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# Cutting the Mobus band

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Problem. What happens if we cut the Mobius band along the middle line?

The Mobuis band is a rectangle with the left edge identified with the twisted right edge. This diagram will help to solve this problem.

Suppose we cut it with $n-1$ cuts, then the rectangle has $n$ bands. Let's call them $B_1, ..., B_n$. Suppose their left edges are $x_1, ..., x_n$, from top to bottom. Then their right edges are $x_n, ..., x_1$, from top to bottom. More precisely, these are $-x_n, ..., -x_1$ indicating the twisted edges.

Now $B_1$ and $B_n$ share edges, both in fact:

$x_1 : B_1 : -x_n$, and $x_n : B_n : -x_1$.

We need to make sure that the edges match. Because of that $B_n$ has to be flipped:

$$-x_n : -B_n : x_1.$$

Now attach them:

$$x_1 : B_1 : -x_n : - B_n : x_1.$$

The ends match! Therefore, this is a cylinder.

In a identical fashion, $B_2$ and $B_{n-1}$ are glued into another cylinder, and so are $B_3$ and $B_2$, etc.

Let $k = \frac{n}{2}$; then if $n$ is even, then $k$ is an integer. In this case the process will use all bands that we cut. The last one made of $B_k$ and $B_{k+1}$. The total is $k$ cylinders.

If $n$ is odd, then there will be one band left after we have made $k=\frac{n-1}{2}$ cylinders. The last one is:

$$x_k : B_k : -x_k.$$

This is another Mobius band.