Wednesday, April 29, 2009

The 0'Factor, episode 10

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(C)Copyright 2009, C. Burke. All rights reserved.

It's that Law of Gravity that's really keeping me down...

By the way, I'm aware that Rich Bond isn't really a money guy. But c'mon! It's just too good a name not to use.

Previous episodes of The 0'Factor: 01, 02, 03, 04, 05, 06, 07, 08, 09.

Monday, April 27, 2009


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(C)Copyright 2009, C. Burke. All rights reserved.

You know they are, but you have to wait for Geometry for the proof.

Sunday, April 26, 2009

Sunday Comic: Super-Stick-Man 52: Super-Stick-Man and the Avalance

Dateline: Stick-World!
In the mountains north of The City, there's always snow, which means that there's always skiing. And where there's skiing, there are skiers in trouble. And where there's trouble, there is Our Hero!

Click on the link to the comic and follow the exciting story!

Saturday, April 25, 2009

Motivational Math Poster

I saw this one, thought about it, decided to post it:

Geocities closing down, closing down, closing down...

Yahoo has announced that they have decided to get rid of all the free Geocities web sites and encourage users to fork over a lot of money for one of the premium webhosting sites. Not right away, of course, but later this year.

How does that affect my blogger account?

Well, it so happens that my first school website was a geocities website. It had a minimum amount of success with my then-current students. However, one of the problems was that there was an advertising bar on the side of the page. It could be closed, but it reopened with every click. It was useful or appropriate for a school website.

However, the page offered 15 Mb of free storage, which didn't have advertising attached. Most of the images that are on this website are actually stored on geocities, which is also why, on occasion, they all disappear for about an hour or two when the bandwidth is too high. (That doesn't happen too often because this page doesn't have that kind of popularity.)

What it means for my page is that I'll have to find someplace else -- cheaper than $10/month -- to store all the images. I seem to have problems with linking directly to the images, so that isn't a solution. But I need to find a solution soon, because I have to edit more than half the posts on this blog.

I will also have to hurry up and copy all the current book reviews on my other book blog Real Soon Now.

UPDATE: Well, two choices have emerged. First, I tried linking an image from Flickr and that seems to work, or I can use a Google page. Yahoo, which owns Geocities, also owns Flickr, which I've been using for a while and already has some of my images. Google, on the other hand, owns Blogger, where this blog resides. Are you thinking what I'm thinking?

I tried using Blogger for my images before and it was horrible, but I believe that was before Google bought it. So I have a decision to make. Any input from anyone that uses either one, or just from my legion of readers (who number almost as many as the Legion of Super-Heroes ... at least, the version from the late 70s) who want to give an opinion.

Besides, I really, really appreciate comments, even the ones from people who aren't related to me or owe me money.

Friday, April 24, 2009


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276 Tesseract
(C)Copyright 2009, C. Burke. All rights reserved.

You are traveling through another dimension, a dimension not only of sight and sound but of bad puns . . .
Yes, the fourth-dimension is a long way to go for a bad pun, but this comic knows no upper bounds (or lower bounds, for that matter).

For more on Henry's and Ian's adventures, click on the following links:

Through a Window Darkly: Part I,II,III, IV, V, VI, TVH, TUC.

Wednesday, April 22, 2009

Identity Property

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(C)Copyright 2009, C. Burke. All rights reserved.

You've never heard of a talking horse?
Well, listen to this.

Tuesday, April 21, 2009

Pythagorean Triples: An Easier Way

Almost done. Promise

There Has to Be an Easier Way

Continued fractions are a pain (to be kind), and my first attempts resulted in errors that were compounded in later steps. (I had 84, 85, 119 at one point. Not quite.)
However, I noticed that the numerators and denominators of the convergents of the continued fraction of follow a pattern

P1 = 1, Q1 = 0, Qn+1 = Pn, Pn+1 = 2 X Pn + Qn

To follow my method, calculate 1 + Q/P as an improper fraction.
As in my prior post, this will give you a triple in the form (a+b)/c, where b = a + 1.

This method was "easier" for me than what Kraitchik (1953) wrote, but it required more iterations to get each Pythagorean Triple.

