An Evening of Math III

I’ve been reading Out of the Labyrinth: Setting Mathematics Free, by Robert and Ellen Kaplan, who are the founders of “the Math Circle.”  As I understand it a math circle is where you get a bunch of people together, ask a couple of provocative questions to get them thinking, and then get out of the way and let them create mathematics.

I decided that I would try something akin to their math circle with my kids.  I settled on using the opening example from the book:  draw a number line with 0 and 1 labeled, ask if there are any other numbers in there, and step back to see what numbers the kids can find.  The idea is that they start marking fractions, get comfortable with those and eventually come up with an irrational number.

A was easily distracted tonight.  She kept wanting to put in numbers beyond 1.  B put in 1/2, then 1/4 in approximately hte right places.  Later he moved them so that he could fit more numbers in.  He ended up with 1/2, 1/3, 1/4, etc. down to 1/13 spaced approximately evenly.  A added 1/14, 1/15, 1/16.  I asked how many more there were.  ” I can’t say how many.  Lots more.  Infinity many.”  Is what B replied.  He ended up writing “infinitely many numbers like these” between 1/16 and 0.  We also ended up with 1/1 somewhere between 1 and 1/2.  For now I’m letting that persist until they figure out that its the same thing as 1.  A wanted to put in 0/2, but B told here that that was the same thing as 0.

A also said that one thousand two hundred thirty four was her favorite number and wrote it out: 1234, with a box around it.

Next week we’ll pull the number line out again and since 0 has so many friends next to it, I’ll ask whether 1 has any.

An Evening of Math I

My wife has taken on the challenge of homeschooling our children this year. My main participation in this is a weekly math session with the kids in the evening on any subject of my choosing!

Tonight was our first session. I decided to do partitions with them. I am priming them to be able to help me with my research on Q1 graphs.

We pulled out cubical blocks and I told the kids to make partitions with them. The hardest part about this is keeping my mouth shut and staying out of their way.

B invented Ferrers diagrams. Meanwhile I set A to work on making all partitions of the small numbers. She found 1 partition of 1, 2 partitions of 2 and 3 partitions of 3. She started working on 4 and found 4 partitions. Then B chimed in with a 5th partition of 4. This upset A and she refused to accept it as a partition, because it didn’t follow the pattern that she had seen. She stormed off, but came back and was ready to accept the 5th partition of 4.

I tried to get them thinking about how we could know that we had got all of them. They haven’t come up with anything along those lines yet. B (of his own volition) started working on an algorithm to generate all of the partitions of a given number. The algorithm needs work, so far only generating the n partitions of n: (1,1,…,1), (2,1,1,..,1), (3,1,1,..,1),…,(n-1,1), (n). He was also working on an algorithm for getting partitions of n+1 from partitions of n. I think that one was pretty incomplete too.

Their minds were still going, but it was getting late, so I sent them to bed. It was a good evening of math.

Partitions of Graphs VII: The Cases of 5 and 7

This is the seventh in a series, the first being here: Part I, the previous here: Part VI.

In the previous post I defined a Q1 graph to be a graph G with q(G)=1.  We found examples of Q1 graphs for every n between 2 and 10 except 5 and 7.

Let’s talk about the case n=1.  There is only one graph with a single node.  Here is a picture of it

o

Obviously p(o)=1 and q(o)=p(1)-p(o)=1-1=0, so there are no Q1 graphs for n=1.  Let’s make the following:

Definition: If P is a partition prohibited by a Q1 graph, we will call P a Q1 partition.

Now for the case n=5.

Theorem 1: There are no Q1 graphs with five nodes.

Proof: Here are the partitions of 5:

5

4+1

3+2

3+1+1

2+2+1

2+1+1+1

1+1+1+1+1

We will eliminate each of these in turn as possible partitions prohibited by a Q1 graph.  In my previous post we saw that 5 cannot be prohibited by a Q1 graph. Let’s state that as a lemma.  Refer to the previous post for a proof.

Lemma 0: Any Q1 graph with n>2 nodes is connected.  In other words,  n is not a Q1 partition.

Also, since 1+1+1+1+1 is allowed by every (five node) graph it cannot be a Q1 partition.  In fact we can extend this result.

Lemma 1: Any partition that has a 1 as a summand is not Q1 (call such a partition a type 1 partition).

The idea behind the proof of Lemma 1 is that any connected graph, say with k nodes, has at least one node that can be cut off by snipping some edges without disconnecting the rest of the graph, to give a connected graph on k-1 nodes (plus a solitary disconnected node).

