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 :)

  • $\begingroup$ Do you have any thoughts about this? Did you try some things to see if and how you could, say, turn the number 2 into 0? $\endgroup$ – Bram28 Apr 21 '18 at 1:59
  • $\begingroup$ I found that you could use Y, Z, Y, Z, Z to do that. And to turn 3 to 0, you could do Y, Z, Y, Z, Z, Y, Z, Z. There was a pattern with the sequence (Y, Z, Z) and you just had to add on a repeat of this to the sequence used to turn 0 to 2, and then this will let you turn 0 to 3. But, I don't know why this works. $\endgroup$ – Brooklyn Apr 21 '18 at 2:12
  • $\begingroup$ That's why I suggest to use induction :) $\endgroup$ – Bram28 Apr 21 '18 at 2:13


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.


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.

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  • $\begingroup$ Can you please elaborate on this as I don't really understand? $\endgroup$ – Brooklyn Apr 21 '18 at 2:13
  • $\begingroup$ @Brooklyn Are you familiar with induction? $\endgroup$ – Bram28 Apr 21 '18 at 2:14
  • $\begingroup$ Unfortunately, I am not. $\endgroup$ – Brooklyn Apr 21 '18 at 2:15
  • $\begingroup$ @Brooklyn Hmm, that could be a problem, as at the end of your post you say that you need to understand how to do this problem before moving on to others ... and I suspect that it is this very proof technique of induction that you'll have to use for those other problems as well. Are you working with a textbook? $\endgroup$ – Bram28 Apr 21 '18 at 2:17
  • $\begingroup$ Only the problem was given (no textbook :/). $\endgroup$ – Brooklyn Apr 21 '18 at 2:19

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