Kraitchik offers a shortcut, which is essentially:
a = 2PQ and b = P2 - Q2, which yields numbers such that
| a - b | = 1

For example:
a = 2(2)(1) = 4, b = (2)2 - (1)2 = 3, c = 5
a = 2(5)(2) = 20, b = (5)2 - (2)2 = 21, c = 29
a = 2(12)(5) = 120, b = (12)2 - (5)2 = 119, c = 169

The same triples were reached in half the steps. The downside is that these numbers get so big as to be unusable. We didn't use calculators when I was in high school, so it is not surprising that I never saw these:

a = 2(29)(12) = 696, b = (29)2 - (12)2 = 697, c = 985
a = 2(70)(29) = 4060, b = (70)2 - (29)2 = 4059, c = 5741
a = 2(169)(70) = 23,660, b = (169)2 - (70)2 = 23,661, c = 33,461
a = 2(408)(169) = 137,904, b = (408)2 - (169)2 = 137,903, c = 195,025
a = 2(985)(408) = 803,760, b = (985)2 - (408)2 = 803,761, c = 1,136,689
a = 2(2378)(985) = 4,684,660, b = (2378)2 - (985)2 = 4,684,659, c = 6,625,109

Last two notes about patterns:
First, notice that every value for c later shows up as P and Q. Keep expanding the list if you don't believe me. I dare you.
Second, notice that the value of b - a alternates between -1 and +1 for each step. What does that mean? I don't know. I just found it curious.

Kraitchik, Maurice (1953). Mathematical Recreations, Second Revised Edition, Dover Publishing.

Monday, April 20, 2009


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(C)Copyright 2009, C. Burke. All rights reserved.

They apparently caught a showing of George Bernard Shaw's Getting Married.

Sunday, April 19, 2009

Sunday Comic: Super-Stickman 51: Super-Stick-Man and the VILLIAN

Dateline: Stick-World!
The Evil Villian is at it again. Find out more!

Click on the link to the comic and follow the exciting story!

Saturday, April 18, 2009

Pythagorean Triples: Consecutive Legs

Almost done now. Honestly. Maybe one more entry after this one. Maybe two.

On page 100 of Mathematical Recreations, by Maurice Kraitchik, Second Revised edition, Dover Publishing, 1953, the author poses this question:

Find a right triangle whose legs are consecutive integers

The answer is not the 3 - 4 - 5 triangle (again!), but an infinite set of such triples (of which 3-4-5 is a member). I have to admit that I didn't follow the logic the first time through and had to re-read the entire section, but I could sum it up with my own method once I discovered that was involved, along with the fraction:
, which converges on .

It made sense to me that would be involved with Pythagorean triples because of the numbers association with right isosceles triangles. As the legs of the triangle increase in size, the ratio of two consecutive numbers comes closer to 1:1.

Followed by:
99/70 and 239/169
If you look at the fractions with the odd denominators, you have a Pythagorean Triple in the form of
(a+b)/c, where a and b are consecutive numbers (i.e., b = a + 1)

7/5 gives you 3, 4, 5.

41/29 gives you 20, 21, 29

239/169 gives you 119, 120, 169

There has to be an easier way to find these triples!

There is ... and that's on the next page of my journal.

Friday, April 17, 2009

Casting Out Nines

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(C)Copyright 2009, C. Burke. All rights reserved.

What's the name of the guy who owns the boat?
No, What's the name of the guy who sold the bait.

I'm still mad at myself. After finishing the picture, I remembered that I wanted to give Ernie a goofy fishing hat. Oops.

Thursday, April 16, 2009

Ancient Klingon Math

Even I am not deranged enough to think this up, and even if I were, I probably couldn't have programmed it.

There is a site called Mr. Klingon, which contains a link to a page on Ancient Klingon Math along with an online calculator and downloadable app.

Sadly, it's little more than a bunch of operations performed in Base 3, without a zero (it uses the "3" instead), but I admire the enthusiasm required to make a page like this in the middle of all that other geekery.

Pythagorean Triples: Are There Other Forms?

continuing my little self-indulgent trip through my math journal...

a2 + b2 = (b + 3)2

My original notes (not in my journal) had a mistake or an omission in them, but my discovery in part:

a2 + b2 = b2 + 6b + 9 and
a2 = 6b + 9 = 3(2b + 3)

So a2, and therefore a, is a multiple of 3.
The omission is proof that b (and therefore c, which is 3 more than b) is also a multiple of 3.
This would mean that the triple can be reduced.
All examples that I can find in this form can be reduced. That isn't proof. I need a new approach to showing that this is always the case.

a2 + b2 = (b + 4)2

This one is actually easier to show than the previous one.

a2 + b2 = b2 + 8b + 16 and
a2 = 8b + 16 = 8(b + 2)

Solving for b yields
b = (1/8)a2 - 2
c = b + 4, so
c = (1/8)a2 + 2

The only way for (1/8)a2 to be a whole number is if a is a multiple of 4.
If a = 4n then
(1/8)a2 = (1/8)(4n)2 = (1/8)(16n2 = 2n2,
which is even.
So a is even, b, which is 2 less than an even number, must be even and c, which is 2 more than an even number, must be even.
Therefore, any Pythagorean Triple in this form can be reduced by a factor of 2.