Lemmas 0 and 1 eliminate all partitions but 3+2 as possibilities for Q1 partitions.  We take care of that with:

Lemma 2: Any partition of the form (m+k)+m+ . . . +m+k+. . .+k (let’s call this a type 2 partition, since it appears in lemma 2) is not Q1.  In other words any partition with summands consisting of a bunch of k, a bunch of m and a single m+k is not Q1.

Proof of Lemma 2: Given a Q1 graph G, it is connected, by Lemma 0.  Next, assuming a type 2 partition P: (m+k)+m+ . . . +m+k+. . .+k is prohibited by G, the partition (m+m+ . . . +m+k+k+. . .+k must be allowed by G.  But since G is connected at least one of the size m subgraphs must be connected via an edge to at least one of the size k subgraphs.  Color the nodes in these two subgraphs the same color to obtain a good coloring of G corresponding to P.  Therefore P is not prohibited, a contradiction.  Therefore P is not Q1. end of proof of Lemma 2.

3+2 is a type 2 partition, therefore it is not Q1.  Since there are no partitions of 5 that are Q1, there are no 5-node Q1 graphs.  This concludes the proof of Theorem 1.

Theorem 2: There are no 7-node Q1 graphs.

Proof: Here are the partitions of 7:

7

6+1

5+2

5+1+1

4+3

4+2+1

4+1+1+1

3+3+1

3+2+2

3+2+1+1

3+1+1+1+1

2+2+2+1

2+2+1+1+1

2+1+1+1+1+1

1+1+1+1+1+1+1

Eliminating the type 1 partitions and the partition 7 we have left:

5+2

4+3

3+2+2

These are all type 2 partitions:

5+2 = (3+2)+2

4+3 = (3+1)+3

3+2+2 = (2+1)+2+2

End of proof of Theorem 2.

We have seen that 5 and 7 do not permit Q1 graphs, but 9 does.  In fact the reason 9 does is that it is divisible by 3.  One might be tempted to guess that prime numbers do not admit Q1 graphs (with 2 and 3 being flukes).  Of course the next number to test this guess on is 11.  The case of n=11 will be the subject of my next post.

Part VIII: The Case of 11 (and 17)

Partitions of Graphs VI: Q1 Graphs

This is the sixth in a series.  The first is found here: Part I.  The previous is found here: Part V.

I have defined q(G) to be:

q(G) = p(n) – p(G), where n is the number of nodes in G.

In other words q(G) is the number of partitions prohibited by G.  We’ve seen graphs, G, with p(G)=1, that is with exactly one allowed partition.  In fact we know that the one allowed partition is:

1+1+1+ . . . +1

and the graph consists of n nodes and no edges.  But let’s say that we want a graph, G, with q(G)=1.  I’ll call such a graph a Q1 graph (in fact if q(G)=k lets call it a Qk graph).  It’s not quite as easy to find Q1 graphs as it was to find graphs with p(G)=1.  For one thing it isn’t at all obvious what the one prohibited partition should be.  A first guess might be that the prohibited partition should be n.  This sounds plausible since in some sense the partition n is the alter-ego of the partition 1+1+1+ . . . +1.  Following this lead we should look for graphs that are not connected, but that allow every partition of n other than n.  Indeed, initially we find success:

     q( o o ) =  1

and

     q( o o-o) = 1

So far so good.  We’ve done n=2 and n=3 (conspicuously skipping n=1, I’ll leave that case for you to ponder), but when we get to n=4 we hit trouble.  If q(G)=1, with the partition 4 prohibited, then 3+1 must be allowed.  That means that for a good coloring there are going to be three nodes that are the same color.  All three nodes have to be part of a connected subgraph, but the other node, the fourth node, can’t be connected to those other three at all.  In other words, the largest connected component of the graph has three nodes.  If it were any bigger it would have four nodes and the whole graph would be connected, which we assumed wasn’t true.  So there is a single solitary node floating all by it’s lonesome in space.  But 2+2 is also allowed (every partition other than 4 is assumed to be allowed) which tells us that each node of G is adjacent to at least one other node (two nodes get colored red, two blue in a good coloring).  But what about our single, solitary node floating in space?  This says that he can’t be single or solitary or floating in space.  You can’t be single, solitary and floating in space when you are closely tethered to your best buddy.  That’s what we call a contradiction, which means that our hypothesis that q(G)=1 with 4 prohibited is impossible.

