What shape is formed by taking a diagonal slice of a four dimensional cube, or a 10 dimensional cube?

It turns out that a very familiar mathematical object, Pascal's triangle, can help us answer this question.

[MUSIC PLAYING] 10 00:00:28,130 --> 00:00:31,560 Pascal's triangle forms a sort of hierarchy.

And we construct the lower levels from the upper ones.

Normally, those levels are comprised of numbers, which we add together.

But we'll look at a geometric analog, where we'll be adding shapes.

But before we get to that, we'll need to define hypercubes and hyperplanes.

Normally, we think of a cube as being three dimensional, like this.

But a square is basically a two dimensional cube.

And a one dimensional cube is just a line segment.

And in four dimensions, well, that's harder to describe intuitively.

People sometimes call a four dimensional cube a tesseract, one of my favorite math words.

And in higher dimensions, we sometimes use the word hypercube.

Let's give a definition for a cube that works in all dimensions.

First, remember that the points in n dimensions are specified by n coordinates.

The vertices of a cube are all the points where the coordinates are either 0 or 1, like 0, 1, 0, 0 in four dimensions or 1, 1, 0, 0, 1, 1, 0 in seven dimensions.

The cube is all those vertices plus all the stuff between the vertices, what's known as the convex hull.

OK, now, what's a hyperplane?

In general, it's a space one dimension smaller than the one you're currently thinking about.

So in one dimension, a line, the hyperplane is zero dimensional, a point.

In two dimensions, a plane, the hyperplane is one dimensional, a line.

In three dimensions, the hyperplane is a regular old 2D plane.

In four dimensions, the hyperplane is three dimensional, and so on.

If we look at the intersection of a hyperplane with an object, like a hypercube-- which is kind of like taking a slice out of it-- we can sometimes understand different features of that object.

Let's start slicing cubes beginning with a square.

We'll draw a diagonal from 0, 0 to 1, 1.

Then we draw the hyperplane that's perpendicular to this diagonal going through 0, 0.

Remember, it's one dimension smaller, so it's a line.

And then we sweep this line up along the diagonal up to the point 1, 1.

Let's pause the picture every time the hyperplane hits a vertex of the square.

That's when we'll be taking slices of the square.

This happens three times-- at the very beginning at 0, 0, in the middle when it hits 1, 0; and 0, 1, and at the end when it hits 1, 1.

Let's look at the intersection of the hyperplane with the square at each of those points.

At the first and last moments, the hyperplane just hits the square at a point.

But in the middle, it's a line segment.

Let's try the whole thing again in three dimensions.

Here's the diagonal line from the corner 0, 0, 0 to 1, 1, 1.

Now, let's draw the hyperplane that's perpendicular to this diagonal going through 0, 0, 0.

This time, it's a 2D plane.

Like before, we sweep this plane along the diagonal up to the point 1, 1, 1.

There are four separate times in which the sweeping plane intersects the vertices of the cube.

First, it starts at 0, 0, 0.

Then it hits three vertices at once-- 1, 0, 0; 0, 0, 1; and 0, 1, 0.

Then it hits three more vertices at once-- 1, 1, 0; 1, 0, 1; and 0, 1, 1.

And finally, it ends at 1, 1, 1.

Again, underneath, we'll record the shape of the intersection of the hyperplane and the cube.

At the first and last intersections, it's a point.

But the middle two intersections are both triangles with one pointing up and one pointing down.

This even works in one dimension.

A one dimensional cube is a line segment.

And the hyperplane is just a point.

In one dimension, we run this point along the line segment.

And it intersects the vertices twice-- once at the beginning and once at the end.

Let's summarize these results.

Here's the shape of the hyperplane intersection when you sweep it across the diagonal of a cube and pause at vertices.

The most logical question to ask now is, what's on the next line?

What happens when you sweep a three dimensional hyperplane through a four dimensional cube?

To help answer this question, we need to talk about Pascal's triangle and explicitly recognize the version of Pascal's triangle that's emerging from this picture.

Pascal's triangle is an infamous object in mathematics that shows up in many unexpected places.

It's defined in terms of the number of unique combinations of things.

This notation indicates the number of distinct ways to choose k objects from a collection of n objects.

We usually say it as n choose k. For example, 5 choose 2 is 10.

If you have 5 puppies and have to select your favorite 2, there are 10 ways to do that.

If you're not allowed to take any puppies home, there's only one way to do that.

So 5 choose 0 is 1.

There's a formula for n choose k. It's n factorial divided by k factorial times n minus k factorial.

But the formula is not important for our purposes.

One way to describe Pascal's triangle is to say that on the n-th row, the entry in the k-th column is n choose k-- as in, the third column on the fifth row is 5 choose 3.

We always start counting with 0-- so this is the 0-th row, and each row has a 0-th column.

So to find the fifth row, third column, we actually count 0, 1, 2, 3, 4, 5 rows and 0, 1, 2 3 entries.

So if we use the formula for choose, Pascal's triangle becomes this.

Compare it with the picture of cube slices.

Amazingly, the number of vertices within each cube slice directly corresponds to Pascal's triangle.

And that's not just a coincidence of the first few rows.

