Since $i=-1/i$, can we write $i=-1/(-1/(-1/\cdots))$?

Question

I am having an argument with a colleague about $i$, the square root of $-1$. My argument goes as follows:

$$ \begin{align*} i &= \sqrt{-1}\\ i^2 &= -1\\ \frac{i^2}{i} &= -\frac{1}{i}\\ i &= -\frac{1}{i} \end{align*} $$

and then through substitution, since both sides are defined and not variables, $$\mathrm{i} = \frac{-1}{\frac{-1}{\frac{-1}{\frac{-1}{\dots}}}}$$

Is this still equal to $\sqrt{-1}$? I argue that it does, although this would mean that there is some real correlation to imaginary numbers.

Answer 1

It is true that $i = \dfrac{-1}{i}$

It is also true that $i = \dfrac{~-1~~}{~\left(\frac{-1}{i}\right)~}$

Further, it is also true that the sequence $a_1 = i$ and $a_{n+1}=-1/a_{n}$ will equal $i$ for all values of $n$ and $\lim\limits_{n\to\infty} a_n = i$ as it is simply a constant sequence. That is to say... if you imagine your tower of division "from the bottom up" then yes, it will equal $i$ as you suggest. That is... if you prefer $(\dots/(-1/(-1/(-1/i))))$

The problem is that this is not how $-1/(-1/(-1/(-1/(\cdots))))$ is generally interpreted. The way you have written it makes it sound as though we are considering this fraction as being defined "from the top down."

Any time we are dealing with an infinite number of operations or the like, we must be very clear how we are defining what it is we are looking at and this is generally done by way of sequences and limits of sequences. If the limit of the sequence converges to a number we can say the expression "equals" that number. The way you have this written now, it makes it sound as though we are considering $b_1 = -1$ and $b_{n+1}=b_n/(-1)$... that each time we consider "more" of the tower of divisions, we are doing so by taking what we already have as the numerator and then dividing with something additional as the denominator. In such an interpretation of what you have written, this sequence does not converge. Instead, it will alternate between the two values of $-1$ and $1$. Since this does not converge, that makes the expression you wrote undefined.

Answer 2

Your writing was correct up to the point where you introduced $(-1)/((-1)/((-1)/\cdots))$.

The problem with this "expression" is that it does not have a value, thus we cannot even ask if its value is $i$.

This comes from the following consideration: in calculus, we normally assign values to infinite expression only if those expressions are limits of some sequences that converge. For example, we do say that $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots=2$ because this infinite sum is, as a convention, taken as a limit of the sequence of the finite sums:

$$1=2-1$$ $$1+\frac{1}{2}=2-\frac{1}{2}$$ $$1+\frac{1}{2}+\frac{1}{4}=2-\frac{1}{4}$$ $$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=2-\frac{1}{8}$$

etc., which obviously get closer and closer to $2$ as we add more terms.

Nothing like that seems to apply to your expression. In the best case, you are looking at whether the following sequence has a limit:

$$-1=-1$$ $$\frac{-1}{-1}=1$$ $$\frac{-1}{\frac{-1}{-1}}=-1$$ $$\frac{-1}{\frac{-1}{\frac{-1}{-1}}}=1$$

and so on... which is in fact the sequence $-1,1,-1,1,\ldots$, which oscillates forever and does not converge to any number at all.