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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:

example_of_application_of_operations

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 :)

Brooklyn
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  • Please edit your question to share what you have attempted and explain where you are stuck. For instance, have you found a method for obtaining $-1$? $-2$? $-3$? – N. F. Taussig Apr 22 '18 at 11:08

1 Answers1

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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.

N. F. Taussig
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