Doing repetetive square root of any non negative rational number is 1

Question

When i tried to do the square root repeatedly of any number greater than 0 the calculator gave me 1(I think it was approximate but doing it infinitely many times should give me 1) at a time, so would the square of any number be the original number taken ?

Answer 1

Consider the Number $1$.
Keep making it half again-and-again iteratively.
You may get a Sequence like this :

$1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64},\frac{1}{128},\frac{1}{256},\frac{1}{512},\frac{1}{1024},\frac{1}{2048}, \cdots$

It will never become $0$ but get arbitrarily close to that.
[ Calculators will eventually show $0$ even though that is not Exact ]

Choose some Positive Number $X$ & raise it to the Power of the Sequence element-by-element.
You may get a new Sequence like this :

$X^{1},X^{\frac{1}{2}},X^{\frac{1}{4}},X^{\frac{1}{8}},X^{\frac{1}{16}},X^{\frac{1}{32}},X^{\frac{1}{64}},X^{\frac{1}{128}},X^{\frac{1}{256}},X^{\frac{1}{512}},X^{\frac{1}{1024}},X^{\frac{1}{2048}}, \cdots$

This new Sequence will never become $X^{0}$ but will get arbitrarily close that that.
We know that $X^{0}=1$ when $X$ is a Positive Number.
In other words, the new Sequence will get arbitrarily close to $1$ but never become that value.
[ Calculators will eventually show $1$ even though that is not Exact ]

This is what you observe by the repetitive Square root on the same Positive Number.
Taking Square root is Equivalent to raising by Power $\frac{1}{2}$ & the Powers keep getting smaller-and-smaller on repetition.

When $X=0$ , we will always get $0$ on Square rooting.
When $X$ is a Negative Number , we will get Imaginary Numbers or Complex Numbers on Square rooting.