I am content to believe at this point that checking any Pythagorean Triple of the form
a2 + b2 = (b + n)2,
where n is a Natural Number and n > 2

would be a waste of time. Particularly because I've crunched hundreds of numbers on an Excel spreadsheet. Not proof, naturally. But if any triples exist, they will be so high as to be useless on an Algebra test.

However, I am not quite done. There is one more kind of triple that I hadn't been aware of until (relatively) recently. And I found it in a fifty-year-old book. I guess I never encountered triangles like these when I was in school because I think I would have remembered!

For my students

Can you guess what the next set of right triangles will look like?
Maybe your previous teachers taught you more than mine did.

Wednesday, April 15, 2009


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(C)Copyright 2009, C. Burke. All rights reserved.

PIrates were among the first to circle the globe.

Did you know that if Planck's constant is reduced by a factor of (2 X Pi), the resulting value is called . . .
(drumroll please)
the reduced Planck constant?

Tuesday, April 14, 2009

Pythagorean Triples: a2 + b2 = (b + 2)2

continuing my little self-indulgent trip through my math journal...

Once I knew that 8, 15, 17 was a Pythagorean Triple, I knew that there had to be others waiting to be found. And, naturally, 4, 3, 5 was just a little too obvious for me to notice. The flip side of this was if, for example, I wanted a triangle where the shortest side was 12, could I find the other sides easily.

a2 + b2 = (b + 2)2
a2 + b2 = b2 + 4b + 4
a2 = 4b + 4
a2 = 4(b + 1)
b=a^2/4 - 1
c = b + 2
c = (1/4)a2 - 1 + 2
c = (1/4)a2 + 1
c=a^2/4 + 1

Because a is even, a2 is even and a multiple of 4. (2n)2 = 4n2.
Therefore (1/4)a2 is a Whole Number.
This works for every value of a that is an even number. However, when (1/4)a2 is an odd number, b and c will also be even numbers. That means that they are not relatively prime and can be reduced by at least a factor of 2.

Only multiple of 4 produce reduced triples.

Since a is a multiple of 4, then a = 4n and
c=a^2/4 + 1
I prefer the latter form it is easier to halve and square then to square a bigger number and then divide by four.

If I see __, 63, 65, I know that I can start with 64, take the square root (which is 8) and double it to get 16.

Granted, I'm not really solving any of these. I'm just looking for problems for my students to solve the old-fashioned way.

What about triples in the form of a, b, b+3 or a, b, b+4 or higher?
That will be the next installment.

Monday, April 13, 2009

Vulcan Economics

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(C)Copyright 2009, C. Burke. All rights reserved.

The biggest Vulcan insult is Drop Dead and Go Broke, represented by inverting the Vulcan salute and wiggling the fingers, showing udder contempt.

Sunday, April 12, 2009

Happy Easter!

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(C)Copyright 2009, C. Burke. All rights reserved.

Have a happy and healthy Easter Sunday,and don't eat too much chocolate.
(Save some for tomorrow.)

Obligatory Math Content:If you ate half of your chocolate and jellybeans today, and have of what was left over tomorrow, and half of what was left over from then the next day, and so on and so on, when would all the chocolate and jellybeans be gone?

Answer: if you were that miserly with your candy, I'd get fed up with you already and finish it for you by Thursday!

Saturday, April 11, 2009

Pythagorean Triples: a2 + b2 = (b + 1)2

So I wanted to find a quick way to find the Pythagorean Triples for right triangles, which had the smallest leg 9, 11, 13, etc. Simply substituting (b+1) in the Pythagorean Theorem in place of c, I could solve for b in terms of a. Then adding 1 to the expression of b yields an expression for c.

a2 + b2 = (b + 1)2
a2 + b2 = b2 + 2b + 1
a2 = 2b + 1
c = b + 1
c = (1/2)(a2 - 1) + 1
c = (1/2)(a2 + 1)

Look at the eqaution a2 = 2b + 1 again.

This spurred my curiosity. Or rather it was the result that I was looking for after a student used an incorrect procedure and yet solved three out of five right triangle problems correctly.

