The taxicab norm on is defined by setting where Show that this is indeed a norm. As usual, we can define a distance function by setting Find 2 non-congruent, non-degenerate triangles with sides of length 1, 1, and 2.

This exercise is related to an 80-year old open problem of Paul Erdös. Consider a unit square . Inscribe in exactly squares with no common interior point. (The squares do not need to cover all of .) Denote by the side lengths of these squares, and define

Show that , and that equality holds iff is a perfect square.

(Erdös problem is to find all for which . Currently, it is only known that for all , , , and that if , then is a perfect square.)

(This exercise comes from Jerry Shurman, Geometry of the quintic, Wiley-Interscience, 1997.)A rigid motion of is a function such that

for all vectors (Here, is the usual inner product on .) The goal of this exercise is to show that these maps are precisely the functions of the form where is an orthogonal matrix and .

By definition, a matrix with real entries is orthogonal iff where is the transpose of . Show that this is equivalent to stating that preserves inner products, in the sense that

for all .

Show that if where is orthogonal, then is indeed a rigid motion.

To prove the converse, assume now that is rigid.

Show first that is a bijection.

Let and set Eventually, we want to show that is a linear transformation. Begin by checking that and that for all .

Show that for all , and conclude that preserves addition.

Argue similarly that preserves scalar multiplication, and is therefore linear.

It follows that for some matrix . Conclude the proof by checking that is indeed orthogonal.

Solve exercises 2-7, 10-14 from Chapter 6 of the textbook.

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Dr. Caicedo,
On question #1, v_1 and v_1 are vectors right? The sum of their magnitudes gives another vector’s length… bold v. I’m just verifying that they’re not scalars…

Also, #2: the max of a sum… does that mean that you take the largest value for the possible sums?

1. What I meant was is a vector in It therefore has two coordinates (with respect to the standard basis), and I am calling the first and the second (So, yes, are scalars.)

2. Yes, that is what I meant, the maximum (or sup, perhaps) is taken over all possible sums.

We have not defined “non-congruent” yet. Are we talking about using the standard distance measure and using trig to show they have different angle measures?

Let’s be as generous as it is reasonable, and take congruent to mean same lengths, angles, areas. What I mean is, we want two “obviously different” triangles whose sides have the same lengths nonetheless. (And non-degenerate is intended to eliminate the example of a segment with a point in the middle.)

Dr. Caicedo,
In the assigned problem 3 #5, To show A is orthogonal, do we prove A^T(A)=I? If so, these two matrices can get very messy with 3X3 real entries… is there another way?

On Assigned problem 3 #1, what is required to show 1 to 1 and onto here?
Thanks,
Amy

Hi Amy,
For 3.5, do you have another characterization of orthogonality? (Perhaps one proved earlier somewhere in problem 3?)
I am not sure I understand your question about 3.1. Can you tell me what you mean?

Hi, I am not sure how to show R is a bijection. I suppose I am still unclear what a rigid motion is. I keep going through the motions of showing one thing after another, but am not sure what the big picture is for R.
Thanks,
Amy

The definition of being rigid is the equation at the beginning, for all

What do you get if ? This says that preserves distances. It implies (very easily) that is 1-1 and that it is continuous.

You may remember from vector calculus that where is the angle between the vectors and . Since preserves distances, the equation defining rigid motions then says that the angle between and is the same as the angle between and

In other words: Rigid motions preserve distances and angles. That is what a rigid motion is: A continuous transformation of with those two properties.

The most direct argument for surjectivity that I have thought of is as follows; there are several claims I make along the way that need to be justified, of course, but this is the skeleton:

Let we want to find some such that Pick points in the range of say

Consider the tetrahedron There are only two tetrahedrons in that are congruent to and have corresponding to Say, and Then must be the image of either or

A better layout of the exercise would have been to show first that preserves inner products and then show that bijects. To show that surjects one could also work in coordinates:

Let be the standard orthonormal basis in Since for all with and then is an orthonormal basis.

Next, for any vector and

So to make hit any point, express the point as
Then let
Then

Maybe this immediately shows that is linear, further shortening the exercise?

Very neat solution. The solution I suggested above and the one below depend a bit too much on geometric ideas that don’t seem relevant to the problem, so I like your approach much better.

Here is another way of proving surjectivity. It was suggested by my friend and colleague Ramiro de la Vega. I think that Dr. Shurman‘s solution above is the best one, but all the solutions we have exhibit different ideas which may prove useful in different contexts.

For this approach, first prove that the image under of a straight line is a straight line. This is the key observation.

Surjectivity is now easy: Show now that the image under of a plane is a plane. For this, consider three noncollinear points in a plane and note that is the union of the lines that go through one of these three points and the opposite side in the triangle they determine.

Finally, consider a tetrahedron, and note that any point in space is in one of the lines going through a vertex and the opposite face.

