There are 2 operations: $Z$ and $Y$.
If$ Z$ is included as an operation, it adds one. This means $7+Z+Z+Z$ would essentially, be $7+1+1+1$ and would, therefore, equal $10$.
If $Y$ is included as an operation, it would turn a number into its negative reciprocal. This means three plus $Y$ would equal $-1/3$ and $-3/1$ plus $Y$ would equal $1/3$.
Here, a proper example is shown:
$$0 \ \xrightarrow{Z}\ 1 \ \xrightarrow{Z}\ 2 \ \xrightarrow{Z} \ 3 \ \xrightarrow{Y}\ -\frac{1}{3}\ \xrightarrow{Z}\ \frac{2}{3}\ \xrightarrow{Z}\ \frac{5}{3}\xrightarrow{Y}\ -\frac{3}{5}\ \xrightarrow{Z} \frac{2}{5}$$
Using such operations, how and why is it possible that any positive integer $t$ can be turned to zero using $3t-1$ operations? Also, how and why is it possible for zero to be turned (using the operations) to any negative integer? An explanation to this would be extremely helpful as I need to understand why and how this occurs before progressing to the other questions.
Thanks :)
HINT
Use induction, and show that for any $t$ you can transform $t+1$ in $4$ operations into $-\frac{1}{t}$, which is the result of the first ($Y$) operation you do to transform any $t$ into $0$ in $3t-1$ steps.
EDIT
If you are not familiar with induction, note that any number $-\frac{1}{t}$ with $t$ a whole number can be transformed into $-\frac{1}{t-1}$ by the sequence of $Z,Y,Z$
So, starting with $t$, do one $Y$ operation to get $-\frac{1}{t}$, then do $t-1$ sequences of $Z,Y,Z$ to get $-\frac{1}{t-(t-1)}=-\frac{1}{1}=-1$, and then do one $Z$ operator to get to $0$, for a total of $1+3(t-1)+1=3t-1$ operations.