Showing $\sqrt{0+\sqrt{0+\sqrt{0+...}}}=0$

Question

How do I show that $a=\sqrt{0+\sqrt{0+...}}=0$ when the solution I get comes to $a^2=a \implies a \in \{0,1\} \:\: \forall a \in \mathbb{R}$? Or is it possible for $\sqrt{0+\sqrt{0+...}}$ to equal 1?

Answer 1

Hint: Using induction $a_{n+1}=\sqrt{0+a_n}$ and $a_1=\sqrt{0}$ You see $a_1$ is zero then you can show if $a_n=0$ then $a_{n+1}=0$. This is the proof by induction.

Answer 2

$a=\sqrt{0+\sqrt{0 +\dots}}$ is usually defined as $$a=\lim_{n\to\infty} a_n$$ where $a_0=\sqrt{0}$ and $a_{n+1} = \sqrt{0+ a_n}$. It is easy to see that $a_n=0$ and thus that $a=0$.

Now, of course you can show that $a$ satisfies the equation $a^2=a$, but that only tells you that $a$ CAN be equal to $1$ according to that particular equation.