Problem Involving Operations 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 is an example:

Using such operations, how and why is it possible for zero to be turned to any negative integer? I have already found a pattern for me to turn 0 into any negative integer. For example, to turn 0 to -1, I would use Z, Y. To turn 0 to -2, I would use Z, Z, Y, Z, Y. To turn 0 to -3, I would use Z, Z, Y, Z, Z, Y, Z, Y. I found that you could just add (Z, Z, Y) to the beginning of a sequence turning 0 into n to turn 0 into n - 1. Except I do not know how to explain this. Help would be appreciated :)
 A: You made the following observations:
$ZY$ produces the sequence $(0, 1, -1)$.
$ZZYZY$ produces the sequence $(0, 1, 2, -1/2, 1/2, 2)$.
$ZZYZZYZY$ produces the sequence $(0, 1, 2, -1/2, 1/2, 3/2, -2/3, 1/3, -3)$.
You asserted that appending ZZY to the beginning of a sequence that produces the integer $-n$ will produce the integer $-(n + 1)$.  However, it is better to think of it this way:  Insert ZZY just before ZY.  Let's look at the next sequence.
$ZZYZZY\color{blue}{ZZY}ZY$ produces the sequence $(0, 1, 2, -1/2, 1/2, 3/2, -2/3, \color{blue}{1/3}, \color{blue}{4/3}, \color{blue}{-3/4}, 1/4, -4)$.
The highlighted terms are the ones that result from inserting ZZY just before ZY.  Notice that the last term before the first highlighted term has the form 
$$-\frac{n - 1}{n}$$
in agreement with the third to last term of the sequence that ends in $-n$.
Apply $\color{blue}{Z}$:  We add $1$.  This produces the first highlighted term 
$$-\frac{n - 1}{n} + 1 = \frac{1}{n}$$
the last term that agrees with the terms in the preceding sequence (since up to that term the sequences are the same).  
Apply $\color{blue}{Z}$:  We must add $1$ to the first highlighted number.  Therefore, the second highlighted term is 
$$\frac{1}{n} + 1 = \frac{n + 1}{n}$$
Apply $\color{blue}{Y}$: The last highlighted term is the negative reciprocal of that number, which gives us 
$$-\frac{n}{n + 1}$$
That still leaves the final ZY.
Apply Z:  Adding $1$ to the last highlighted number produces 
$$-\frac{n}{n + 1} + 1 = \frac{1}{n + 1}$$
Apply Y:  We take the negative reciprocal of that number, which produces
$$-(n + 1)$$
as claimed.
