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!

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TopNewestNot 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. – Clive Chen · 1 year, 9 months ago

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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. – Abhishek Sinha · 1 year, 9 months ago

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– Vincent Miller Moral · 1 year, 9 months ago

It's some kind of elvish. I can't read it. :)Log in to reply

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– Abhishek Sinha · 1 year, 9 months 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, \(y=1\).Log in to reply

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 – Caio Garcez · 1 year, 8 months ago

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