties will equal $2$... etc
We can determine that the function is onto as every number in the range will be matched with at least one number in the domain, but the function isn't one-to-one because $f(0) = 0$ and $f(1) = 0$

*

*The range of the function would be $\Bbb Z$.


*As for the cases, there's only one case: $x = 0$, $f(0) = 0$

This is the answer I ended up with. I am mainly uncertain when it comes to finding all cases where $f(x) = x$. A confirmation/feedback would be highly appreciated. Thank you!
 A: Your answer for A is perfectly alright. Your answer for B however, has some mistakes that need to be addressed.

If $f: \Bbb N\to \Bbb N$, then any single digit natural number will not have an image in $\Bbb N$ ($0 \notin \Bbb N$) and thus the function cannot be defined $\Bbb N\to \Bbb N$. The correct co-domain, therefore must at least be $\Bbb N 
\space \cup \{0\}$.
Now, for $f: \Bbb N\to \Bbb N \space \cup \{0\} $, there is no $x$ that would satisfy $f(x)=x$. To understand why, simply note any number with more than $1$ significant digit will yield a number with $1$ lesser significant digit. These two numbers can obviously not be equal. The single digit numbers on the other hand, yield $0$ which does not even belong to the domain.
If you wish to include $0$ in your domain, you can define $f: \Bbb N \space \cup \{0\} \to \Bbb N \space \cup \{0\} $ which now yields $x=0$ as the only solution to $f(x)=x.$
A: If $f(x)=x$,
$f:N\rightarrow N$ is one to one,
onto
$f:I \rightarrow R$ is one to one, into
$f:R \rightarrow I$ is not a not a map.
Here R is set of all real numbers, I is set of all integers, N is set of all natural numbers.
