You can’t fold a sheet of paper in half more than seven or eight times, no matter how large the sheet or thin the paper may be.

How often have you heard that statement? Perhaps you’ve even put this assertion to the test. And, indeed, it *is* difficult to get beyond about seven or eight folds when you start with, say, a sheet of newspaper.

So, is seven or eight folds a reasonable rule of thumb for paperfolding in general?

Suppose you start with a sheet that has thickness *t* and width *w* (see http://mathforum.org/library/drmath/view/60675.html). If you always fold in the same direction, you end up with a narrow strip with a constant thickness. With each fold, the thickness doubles, so the thickness after *N* folds would be 2^{N} times *t*. The resulting width would be 2^{–N} times *w*. The ratio of thickness to width would be 2^{2N} times *t*/*w*.

If you start with a sheet of paper that’s 11 inches wide and about 0.001 inch thick, after seven folds, the ratio of thickness to width is 1.6. This means that the folded wad is thicker than it is wide!

If you fold in alternating directions, using up a sheet’s width and height, you might be able to squeeze out an extra fold. In this case, the thickness would still double with each fold, but the width would be cut in half only every second fold.

Several years ago when she was in high school, Britney Gallivan faced the challenge of folding a sheet in half an unheard-of 12 times. She had to solve the problem to get extra credit in one of her classes.

After a lot of experimenting, Gallivan succeeded, folding a thin film of gold 12 times. What about folding paper?

Gallivan worked out her own equations for the limit on the number times it’s possible to fold a sheet of paper of a given size. For the case of folding in a single direction:

where **L** is the minimum possible length of the material, **t** is the material thickness, and **n** is the number of folds possible in one direction (**L** and **t** must be expressed in the same units).

For a given thickness and length, the equation **a**(**n**) = (2^{n} + 4)(2^{n} – 1)/6 tells how much paper has been lost after **n** folds.

Starting with **n** = 0, the sequence takes on the following values: 0, 1, 4, 14, 50, 186, 714, 2794, 11050, 43946, 175274, 700074, 2798250, . . .

This means that for the 12th folding of paper in half, 2798250 times as much material has been lost to potential folding as was lost on the first fold.

Gallivan described how she obtained her equations and her efforts to solve the folding problem in a booklet published by the Historical Society of Pomona Valley, to which Gallivan belongs (see http://pomonahistorical.org/12times.htm).

In January 2002, Gallivan herself demonstrated how to fold a large sheet of paper in half 12 times.

It pays to question received wisdom, at least as far as paperfolding is concerned.