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Is there some way to determine how many times one must root a number and its subsequent roots until it is equal to the square root of two or of the root of a number less than two?

sqrt(16)=4

sqrt(4)=2

sqrt(2) ... 3

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sqrt(27)=5.19615...

sqrt(5.19615...)=2.27950...

sqrt(2.27950...)=1.509803...

sqrt(1.509803...) ... 4

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Also, using the floor function...

sqrt(27)=floor(5.19615...)

sqrt(5)=floor(2.23606...)

sqrt(2) ... 3

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3 Answers 3

up vote 3 down vote accepted

Let $x$ be our (positive real) number. Then $$\underbrace{\sqrt{\sqrt{\cdots\sqrt{x}}}}_{n\text{ times}}=x^{1/2^n}\leq \sqrt{2}$$ if and only if $x\leq 2^{2^{n-1}}$, which is the case if and only if $$\log_2(\log_2(x))\leq n-1,$$ or equivalently $\log_2(\log_2(x))+1\leq n$. The smallest integer $n$ with this property is, by definition, $$\lceil\log_2(\log_2(x))+1\rceil$$

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Excellent, thank you. –  Char Apr 8 '12 at 23:48
    
Glad to help! If my answer is satisfactory, you can officially accept it by clicking the checkmark (see here for explanation). –  Zev Chonoles Apr 8 '12 at 23:50

Let's do this by example, and I'll let you generalize.

Say we want to know about $91$, one of my favorite numbers because it is the lowest number that I think most people might, at first thought, say is prime even though it isn't (another way of saying it isn't divisible by the 'easy-to-see' primes).

Well, I note that $16 = 2^4 < 91 < 256 = 2^8$

So if we take 2 square roots, then our number will be bigger than 2, as it's greater than $2^{4/4} = 2$. But If we take 3, as our number is less than $2^{8}$, then it's 3rd iterated square root will be less than $2^{8/8} = 2$.

So the third square root of $91$ will be between $\sqrt 2$ and $2$. So the 4th will place it below $\sqrt 2$.

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Taking the square root $n$ times is taking the $2^n$-th root. If the $2^n$-th root of $x$ is $\le\sqrt2$, but the $2^{n-1}$-st root of $x$ is $>\sqrt2$, then

$$x^{1/2^n}\le \sqrt2<x^{1/2^{n-1}}\;,$$

which, after raising everything to the $2^n$ power, is equivalent to

$$x\le (2^{1/2})^{2^n}<x^2\;,$$

or $$x\le 2^{2^{n-1}}<x^2\;.$$ Now take logs base $2$ (which I write $\lg$):

$$\lg x\le 2^{n-1}<2\lg x\;,$$ which after a little massage can be written $$2^{n-2}<\lg x\le 2^{n-1}\;.$$

Take your examples: $\lg 16=4$, so $2^1<\lg 16\le 2^2$, and we must have $n=3$, while $\lg 27$ is clearly between $4$ and $5$, so $2^2<\lg 27\le 2^3$, and we must have $n=4$.

Taking the floor of the square root each time complicates matters slightly, but not much. $\lfloor\sqrt x\rfloor\le\sqrt2$ iff $x<2^2$ iff $\lg x<2$, $\lfloor\sqrt x\rfloor<2^2$ iff $\sqrt x<2^2$ iff $x<2^{2^2}$ iff $\lg x<2^2$, $\lfloor \sqrt x\rfloor\le 2^{2^2}$ iff $\sqrt x<2^{2^2}$ iff $x<2^{2^3}$ iff $\lg x<2^3$, and so on. If $\frac12<\lg x<2$, one step is required. If $2\le\lg x<2^2$, two steps are required. In general, $n$ steps are required if $2^{n-1}\le\lg x<2^n$.

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