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I was playing with some iterated functions/fractions when I stumbled across this one $$\frac{1+\frac{1+\frac{1+...}{1-...}}{1-\frac{1+...}{1-...}}}{1-\frac{1+\frac{1+...}{1-...}}{1-\frac{1+...}{1-...}}}$$ which I found to be equal $i$. Am I correct? And if so, does it mean real numbers are not closed under arbitrary basic arithmetic (assuming there is no zero-division somewhere in it)?

eagr
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As any field, the real numbers are definately closed under basic arithmetic operations. Namely for any two reals $x$ and $y$, $x+y$, $x-y$, $x\cdot y$ and, given that $y\neq 0$, $\frac{x}{y}$, are real numbers.

The reals are also complete, and from this it follows that any sequence of reals that converges in the complex numbers must converge to a real number. So I don't see how $$ a_{n+1}=\frac{1+a_n}{1-a_n} $$ can converge to $i$.

Anguepa
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  • I'm not sure your $a_n$ is the same as my fraction since you must have $a_0$ which I don't have. And if I'm wrong, can anybody please tell me what my expression equals? – eagr Oct 26 '18 at 15:14
  • Yeah I would need to know what $a_0$ is. To me that expression the way you've written it is not well defined. – Anguepa Oct 26 '18 at 15:35
  • Since it's infinite I thought I can substitute $x=\frac{1+x}{1-x}$, and that's how I got $i$ (and $-i$ ). – eagr Oct 26 '18 at 16:06
  • You need to define clearly what the expression is. If it's the limit of a sequence then you need to say what $a_0$ is. As I've said for any real $a_0$ the limit of the sequence if it exists must be real too. If however $a_0=i$ or $a_0=-i$ then indeed the limit would be $i$ and $-i$ respectively. – Anguepa Oct 26 '18 at 16:13
  • I thought I can find its value like finding the value of $\sqrt{1+\sqrt{1+\sqrt{1+...}}}$ by having $x=\sqrt{1+x}$ and getting $x=\frac{1\pm\sqrt{5}}{2}$ – eagr Oct 26 '18 at 16:23
  • @GetMathed again you have to fix a valued for $a_0$, see here. For example the sequence in your comment seems to converge only for certain initial values. – Anguepa Oct 27 '18 at 14:19