# Celebrating Square Root Day?

Today is square root day. What a weird day. To "celebrate," I will prove that consecutively applying the square root operator to any number greater than or equal to one will yield a number increasingly closer to one.

In other words, if $x_1\geqslant1$ is any number, and if $x_{n+1}=x_n^{1/2}$ for all integers $n$ greater than zero, then $\lim_{n\to\infty}x_n=1$.

Let $f:\left[1,\infty\right)\to\left[1,\infty\right)$ such that $x\mapsto x^{1/2}$, and let $x,y\in\left[1,\infty\right)$. Then

$$\left|x^{1/2}-y^{1/2}\right|=\frac{1}{\left|x^{1/2}+y^{1/2}\right|}\left|x-y\right|\leqslant\frac12\left|x-y\right|,$$

which implies that $f$ is Lipschitz with constant $1/2$, which in turn implies that $f$ is a contraction on $\left[1,\infty\right)$. Moreover, observe that $f\left(1\right)=1$, which implies that $1$ is a fixed point of $f$. Since $\left[1,\infty\right)$ equipped with the usual euclidean metric is a complete metric space, our sequence $\left\{x_n\right\}_{n=1}^\infty$ converges to $1$ for any $x_1\in\left[1,\infty\right)$.

In other words, if $x_1\geqslant1$ is any number, and if $x_{n+1}=x_n^{1/2}$ for all integers $n$ greater than zero, then $\lim_{n\to\infty}x_n=1$.

Let $f:\left[1,\infty\right)\to\left[1,\infty\right)$ such that $x\mapsto x^{1/2}$, and let $x,y\in\left[1,\infty\right)$. Then

$$\left|x^{1/2}-y^{1/2}\right|=\frac{1}{\left|x^{1/2}+y^{1/2}\right|}\left|x-y\right|\leqslant\frac12\left|x-y\right|,$$

which implies that $f$ is Lipschitz with constant $1/2$, which in turn implies that $f$ is a contraction on $\left[1,\infty\right)$. Moreover, observe that $f\left(1\right)=1$, which implies that $1$ is a fixed point of $f$. Since $\left[1,\infty\right)$ equipped with the usual euclidean metric is a complete metric space, our sequence $\left\{x_n\right\}_{n=1}^\infty$ converges to $1$ for any $x_1\in\left[1,\infty\right)$.

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