Of course in a previous post we’ve seen that

            o
           /
     q( o-o-o   ) = 1

with the prohibited partition being 2+2 (This is S4, I’ve drawn it slightly differently here).  It took me some time to see that this is just the first of a whole family of graphs with q(G)=1.

Remark: If n is even, then

                      o
                     /
     q( o-o-o-...-o-o-o   ) = 1  (n even)

and the prohibited partition is 2+2+2+ . . . +2.  If n is odd then

                      o
                     /
     q( o-o-o-...-o-o-o   ) = 0  (n odd)

Proof: We need to show that if there are an even number of nodes then 2+2+2+ . . . +2 is prohibited and all other partitions are allowed.  To see that 2+2+2+ . . . +2 is prohibited is pretty easy.  What would a good coloring look like?  The node farthest to the right and on the bottom would be some color, let’s say it’s purple.

                   o
                  /
     o-o-o-...-o-o-o

There must be two purple nodes and they have to be adjacent so the next node to the left has to be purple also.

                   o
                  /
     o-o-o-...-o-o-o

But where does that leave the top node?  High and dry, without a friend in the world.  In other words it has to be a color other than purple, let’s say orange and it has to have an orange buddy, but there are none to be found.

                   o    (Which other node should be orange?)
                  /
     o-o-o-...-o-o-o

So 2+2+2+ . . . +2 is prohibited.  Now we prove that every other partition is allowed.  If P is a partition whose summands are not all 2‘s then either P has a 1 as a summand or P has a summand that is 3 or greater.  We consider thees two cases.  First if P has a 1 as a summand then say that 1 gets colored purple.  Then we color the node on the top purple

                   o
                  /
     o-o-o-...-o-o-o

no other node is colored purple, and what remains of the graph is a path, so we will be able to find a good coloring for it.  Here is an example of what I’m talking about.  Say n=8 and P=4+3+1.  We color P: and use this as instructions for coloring the graph.  First we color the top node purple and then we color the first four of the path red and the next four green.

                 o
                /
     o-o-o-o-o-o-o

Here’s another example.  A good coloring of 2+2+2+2+2+1+1:

                         o
                        /
     o-o-o-o-o-o-o-o-o-o-o

Let’s address the other case of there being a summand of P greater than or equal to 3.  Say the partition is 6+5+5+2+2+2, (n=22).  We color P: 6+5+5+2+2+2 .  Now we pick one of the summands which is 3 or greater.  Let’s say we pick one of the 5‘s.  We color the 5 nodes farthest to the right the color of that 5.  Let’s say we picked the green 5.  What’s left over is a path.

                                             o
                                            /
     o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o-o

In general pick any summand larger than 2 and color the right most nodes that color.  What’s left will be a path and allow a good coloring no matter what the other summands are.  This proves that all partitions other than the one consisting entirely of 2′s are allowed.  Finally odd numbers do not have a partition consisting entirely of 2′s, so every partition of an odd number is allowed. QED.

So we have this family of Q1 graphs:

                   o
                  /
     o-o-o-...-o-o-o

for even numbers, n.  It turns out there is a similar family of Q1 graphs when n is divisible by 3.  Here it is:

                   o
                  /
     o-o-o-...-o-o-o-o

The proof that this is a Q1 graph when n is divisible by 3 (and a Q0 graph otherwise) is similar to the above proof.  The prohibited partition is 3+3+3+ . . . +3.  If P is a partition whose summands are not all 3, then there are three cases to consider: P has 1 as a summand, P has 2 as a summand and P has a summand that is 4 or larger.  Here are three pictures illustrating the three cases respectively.

                   o
                  /
     o-o-o-...-o-o-o-o

                   o
                  /
     o-o-o-...-o-o-o-o

                   o
                  /
     o-o-o-...-o-o-o-o

In each case what is left over is a path, so the rest of the graph can be given a good coloring.

We’ve found two families of Q1 graphs that give us examples for n even and n divisible by 3.  So we have Q1 graphs for all n between 2 and 10 except n=5 and n=7.  In my next post I will address the cases n=5 and n=7.  In subsequent posts n=11 and n=13 will be addressed.

Part VII: The Cases of 5 and 7

Yet more sentences

A: Sentences B and C together have 2 1s, 6 2s, 5 3s, 4 4s, 6 5s, 4 6s, 3 7s, and 2 8s.

B: Sentences C and D together have 2 1s, 6 2s, 4 3s, 5 4s, 7 5s, 3 6s, 3 7s, and 2 8s.

