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If $G$ is the golden ratio, then $\lim_{n \to \infty}G^n$ tends ever nearer to integer values that approach $\infty$. Can it therefore be proved that $\infty$ is itself an integer? If not, why not?

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What does it mean for $\infty$ to be an integer? You may be interested in reading the discussion at math.stackexchange.com/questions/36289/is-infinity-a-number . –  Qiaochu Yuan May 27 '11 at 12:48
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That is more to do with whether $\infty$ is a number. I am certain it is because, given any really big number n, we know that n+1 is one step closer to our goal of $\infty$ and all the steps we take to get there (albeit an infinite number of steps), are all considered numbers. Even if we never reach our goal of $\infty$ we can certainly say that we have only ever encountered numbers on the way. Surely our destination will ultimately be a number even if we never reach it! –  Graphic Equaliser May 27 '11 at 13:02
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Why are you assuming that "integerness" respects limits? (What does this discussion even mean? Mathematics is nothing without definitions and you have not provided a definition of anything.) –  Qiaochu Yuan May 27 '11 at 13:30
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Let's take a look at your argument that $\infty$ is a number. In your argument, I believe your use of "number" is equivalent to "real number" and the real numbers can be identified 1-1 with points on a Euclidean line. The process you describe is moving in some direction along that line. For every point on that line, there is at least one point on each side of it, so for every real number, there is at least one real number greater than it (and one less than it). If the "goal" is $\infty$, then there cannot be any real number past $\infty$, so $\infty$ cannot be a number. –  Isaac May 27 '11 at 13:56
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@lhf We're attempting to get rid of the [number] tag. I am forced to revive some dead questions as a part of this. :-/ –  Srivatsan Dec 1 '11 at 0:04

4 Answers 4

At least your observation that the powers of the golden ratio $G$ seem to approach large integers is true and interesting. The fundamental equation defining $G$ is $G^2=G+1$. From it you can deduce that $G^n = G \cdot F_n + F_{n-1}$, where $F_n$ is the $n$-th Fibonacci number. Since for large $n$ we have $G \approx F_{n+1}/F_n$, we get $G^n \approx F_{n+1}+F_{n-1}$, an integer. A more precise statement of this approximation is $G^n+H^n = (G+H)F_n +2F_{n-1}=F_{n+1}+F_{n-1}$, where $H=-1/G=1-G$ is the other root of the equation defining $G$. Since $|H|<1$, its powers go to zero.

Wikipedia mentions this in almost integer. The golden ratio is a Pisot–Vijayaraghavan number, whose characteristic property is that their powers approach integers at an exponential rate.

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What does the golden ratio have to do with it? The integer values 1, 2, 3, ..., approach infinity, too, but that doesn't prove infinity is an integer.

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Because the numbers 1.5, 2.5, 3.5, ... also approach $\infty$ but they are not integer. However, $G^\infty$ is definitely $\infty$ and also approaches an integer value the nearer it gets to $\infty$ so you can see my argument that $\infty$ is an integer. –  Graphic Equaliser May 27 '11 at 12:57
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@Graphic: $\infty$ is not a number. You cannot say that it is integer, rational, real, imaginary. –  Beni Bogosel May 27 '11 at 13:04
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@Graphic: I don't understand your argument. What do you make of $\lim_{n\to \infty} (G^n+.5)$? It tends to $\infty$ too, but is not integral. –  Ross Millikan May 27 '11 at 13:04

$\infty$ can't be a natural number. The natural numbers satisfy:

  1. There is exactly one element called '$0$' with no predecessor($-1$ isn't a natural number).
  2. Every element has a unique successor, different from itself.

You could argue that we excluded by definition in (1) by using 'exactly one' and change it, but you don't get the natural numbers - you get your own custom-made structure. The integers can be defined very conservatively using the natural numbers, so you would have to change this if you wanted "\infty" to be a number.

But if you change (1), you lose out on the Principle of Induction which is needed to prove anything useful about the natural numbers. And if you change (2) to remove "different from itself", then {0,1,2,3,4,5,6} with 6+1 = 6 could be considered "the natural numbers" because it satisfies our defining properties. It's not easy to change the definitions to allow $\infty$, and it doesn't seem to give much benefit.

It's perfectly possible to define your own structure where $\infty$ exists, but you're asking whether it meets the criteria for a specific already-existing structures and it doesn't.

Regarding your limiting argument:

3,3.1,3.14,3.141,3.1415, etc. is a sequence of rational numbers tending to $\pi$, but $\pi$ is not rational.

You can't always argue that because $P(n)$ holds for every $n$, that it must also hold for the limit.

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The values $1.5, 2.5, 3.5, 4.5, 5.5, ...$ don't tend ever nearer to integer values as they approach $\infty$. Does this prove $\infty$ is not an integer?

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