I learned of this problem through Su Gao, who heard of it years ago while a post-doc at Caltech. David Gale introduced this game in the 70s, I believe. I am only aware of two references in print: Richard K. Guy. Unsolved problems in combinatorial games. In Games of No Chance, (R. J. Nowakowski ed.) MSRI Publications 29, Cambridge University Press, 1996, pp. […]

Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see here. What would be an explicit example of a number $r$ with this property? Short of an explicit example, are there any references addressing this question? A natural approach would […]

Suppose $M$ is an inner model (of $\mathsf{ZF}$) with the same reals as $V$, and let $A\subseteq \mathbb R$ be a set of reals in $M$. Suppose further that $A$ is determined in $M$. Under these assumptions, $A$ is also determined in $V$. The point is that since winning strategies are coded by reals, and any possible run of the game for $A$ is coded by a real, […]

Yes. This is obvious if there are no such cardinals. (I assume that the natural numbers of the universe of sets are the true natural numbers. Otherwise, the answer is no, and there is not much else to do.) Assume now that there are such cardinals, and that "large cardinal axiom" is something reasonable (so, provably in $\mathsf{ZFC}$, the relevant […]

Please send an email to mathrev@ams.org, explaining the issue. (This is our all-purpose email address; any mistakes you discover, not just regarding references, you can let us know there.) Give us some time, I promise we'll get to it. However, if it seems as if the request somehow fell through the cracks, you can always contact one of your friendly edit […]

First, prove that the map $\alpha\mapsto\omega^\alpha $ is normal, that is, strictly increasing and continuous at limits. Use this to show that for any $\alpha $ there is a least $\beta $ such that $\alpha

Consider any club subset of $\kappa $. Check that it has order type $\kappa>\lambda $, and that its $\lambda $th element (in its increasing enumeration) has cofinality $\lambda $.

A very nice introduction to this area is MR0891258(88g:03084). Simpson, Stephen G. Unprovable theorems and fast-growing functions. In Logic and combinatorics (Arcata, Calif., 1985), 359–394, Contemp. Math., 65, Amer. Math. Soc., Providence, RI, 1987. Simpson describes the paper as inspired by the question of whether there could be "a comprehensive, self […]

There are continuum many (i.e., $|\mathbb R|$) such functions. First of all, there are only $|\mathbb R|$ many continuous functions, so this is an upper bound. On the other hand, for any real $r$, $f(x)=x+r$ satisfies the requrements, so there are at least $|\mathbb R|$ many such functions.

Dr. Caicedo,

On question #1, v_1 and v_1 are vectors right? The sum of their magnitudes gives another vector’s length… bold v. I’m just verifying that they’re not scalars…

Also, #2: the max of a sum… does that mean that you take the largest value for the possible sums?

I apologize, I meant v_1 and v_2 on my first question.

Thanks,

Amy

Hi Amy,

1. What I meant was is a vector in It therefore has two coordinates (with respect to the standard basis), and I am calling the first and the second (So, yes, are scalars.)

2. Yes, that is what I meant, the maximum (or sup, perhaps) is taken over all possible sums.

Hey Dr. Caicedo,

I think you meant to say in your last comment that .

Nick

[*Sigh* Yes, thanks. Corrected. -A]We have not defined “non-congruent” yet. Are we talking about using the standard distance measure and using trig to show they have different angle measures?

Let’s be as generous as it is reasonable, and take congruent to mean same lengths, angles, areas. What I mean is, we want two “obviously different” triangles whose sides have the same lengths nonetheless. (And

non-degenerateis intended to eliminate the example of a segment with a point in the middle.)Dr. Caicedo,

In the assigned problem 3 #5, To show A is orthogonal, do we prove A^T(A)=I? If so, these two matrices can get very messy with 3X3 real entries… is there another way?

On Assigned problem 3 #1, what is required to show 1 to 1 and onto here?

Thanks,

Amy

Hi Amy,

For 3.5, do you have another characterization of orthogonality? (Perhaps one proved earlier somewhere in problem 3?)

I am not sure I understand your question about 3.1. Can you tell me what you mean?

Hi, I am not sure how to show R is a bijection. I suppose I am still unclear what a rigid motion is. I keep going through the motions of showing one thing after another, but am not sure what the big picture is for R.

Thanks,

Amy

The definition of being rigid is the equation at the beginning, for all

What do you get if ? This says that preserves distances. It implies (very easily) that is 1-1 and that it is continuous.

You may remember from vector calculus that where is the angle between the vectors and . Since preserves distances, the equation defining rigid motions then says that the angle between and is the same as the angle between and

In other words: Rigid motions preserve distances and angles. That is what a rigid motion is: A continuous transformation of with those two properties.

I’ll write about surjectivity a bit later.

The most direct argument for surjectivity that I have thought of is as follows; there are several claims I make along the way that need to be justified, of course, but this is the skeleton:

Let we want to find some such that Pick points in the range of say

Consider the tetrahedron There are only two tetrahedrons in that are congruent to and have corresponding to Say, and Then must be the image of either or

A better layout of the exercise would have been to show first that preserves inner products and then show that bijects. To show that surjects one could also work in coordinates:

Let be the standard orthonormal basis in Since for all with and then is an orthonormal basis.

Next, for any vector and

So to make hit any point, express the point as

Then let

Then

Maybe this immediately shows that is linear, further shortening the exercise?

HTML ate my inner product symbols.

[Edited: I fixed it. Thanks! -A.]Hi Jerry,

Thanks a lot!

Very neat solution. The solution I suggested above and the one below depend a bit too much on geometric ideas that don’t seem relevant to the problem, so I like your approach much better.

There are a few statements that probably need more details (if this were to be turned in as homework), but this is really an effective proof.

Thanks for sharing.

Here is another way of proving surjectivity. It was suggested by my friend and colleague Ramiro de la Vega. I think that Dr. Shurman‘s solution above is the best one, but all the solutions we have exhibit different ideas which may prove useful in different contexts.

For this approach, first prove that the image under of a straight line is a straight line. This is the key observation.

Surjectivity is now easy: Show now that the image under of a plane is a plane. For this, consider three noncollinear points in a plane and note that is the union of the lines that go through one of these three points and the opposite side in the triangle they determine.

Finally, consider a tetrahedron, and note that any point in space is in one of the lines going through a vertex and the opposite face.