The pattern holds for all the rows below.

To explain why this happens, let's first focus on the three dimensional case.

The slices of the 3D cube are a point, a triangle, another triangle, and another point.

So the number of vertices these slices have are 1, 3, 3, 1, which corresponds to the third row in Pascal's triangle.

The first diagonal slice intersects one vertex-- 0, 0, 0.

It corresponds to 3 choose 0, because there are three coordinates and zero of them are 1's.

The next diagonal slice goes through 1, 0, 0; 0, 1, 0; and 0, 0, 1.

It corresponds to 3 choose 1, because there are three coordinates and one of them is a 1.

The vertices correspond to all the ways to choose the position of one 1 among the three coordinates.

The next diagonal slice corresponding to 3 choose 2 contains all the vertices with exactly two 1's-- 1, 1, 0; 1, 0, 1; and 0, 1, 1.

The final slice corresponding to 3 choose 3 contains all the vertices with exactly three 1's, which is just 1, 1, 1.

What does that tell us about the four dimensional case?

The first diagonal slice will only contain the vertex 0, 0, 0, 0.

The next diagonal slice corresponding to 4 choose 1 will contains all the vertices with 1, 1.

The middle diagonal slice, 4 choose 2, will contain six vertices each with two 1's.

The next diagonal slice, 4 choose 3, will contain four vertices each with three 1's.

And the final slice is just the vertex with all 1's-- 1, 1, 1, 1.

In general, the number of vertices of the diagonal slices of an n dimensional cube are n choose 0, n choose 1, and so on up to n choose n. The n choose k entry corresponds to the slice which goes through all the vertices whose coordinates have k many 1's and n minus k 0's.

So this analogy with Pascal's triangle tells us how many vertices the k-th diagonal slice of an n dimensional cube will have.

But can it also tell us what shape those vertices make?

For example, we just learned the middle slice of the 4D cube has six vertices.

We also know those six vertices will make some three dimensional object, since it's a 3D slice of a 4D cube.

But there's a lot of ways to arrange six vertices in three dimensions-- like this or this.

How do we know what shape they make?

Here's where Pascal's rule can help us.

Pascal's rule tells us that each entry of Pascal's triangle is the sum of the entry to its upper right and its upper left.

For example, 4 choose 2 is equal to 3 choose 1 plus 3 choose 2.

Here's the general statement.

There's actually an analogous version for cube slices.

We can obtain a slice by geometrically adding the two slices above it.

For example, the middle slice of a 2D cube is a line segment.

The slices above it are both single points.

Here's how we add the single points to get the line segment.

Place the single points directly on top of each other and fill in the space between them.

This produces a line segment.

Let's try it again.

The second slice of a three dimensional cube is a triangle.

It's produced by geometrically adding the line segment and vertex above it.

The top and bottom face of a 3D cube are both 2D cubes.

Place the two slices we are combining on these two faces.

Now, draw lines between the vertices and fill it in.

This is the convex hull, which in this case is a triangle.

So back to the question from the very beginning-- what do the diagonal slices of a 4D cube look like?

We can repeat the same procedure we just did.

The top and bottom faces of a 4D cube are 3D cubes.

We already know this slice is made from four vertices.

But now, we can tell exactly what it looks like.

It's the convex hull of a triangle and a vertex placed on their respective locations within the 3D faces.

So it makes a regular tetrahedron.

Similarly, we already knew this slice is made from six vertices.

But now, we know it's the convex hull of the two triangles each properly positioned within the top and bottom face.

This makes a regular octahedron.

So taking the three dimensional hyperplane and sweeping it diagonally across a four dimensional cube, we encounter these five slices-- a vertex, a tetrahedron, an octahedron, another tetrahedron, and another vertex.

Now, you have all the tools necessary to construct the four dimensional slices of a five dimensional cube.

The hard part is visualizing the shapes in four dimensions.

See you next time on Infinite Series.

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Hello.

There were so many wonderful comments in response to our Devil's staircase video.

I want to first start with a clarification.

I said that Cantor set includes all of the numbers whose base 3 expansion contains no 1's.

That's correct, but we have to remember that you can write any terminating 1 as a repeating 2.

So like in base 10, a lot of people know you can write 1 as 0.9 repeating.

You can do the same thing in base 3, except instead 0.90 repeating, you use 0.2 repeating.

And using that little modification, we can say that the Cantor set is all of the numbers whose base 3 expansion contains no 1's.

So thanks for pointing that out in the comments.

Next, a lot of people ask whether the Cantor function has a derivative.

That's a great question.

It is a continuous function, but not all continuous functions have derivatives.

And in fact, the Cantor function does have a derivative at all the points that aren't in the Cantor set.

And there, it's derivative is 0.

On the points that are in the Cantor set, the Cantor function does not have a derivative.

It's just not defined.

Finally, for our challenge winner, we had [INAUDIBLE] who gave a great proof of the uncountability of the Cantor set.

And what I like about this proof is that it spawned a ton of discussion.

And there were some corrections made, and people worked through the problem together.

And it's a really nice discussion, not just the comment itself, but everything that comes under it.

So I highly recommend you read it.

OK, have a good week.