Algorithms for approximating $\sqrt{2}$ Well, "Solving" is the wrong term since I am speaking about irrational numbers. I just don't know which word is the correct word... So that can be part $1$ of my question... what is the correct word since you obviously can't "solve" an irrational number because it goes forever.
Part $2$ (my real question) are there algorithms for figuring out the answer to a problem like the square root of $2$ other than guess-and-checking your way to infinity? Again, I'm obviously not asking for an algorithm to give me the never ending answer because that's crazy... but for example if I wanted to know what the $15^{th}$ decimal place of the square root of $2$ was, is there an algorithm for that?
Thank you! (I'm new here and know nothing about how to format math questions so any help or links would be appreciated as well, thanks!)
 A: You can use newton's method to compute the digits of $\sqrt{(2)}$: 
Let:
$$
f(x) = x^2 -2
$$
Define the iteration:
$$
x_0 = 1\\
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
$$
This will converge to $\sqrt{2}$ quadratically. 
If you want to compute other square roots: 
Consider: $$g(x) = x^2 - a$$ 
Which has the iterants:
$$
x_{n+1}=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right)
$$
As mentioned below. 
There's also what's called the continued fraction expansion of an algebraic number. You can use a finite continued fraction expansion. 
As an example:
$$
x_0 = 1 \\
x_1 = \frac{1}{2}\left(x_0 + \frac{2}{x_0}\right) =\frac{1}{2}\left( \large \textbf{1} + \frac{2}{ \large \mathbf{1}}\right) = \frac{3}{2}\\
x_2 = \frac{1}{2}\left(x_1 + \frac{2}{x_1}\right) = \frac{1}{2}\left( \large \mathbf{\frac{3}{2}} + \frac{2}{ \large \mathbf{\frac{3}{2}}}\right), \text{ etc. } 
$$
Added
Since we are using Newton's method, and you are wondering why it converges to the root of $f(x)$,   Note the following: 
 $\textbf{Theorem} $: 
Suppose that the function $f$ has a zero at $\alpha$, i.e., $f(\alpha) = 0$
If $f$  is continuously differentiable and its derivative is nonzero at $\alpha$, then there exists a neighborhood of $\alpha$ such that for all starting values $x_0$ in that neighborhood, the sequence ${x_n}$ will converge to $\alpha$.
So if we choose our starting guess appropriately, Newton's method always converges to the root of the equation if $f$ has these properties . 
A: A related problem. Another way to go is the Taylor series. Derive the Taylor series of the function $\sqrt{x}$ at the point $x=1$  
$$ \sqrt{x} = 1+{\frac {1}{2}} \left( x-1 \right) -{\frac {1}{8}} \left( x-1
 \right) ^{2}+{\frac {1}{16}} \left( x-1 \right)^{3}-{\frac{5}{128}
} \left( x-1 \right)^{4}+O\left(  \left( x-1 \right) ^{5} \right). $$
If you plug in $x=2$, you get an approximate value for the $\sqrt{2}\sim 1.398437500$. Increasing the number of terms in the series improves the approximation.
Added: We can write the Taylor series of $\sqrt{x}$ explicitly by finding the $n$th derivative of $\sqrt{x}$ as

$$ \sqrt{x} = \sum _{n=0}^{\infty }\frac{\sqrt {\pi }}{2}\,{\frac {{a}^{\frac{1}{2}-n} \left( x-a\right)^{n}}{\Gamma\left( \frac{3}{2}-n \right)n! }}.$$

Substituting $a=1$ in the above formula gives the Taylor series at the point $a=1$:
$$\sqrt{x} = \sum _{n=0}^{\infty }\frac{\sqrt {\pi }}{2}\,{\frac { \left( x-1\right)^{n}}{\Gamma\left( \frac{3}{2} - n \right)n! }}.$$
Putting $x=2$ in the above equation, we have:
$$\sqrt{2} = \sum _{n=0}^{\infty }\,{\frac {\sqrt{\pi}}{2\,\Gamma\left( \frac{3}{2} - n \right)n! }}. $$
A: You can also compute square roots using continued fractions.  For example for $\sqrt{2}$ you have
$$ \sqrt{2}=1+(\sqrt{2}-1)=1+\frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1}=1+\frac{1}{\sqrt{2}+1} $$
where $1$ is the integer part of $\sqrt{2}$.  Then repeat the process for $\sqrt{2}+1$ whose integer part is $2$:
$$
\sqrt{2}+1=2+(\sqrt{2}-1)=2+\frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1}=2+\frac{1}{\sqrt{2}+1}
$$
therefore by repeating the process we have
$$
\sqrt{2}=1+\frac{1}{2+\frac{1}{\sqrt{2}+1}}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots}}}}
$$
