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(2) ... 3





sqrt(1.509803...) ... 4


Also, using the floor function...



sqrt(2) ... 3


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, $$\lceil\log_2(\log_2(x))+1\rceil$$

  • 1
    $\begingroup$ Excellent, thank you. $\endgroup$ – Char Apr 8 '12 at 23:48
  • 1
    $\begingroup$ Glad to help! If my answer is satisfactory, you can officially accept it by clicking the checkmark (see here for explanation). $\endgroup$ – Zev Chonoles Apr 8 '12 at 23:50

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 $2^{n-1}\le\lg x<2^n$.


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 $\sqrt 2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.