C: Sentences D and E together have 2 1s, 5 2s, 6 3s, 5 4s, 5 5s, 4 6s, 3 7s, and 2 8s.

D: Sentences E and F together have 2 1s, 5 2s, 7 3s, 4 4s, 4 5s, 5 6s, 3 7s, and 2 8s.

E: Sentences F and G together have 2 1s, 6 2s, 6 3s, 4 4s, 5 5s, 3 6s, 3 7s, and 3 8s.

F: Sentences G and H together have 2 1s, 5 2s, 7 3s, 4 4s, 6 5s, 2 6s, 3 7s, and 3 8s.

G: Sentences H and I together have 2 1s, 4 2s, 8 3s, 5 4s, 5 5s, 2 6s, 3 7s, and 3 8s.

H: Sentences I and J together have 2 1s, 5 2s, 7 3s, 5 4s, 4 5s, 3 6s, 3 7s, and 3 8s.

I: Sentences J and K together have 2 1s, 5 2s, 8 3s, 4 4s, 3 5s, 4 6s, 3 7s, and 3 8s.

J: Sentences K and L together have 2 1s, 5 2s, 7 3s, 6 4s, 2 5s, 4 6s, 3 7s, and 3 8s.

K: Sentences L and M together have 2 1s, 6 2s, 4 3s, 8 4s, 3 5s, 3 6s, 3 7s, and 3 8s.

L: Sentences M and N together have 2 1s, 7 2s, 4 3s, 6 4s, 4 5s, 2 6s, 4 7s, and 3 8s.

M: Sentences N and O together have 2 1s, 8 2s, 3 3s, 4 4s, 5 5s, 4 6s, 4 7s, and 2 8s.

N: Sentences O and P together have 2 1s, 7 2s, 3 3s, 4 4s, 7 5s, 5 6s, 2 7s, and 2 8s.

O: Sentences P and A together have 2 1s, 6 2s, 4 3s, 5 4s, 6 5s, 5 6s, 2 7s, and 2 8s.

P: Sentences A and B together have 2 1s, 6 2s, 5 3s, 5 4s, 4 5s, 5 6s, 3 7s, and 2 8s.

A couple more sentences.

Sentence A: Sentence B has 1 1, 3 2s, 2 3s, 3 4s, 2 5s, 3  6s, 1 7, and 1 8.

Sentence B: Sentences A and B together have 5 1s, 6 2s, 6 3s, 4 4s, 3 5s, 4 6s, 2 7s, and 2 8s.

Sentence C: Sentence D has 5 1s, 2 2s, 1 3, 3 4s, 1 5, 1 6, and 1 7.

Sentence D: Sentence E has 4 1s, 4 2s, 1 3, 1 4, 2 5s, 1 6, and 1 7.

Sentence E: Sentence C has 5 1s, 2 2s, 2 3s, 1 4, 2 5s, 1 6, and 1 7.

An alien is thinking of a number…

…and you have to guess what it is.

You were captured by a malevolent alien, who would really like to eat you for dinner. He told you his name, but since it is unpronounceable to humans, you like to call him Evil Eddie. Evil Eddie has a morbid sense of fairness and so, before roasting you with an apple stuck in your mouth and serving you with a side of glowing, purple, French-cut string beans, has locked you in a chamber carved out of the bedrock of a strange world. Before the polycarbide door slides shut with a hiss, Eddie tells you that you he is thinking of a number. If you can guess it, he will reluctantly release you. If not, well…let’s just hope that you can guess the number.

As you look around the cavern you see, in the dim and flickering light (does it come from torches? You don’t see any.), an enormous stone bowl, filled to the brim with glass marbles. How many are there? Thousands at least. Tens of thousands, perhaps. On the other side of the chamber is an identical bowl, but this one is empty. In the middle of the room, hanging from the ceiling is a rope. You hear Eddie’s voice thundering into the room “when you think that you know the number, put that many marbles in the empty bowl. Then pull the rope.”

Fortunately you have a friend. In an adjacent chamber is a benevolent alien. You don’t know his name, so you decide to call him Manvel. By some contrived plot device that the author has not taken the time to think of, you know that Manvel:

1) can read Evil Eddie’s mind (hence, knows the number),

2) is going to try to transmit this number to you, and

3) is going to do so by tapping out the digits (through the wall) in some base (could be base 10 (that’s what we normally use), could be base 2 (called binary. We think of computers as using base 2), could be any whole number base.

You listen and hear …tap… long pause …tap tap tap tap tap… long pause …tap tap tap… then silence.

