Help with a proof that sequence of rational numbers $ a_n = \frac {a_{n-1} + \frac {2}{a_{n-1}}}{2}$ converges to an irrational, $\sqrt2$ I know that there are sequences of rational numbers with irrational limits. 
One in particular I've seen is $$ a_n = \frac {a_{n-1} + \frac {2}{a_{n-1}}}{2}$$ with $a_0 =1$, This is clearly rational for any $n$, but converges to $\sqrt2$ .
Can anyone provide a mathematical reasoning for this?

If this converges, then the limit $L$ will satisfy $L=\frac{L+2/L}{2}$.  We rearrange to $2L=L+2/L$ or $L=2/L$.  Cross-multiplying we get $L^2=2$ or $L=\pm \sqrt{2}$.  Since every term is positive, we must have $L=\sqrt{2}$.
There are various ways to prove convergence.  One powerful tool comes from dynamical systems.  Consider the function $f(x)=\frac{x+2/x}{2}$.  We have $f'(x)=\frac{1}{2}-\frac{1}{x^2}$.  We evaluate this at $x=\sqrt{2}$, the fixed point of the iteration, to get $f'(\sqrt{2})=0$.  If the result is in the interval $(-1,1)$ then iteration will converge, i.e. an attracting fixed point.
To prove convergence directly, we can prove that all terms (except $a_0$) are greater than $\sqrt{2}$, and monotonically decreasing.  [Note: this method is custom-tailored to this particular sequence, and different methods will be required for other sequences.] We calculate $$\frac{a_n}{a_{n-1}}=\frac{a_{n-1}+2/a_{n-1}}{2a_{n-1}}=\frac{1}{2}+\frac{1}{a_{n-1}^2}$$
So long as $a_{n-1}>\sqrt{2}$, $\frac{1}{a_{n-1}^2}<\frac{1}{2}$ and thus the ratio $a_n/a_{n-1}<1$ so $a_n<a_{n-1}$.  Since $a_1=2>\sqrt{2}$ the induction gets going properly starting with $a_1$ (not $a_0$).  Now, we have a monotonically decreasing sequence of real  numbers.
We need to prove this sequence is bounded below by $\sqrt{2}$.  We assume $a_{n-1}>\sqrt{2}$, so $$0<(a_{n-1}-\sqrt{2})^2=a_{n-1}^2-2a_{n-1}\sqrt{2}+2$$
$$2a_{n-1}\sqrt{2}<a_{n-1}^2+2$$
$$\sqrt{2}<a_{n-1}/2+1/a_{n-1}=a_n$$
Hence $a_n>\sqrt{2}$.  Now we use the completeness of the real numbers to finish the proof of convergence.
