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!

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.

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.

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

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, \(y=1\).

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply

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.

Log in to reply

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

Log in to reply

Log in to reply

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

Log in to reply