You have the digits! They are 153. As stated above, you now know the number is:

You just don’t know what is. (I’ve glossed over the issue of big endian versus little endian order, but we are assuming, by aforementioned plot device, that big endian order has been used by Manvel).

So…how many marbles do you put in the bowl?

to be continued…

Baseball

Happy Ruth-Aaron days!

Sequences and Creative Math for Kindergartners

I visited my kids’ classes for career day in May. I have a son in Kindergarten and a daughter in pre-K. I also visited two other kindergarten classes. I told the kids that I am a research mathematician: mathematician means I do math. Research means I do math that nobody has ever seen before, math that I make up.

I mentioned that math is lots of things: It’s thinking about shapes and numbers and patterns, and other stuff. This day we were going to do numbers and patterns.

I showed them a board covered with three lines of (blank) sticky notes. Under each note was a number.The point is to try to guess which number is coming next.

We started with the blue line of numbers.

Once I had uncovered the 1 and 2 almost everyone expected the next number to be 3.

When I showed them that it was a 1, they caught on pretty quickly.

They had fun shouting out the 1-2-1-2-1-2 pattern as I frantically tried to keep up with them on the board.

Next, the pink ones.

I started by showing the 1-2-1 with a warning that I was being tricky. A 2 should come next, right? Wrong! But I didn’t tell them what it was yet. I put the sticky notes back on and explained that I was going to pull them off in a different order to see if it gave them a hint. (At this, some kids suggested I pull them off starting at the right instead of the left.)

I pulled off every other sticky: 1-skip a sticky-1-skip a sticky-1-skip… Okay, done with the ones. Next we have:

2-skip a sticky-2-skip a sticky-2-skip… (pulling off every other sticky that is left after the first pass). Okay, done with the twos. So the next one should be (the one that everyone thought should be a 2):

It’s a 3! And most of the students guessed that it was a 3!

And so it goes: 3-skip a sticky-3-skip a sticky-3 skip…

Next we have the fours: 4-skip a sticky (at this point we hit the end of the board, but if the board were longer…we would continue 4-skip a sticky-4-skip a sticky-4-skip…

Only one pink number left…

Everyone knows it’s a 5 by now! If we had a long enough board we would have 5-skip a sticky-5-skip a sticky-5-skip…

The pink numbers are:

1-2-1-3-1-2-1-4-1-2-1-3-1-2-1-5-1

Some interesting things (I think) about these pink numbers: Once you’ve done the ones and twos the row looks exactly like it would if you had pulled off the same patterns on the blue row! Nevertheless, most kids pick up on the fact that 3 should come next (well…I did give a big clue “done with the ones! . . . Done with the twos!”)

These pink numbers are called the ruler sequence, which comes up in the towers of Hanoi problem.

Now for the greens (the stickies look yellow in these pictures, but they are really green (yellowish-green)).

What do you think I will start with? Most kids had figured that I am into ones by now. So naturally, I start with a one again.

This row is really sneaky (but usually the kids understand it before their teachers! A testament to the freedom that young minds who haven’t had mathphobia drilled into them are capable of!), so I give a big hint. What do we have so far (on the green row)? One. How many ones? One one.

So that is what comes next. You start with a 1 then you say “look, it’s 1 one,” and that’s what you put next 1-1.

Now what do you have? 1-1-1. You have 3 ones. So next comes:

3-1.

Now you have 1-1-1-3-1. Four ones and one three. So next is:

4-1-1-3.

Now you have:

1-1-1-3-1-4-1-1-3. That is, six ones, two threes, and one four. So next will be:

6-1-2-3-1-4.

And now you have:

1-1-1-3-1-4-1-1-3-6-1-2-3-1-4. That is eight ones, one two, three threes, two fours and one six.

There is only room on the board to put 8-1. But if we had more room we would keep going. If we had paper going all the way around the school… I asked the kids if they thought we would ever get a 5. We haven’t so far! (At this someone always pointed out that we got a 5 on the pink line. A good observation! But we are just looking at the green line now.) What about 10? If we kept going would we get a 10? Would we ever get a 20? Or 100? Would we ever get a thousand or a million or a billion?

I’m pretty sure that you do get a 5 and a 10. I don’t know about 20, 100, 1,000, 1,000,000 or 1,000,000,000. I’m guessing that you do get them… Why? well, my reasoning is this: in some sense you have infinitely many chances to hit those numbers. Maybe at one point you will have 20 ones. If not, maybe you will get 20 twos. Of course maybe you will skip right over 20 twos. So maybe you will get 20 threes or 20 fours or 20 fives or…, you get the idea.

