Visualizing the square root of 2

Try explaining it with a number line:

Divide the interval between 1 and 2 into ten equal parts and number them 0-9. Note that these are tenths, and so numbers of the form 1.4... fall into division four.

Now divide the interval between 1.4 and 1.5 into ten equal parts and note that these are hundredths, so numbers of the form 1.41... fall into division one.

Continuing this way demonstrates that at each step, adding more digits gives us a smaller and smaller interval that the number can be in and so increases accuracy rather than just making the number grow.

I would start off with a much simpler example. Suppose we wanted to locate $1/3$ on a number line, but the number line is such that every interval is divided into tenths, not thirds (e.g., like a metric ruler). How would you locate such a number? Well, you would start off by moving $3/10$ units from $0$, so now you're at $0.3$. Then between $3/10$ and $4/10$, the ruler is again divided into ten equal sub-intervals, namely $0.31, 0.32, 0.33, \ldots, 0.39, 0.40$. We again choose the third tick mark, so now we're at $0.33$. By now we should see what to do next: between $0.33$ and $0.34$, there are ten more intervals, and we choose the third tick mark, which is $0.333$. This process never ends. We have gone from $0.3$ to $0.33$ to $0.333$ to $0.3333$, and so on. The position we are at after an infinite number of steps is exactly $1/3$.

Now you might say, "why does the ruler have to be divided into tenths? If the ruler had simply been divided into exact thirds, we would not need such an elaborate, infinite process." And that's correct. But the point is that the division of the ruler is analogous to the decimal representation of a number, where each successive digit can only be obtained by moving from one tick mark to the next. Otherwise, if we wanted to "measure" a number like $\sqrt{2}$ on a number line, we could just get ourselves a "special ruler" where there are just two tick marks: one at $0$ and the other one at exactly $\sqrt{2}$. But that doesn't illustrate that the decimal expansion of $\sqrt{2}$ is $1.41421356237309504880168872421\ldots.$ Each real number is identified with a unique point on the number line. But if we want to regard such numbers as being some kind of decimal expansion, then we must resort to using the ruler divided into successive tenths.

Here is another way of approximating the square root of two by rational numbers which doesn't depend on the decimal system.

Suppose $p^2-2q^2=\pm 1$ so that $\left(\cfrac pq\right)^2=2\pm\cfrac 1{q^2}$, then the larger we can make $q$ the closer $\cfrac pq$ is to $\sqrt 2$.

Consider now $(p+2q)^2-2(p+q)^2=p^2+4pq+4q^2-2p^2-4pq-2q^2=2q^2-p^2=\mp 1$ so that $\cfrac {p+2q}{p+q}$ is a better approximation.

From this we obtain the approximations $$\frac 11, \frac 32, \frac 75, \frac {17}{12}, \frac {41}{29},\frac {99}{70} \dots$$

The Wikipedia entry gives also that if $r$ is an approximation, $\frac r2+\frac 1r$ is a better one, which picks out $1, \frac 32, \frac {17}{12}, \frac {577}{408} \dots$ which converges very quickly, picking out a subsequence of the previous one..

This takes $$\frac pq \text{ to } \frac {p^2+2q^2}{2pq}$$

It also gives a geometric proof which is quite visual and may help - the problem with these kinds of proofs is that they often work by some form of descent, and therefore terminate, so don't give a sense of never-ending.

1/3 of a pie is clearly finite. The only infinite part is the number of digits required to express absolute accuracy (i.e. you can't) when written in Base 10. In Base 3, it'd be 0.1. It is finite, but it also has absolute accuracy.