When given a leg and the hypotenuse, the student knew that to find the other leg, squaring and subtraction were involved, but didn't have the idea down pat.

To solve 5-__-13, he squared 5, getting 25, and subtracted 13, getting 12, which is the correct answer, He did the same thing for four other problems.

The "method" works for triangles in the form a2 + b2 = (b + 1)2, when looking for b or b + 1.

If b = a2 - c (which was the method used)
b = a2 - (1/2)(a2 + 1
and therefore
which is the condition that must be true for the method to work.

Okay, I admit, I'm rambling a bit, but when the "method" worked 3 out of 5 times, I needed to know why it had worked then and under what conditions it would work again. And given what I know now, it would be hard to fault the "logic" behind it if one recognized the pattern in the numbers. I'm fairly certain that the student didn't see any pattern here, which explains the two wrong answers, but it did point me back in the direction of finding better examples to use to better ascertain if they understood the work.

Friday, April 10, 2009

Life is But a Dream

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(C)Copyright 2009, C. Burke. All rights reserved.

And if it is a dream, can you wake up from it?
Or is it like that Twilight Zone episode?

Wednesday, April 08, 2009

What's Up?, Hypertext edition

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(C)Copyright 2009, C. Burke. All rights reserved.

Sup is actually a precursor of wassup, not a derivative of it.

See also How Many Times Has Tokyo Been Destroyed?

Tuesday, April 07, 2009

Pythagorean Triples: Introduction

Sometimes I get a bug in my head about something and it won't go away until I write it down and work it out. This time I recorded it in a journal, so I decided to upload it here on a non-comic days. Comments are welcome, especially from my students.

A few years back, I got tired of every right triangle problem involving a 3-4-5, 6-8-10, 30-40-50 or 5-12-13 triangle. I started creating a list of other triples that I could use when avoiding irrational numbers.

First discovery: Triples are either E2 + O2 = O2 or E2 + E2 = E2
(where E is an even number, and O is an odd number), and the latter form could always be reduced to the first form. For need of a term, let's say that the triples that are reduced are in the simplest form, which means that the numbers are relatively prime.

Second discovery: Many (a, b, c) triples could be reduced to (a, b, b + 1), when a was odd, or (a, b, b + 2), when a was even. (There are some that don't fit either, but those will have to wait.)

Squares can be made by summing consecutive odd numbers. When that addend is a perfect square, the result is a Pythagorean Triple.
9 = 32 ==> 32 + 42 = 52
25 = 52 ==> 52 + 122 = 132

Was there a quicker way to find the triples for 7, 9, 11, etc?
I'll get to that.

Also notice that 31 + 33 = 64, which is 82, so 82 + 152 = 172.
And it had been under my nose all the time that 42 + 32 = 52.
The simplest Pythagorean Triple of them all, fits both these models.

(to be continued)

Questions for my students

Extra credit for answering or for participating in the discussion.

  1. Can you explain in your own words what I meant by "relatively prime"? What do you know about prime numbers that might give you a hint?

  2. Can you find Pythagorean Triples where the smallest side is 9, 11, and 13?
    Remember: 9-12-15 doesn't count. It can be reduced to 3-4-5.

  3. Why can't there be any Triples of the form: Even2 + Even2 = Odd2?

Monday, April 06, 2009

How Many Times Has Tokyo Been Saved?

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(C)Copyright 2009, C. Burke. All rights reserved.

Godzillion vs. Mathra
A classic confrontation.

See also How Many Times Has Tokyo Been Destroyed?

Sunday, April 05, 2009

Sunday Comic: Super-Stickman 50: Super-Stick-Man and the Masked Man

Dateline: The Old West!
In the eye of a temporal storm, how many cultural references of 50s TV, based on movie serials and old-time radio programs of the 30s and 40s, featuring characters from the Old Western days of the late 19th century could one high school kid back in the 80s stick into a comic that he didn't know back then would appear on the Internet in 2009?

Click on the link to the Fabulous 50th comic and follow the exciting story!

Friday, April 03, 2009

A Knotty Theory

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(C)Copyright 2009, C. Burke. All rights reserved.

And 'that's knot all' ... or maybe, it's 'Knot a Theory'...

Wednesday, April 01, 2009

How Many Times Has Tokyo Been Destroyed?

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(C)Copyright 2009, C. Burke. All rights reserved.

Megalonumerophobia -- Fear of large numbers -- is more common than you might imagine.
Googol it.

Remember: A "gazillion" is an awfully big number.
"Godzillion" is an awful big number.