I told the kids that I didn’t know if you ever get 20 or 100 or 1,000. All I know is that you keep getting bigger and bigger numbers (can you see why?). I pointed out to the kids that this is a math question that I don’t know the answer to, and (as far as I know) neither does anybody else. This is math research, and it’s math research that they can do! They can figure out if we ever get a 5 or a 10. A million…probably they won’t figure that out until they learn some computer programming…

The kids were perfectly willing to accept that this is a math question that has an answer, but nobody knows that answer yet.

As for me, I’m conjecturing (conjecture=educated guess) that if you continue with the blue numbers, you will eventually hit any positive, whole number that you might care about. I already told you my reasons for making this conjecture. Can you prove it? Can you prove me wrong? If you’re a teacher, give the problem to your students. Maybe they’ll have an idea about it. Then let me know.

Finally, I wanted to let the kids make up their own number sequences. I handed out sheets of paper with five slots for numbers and five sticky notes. I told them that they could write any numbers they want in the blanks. They could be tricky, or they could be easy (like 1-1-1-1-1). Then, cover them up, and see if someone can guess the numbers.

Do they have to be able to articulate some reasoning for the patterns behind their numbers? No. But if they can, that’s great. There doesn’t have to be any reasoning at all. The point of this exercise was for the kids to do some free-form creative math. Just release your inhibitions (actually the kindergartners don’t have many inhibitions. The grown-ups on the other hand…) and write down some silly or crazy or boring or weird or hard-to-guess or happy, or sad pattern. The kids loved this part. They especially loved seeing if I could guess their numbers.

Let me know what you think about these sequences and this activity!

Thinking about infinity

As promised, a post about how natural it is for humans to think of infinity (see also My Grandma).

Let me say first that there are some very big and important questions about infinity that mathematicians and philosophers are still trying to answer (and I’m just talking about mathematical infinity here!). Unfortunately I have nothing to say here about those questions. This post is very naive.

I hadn’t planned on doing it so soon but John Lienhard did a piece on infinity on the Engines of Our Ingenuity. So it seemed like an appropriate time.

As I said, the question that I’m interested in here is: How natural is it for people to think about infinity? I want to consider three cases: my son, myself and my grandmother.

My Son

One thing that has me thinking about thinking of infinity is that my son has been talking about infinity lately. He is 6 years old and in kindergarten. He has asked me several times “Dad, is infinity a real number.” This catches me off guard and I usually mumble “umm…no. It’s an extended real.” Of course this is not at all helpful and usually prompts a “Huh?” from him.

I just asked my 4-year-old daughter what infinity is. She said that her teacher hasn’t told her yet.

What I wonder is, where did he even learn the word infinity? Where and when did he learn what it means? I don’t remember ever specifically teaching him what infinity is. Then again, I do remember him asking me questions about what the biggest number is. I told him that if you had a number that you thought was biggest you could always add one to it. . . which leads me to. . .

Myself

I don’t remember where or when I learned the words infinity or infinite. On the other hand I remember exactly where and when I learned the word finite. It was in the fifth grade. I immediately understood what it meant: not infinite. The word finite itself seemed very funny to me.

I know that I had started thinking about infinity much sooner. When I was, oh probably 6 or 7, I guess. I remember trying to come to grips with the following paradox: it was absolutely impossible for me to comprehend space going on forever and it was absolutely impossible for me to comprehend space “stopping” at some point (what would be just beyond that point?). It seemed that one of these alternatives had to be true (I hadn’t yet learned that there are compact, boundaryless 3-manifolds).

So how natural is it for us to think about infinity? First of all, from studies that psychologists have done with newborns it appears that we are born thinking about numbers (at least the numbers “one.” “two,” and “three or more.” I don’t think that thinking about infinity comes that early.

So here is my hypothesis (prompted by the above two anecdotes): at some point around 6 years of age it is natural for people to think “do numbers (or space) ever end? If they do, why can’t you just go one step further? But it seems impossible that they could go on forever. The resolution of this paradox, for most of us at least, is to construct for ourselves and accept the concept of infinity.

I could be totally wrong on this. I am no psychologist or philosopher or anything or the sort.

My Grandma

Of course the one monkey-wrench in the works of my hypothesis is my grandma. I can’t be sure whether she understood the concept of infinity or not (see My Grandmother).

So this post is more of a question and request: What are your fist memories of thinking about infinity? How natural do you think it is for us to think about infinity? Please share your ideas.