# Infinite Square Roots

If you square root a number infinite amount of times, will it eventually become 1, or some sort of whole number? Can you please give proof as to why you got your answer? Thanks!

Note by Jonathan Hsu
3 years ago

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Not really a proof, but if we look at $$\sqrt{x}$$, we know the result will be less than $$x$$ (unless of course $$x$$ is less than or equal to 1.

When we square-root a number an infinite number of times, the number $$x$$ must lower in value an infinite number of times, until 1 is reached.

Interestingly, when we square-root a number greater than -1 but less than 1 an infinite number of times, we get back to 1. When we square (not square-root) a similar number an infinite number of times, we get back to 0.

For negatives, squaring gets us infinitely high (or infinitely low), and square-rooting gets us imaginary numbers.

- 3 years ago

If the initially chosen number is $$N\geq1$$ the sequence $$\{x_n\}$$ defined as $x_1=N, x_{n+1}=\sqrt{x_n}, \forall n\geq 1$ converges to $$1$$. This follows from Banach's Fixed Point Theorem because the map $$f:[1,N] \to [1,N]$$ defined by $$f(x)=\sqrt{x}$$ is a contraction map, with the unique fixed point $$1$$, in the metric space $$[1,N]$$, with the usual euclidean metric.

- 3 years ago

It's some kind of elvish. I can't read it. :)

- 3 years ago

  1 2 3 4 5 6 7 8 9 10 11 12 13 Let y = ... sqrt ( sqrt ( sqrt ( x Square both sides: y**2 = ... sqrt ( sqrt ( sqrt ( x OR y**2 = y y**2 - y = 0 y( y - 1 ) = 0 y = 0 or 1 y cannot be 0 since the input number (x) is >0 so y=1 

- 3 years ago

You need to show that the sequence converges to $$1$$. Finding a fixed point is not sufficient for convergence, e.g., instead of taking square roots, if we square the numbers, for any initially chosen number $$N>1$$, the sequence diverges to $$\infty$$. But you still have the equation $$y=y^2$$, hence$$y=0$$ or, $$y=1$$.

- 3 years ago

sqr(sqr(sqr(...sqr(n) = x , X^2= sqr(sqr(...sqr(n) , X^2=X , X= 1 or 0 . I think that would be it

- 3 years ago