Iterated functions

Question

A function F(x) when composed in itself number of times is called an iterated function.

Let $F(x)=\frac{1}{1+x}$

$F(\frac{1}{2})=\frac{2}{3}$

$F(\frac{2}{3})=\frac{3}{5}$

$F(\frac{3}{5})=\frac{5}{8}$

Continuing in this way to infinite number of times it turns out that we get $\frac{\sqrt{5}-1}{2}.$

But we see above that we will continue to get rational fractions however long we continue.

So $\frac{\sqrt{5}-1}{2}$ or simply $\sqrt{5}$ is rational.

Are we incorrect in assumption or we haven't defined irrational numbers properly?

Answer 1

Think of a sequence of rational numbers that approximate an irrational number closer and closer:

$$3,\\ 3.1, \\3.14, \\3.141, \\3.1415, \\3.14159, \\3.141592, \\3.1415926,\\\cdots$$

Such a rational sequence converges... to an irrational number.

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Intuitively, this is possible because the denominators get larger and larger, and "in the end" the denominator becomes infinite.

Answer 2

As $n\rightarrow\infty$ your expression can be written as $$F^n(x)=\frac{1}{1+F^n (x)}$$

or $$F^n(x)+\left (F^n(x)\right)^2=1$$ which approaches the golden ratio, but notice that this is just a limit. Just like $$\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}=\frac{\pi}{4}$$ is a limit of a rational sum.

Answer 3

But we see above that we will continue to get rational fractions however long we continue.

That's true, as long as "however long we continue" is a finite number of iterations. We can imagine iterating the function a finite number of times, even if that number is really large and it isn't physically possible to do the calculations that many times, and still getting a rational number.

However, we can't possibly imagine iterating infinitely many times and arriving at an answer. That process is impossible, even in theory.

Thus, this claim isn't entirely true:

Continuing in this way to infinite number of times it turns out that we get √5-1/2.

We can't continue infinitely many times and get a result.

What we can say is that the more we continue (i.e., the "closer to infinity" we get in terms of number of iterations), the closer our answer gets to $\frac{\sqrt5-1}{2}$. That means $\frac{\sqrt5-1}{2}$ is the limit of a sequence of rational numbers. As the other answers have already pointed out, our definition of irrational numbers doesn't preclude them from being limits of sequences of rational numbers.

Answer 4

If a function $f(x)$ is iterated in itself then it will converge to a point but this will happen if we assume infinite iterations which is just a concept and thats why we solve $f(x)=x$ which gives us expected result but that does not make result rational