Play a game, meet John Nash

This blog post falls into the category of “pure fun,” at least for people who enjoy board games.

Here’s a board game that is easy to learn, fun to play, enjoyed by people of all ages, and best of all — invented by two world-class mathematicians: Piet Hein and John Nash.

The game is generally called Hex*, and you can even play it online.

The rules of Hex are so easy that anyone can grasp them fast. They are: use game tokens to create an unbroken path that goes from one side of the game board to the other. You get to put down one token on every turn, and players alternate putting them down. Tokens, once put down, cannot be taken off the board. Tokens do not kill other tokens, jump or move in any other way. Their sole purpose is to be part of a path that traverses the board.

Of course, as one player tries to go from his side of the board to the other, the other player tries to do the same, but from her side of the board to the other. As you defend against the other player’s path, you start to make your own path. Very interesting.

The best thing to do with this game, though, is just play it. To that end, go to any site online that has the game. I recommend either of these sites:

This one allows you to play on a 7×7 board, and it has background info on the game and some strategy tips.

This one lets you play on a board that is 11 x 11, and it allows you to record your moves and get a record of your moves, for study.

I have brought Hex — along with checkers, go and chess — to various “Games Nights” that I have organized and run at local schools and rec centers. Invariably more people end up playing Hex than any of the other games. It has a sort of magnetic attraction quality to it.

Try it and see for yourself.

* One note about the name. The story goes that Nash invented the game one night while sitting on the “john” in a bathroom at Princeton. The floor of the bathroom, tiled in hexagons, gave Nash the idea for the game. So the game is sometimes affectionately called “John” in honor of both the inventor and place of discovery.

P.S.: For a good book about the game, check out Hex Strategy by Cameron Browne. I got my copy through Amazon. The book also has templates that allow you to create your own physical game boards. It also brings out many of the mathematical ways of looking at the game.

See a world-class mathematician

It’s not every day that you can see a great mathematician giving a lecture. But thanks to the power of today’s technology, you can do that right now.

Terence Tao, winner of the prestigious Fields Medal (like a Nobel for math), has been compared to some of the greatest mathematicians of all time, like Carl Friedrich Gauss and Leonhard Euler. And as you will see, he has so much of his life yet to live.

Here is a lecture Terence Tao gave at UCLA on the distribution of prime numbers, an interesting area of number theory.

If you’d like to learn more about the hunt for very large prime numbers,
click here.

Turtle Talk – January 2010

Note: Below is a copy of my January 2010 ezine, Turtle Talk.

If you’d like to subscribe to Turtle Talk to get it as soon as it gets published, just go to this site:

http://singingturtle.com/pages/turtle_talk.html

There is a cute animation of moving turtles.

Turtle Talk
— a newsletter —
January 2010
Vol. XIII, Issue #1

QUOTE OF THE MONTH –

“If you think dogs can’t count, try putting three dog biscuits
in your pocket and then giving Fido only two of them.”
— Phil Pastoret

MathChat Blog Update:

In case you did not know, I have been spending a fair amount of time
writing on my blog, and there are many articles and ideas over there.
I have articles on many topics. Here’s just a small sampler:

Multiplication tricks
Mental math shortcuts
Dividing a fraction by a fraction using the “bologna-cheese” sandwich
Using color to elucidate ideas in geometry
Using colors to clarify concepts in algebra
How to solve algebraic mixture problems
Using “master equations” to make word problems less scary

To check out the blog, just visit:

http://www.mathchat.wordpress.com

Feel free to leave comments, too.

Twitter presence too:

I send out tweets about blogs and other issues of interest
to math teachers and students. If you’d like to FOLLOW ME,
just go to twitter, type in joshmathguy and sign up to follow.

January Problem of the Month

Samantha gets 12 out of 14 problems correct on a test.
Then she gets half of the remaining problems correct.
If every problem is worth the same number of points,
and Samantha ends up getting a score of 60% on the test,
how many problems are on the test altogether?

To get credit you must show your work as well as get the problem right.

First person to send in the right answer gets a FREE COPY of any
Singing Turtle Press book.

Send your answers to: josh@SingingTurtle.com

In the next newsletter I’ll name the first 9 others who get this
right.

DON’T FORGET: when you submit answers for the Problem
of the Month, please write
POTM
on the Subject line of your email’s header and, knowing that
it is optional, feel free to share:

— your age (if a child)
— your state or hometown
— status: student, teacher, tutor, etc.
— whether in school or homeschooled
— anything else about you of interest

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Discussion: A Way to Help Students Notice
& Understand the Mistakes they make in Algebra

Have you ever noticed that students often have a hard time spotting and understanding what is actually wrong — when they make mistakes in algebra?
One reason is that — when students enter the “Land of Xs and Ys,” they have little intuition as to what should happen. It’s as if they know that “they’re not in Kansas anymore,” so as far as they’re concerned, there could be “munchkins” around the corner. With such a level of uncertainty, they have little idea what to expect.
And if students don’t know what SHOULD happen, they also don’t know what SHOULD NOT happen. However, it’s helpful to remember that there is still one way to reach such early algebra students — appeal to their understanding
of how numbers work.
Here’s an example of how you can use this to your advantage:
Suppose a student makes the following mistake when solving an equation:
3x – 4 – 2x = 12
+ 2x + 2x
5x – 4 = 12
What’s the student is doing wrong? Adding 2x to the SAME SIDE of the equation TWICE, instead of adding it to BOTH SIDES of the equation ONCE.
Of course, if you asked the student why s/he did this, the student would probably offer up some half-true, yet misguided phrase like, “you have to add 2x on both sides.”
What you do here is get a “change of venue.” Move the problem from the field of algebra to the field of arithmetic.
Simply substitute numbers for letters. Ask the student if the following steps would make sense:
10 – 3 = 7
 + 3 + 3
13 = 7
Generally, students will see right away that the answer is wrong. And that will alert them that something is wrong up above. And usually students can see that it is weird to add 3 twice on the left side.
Once they get that idea — either on their own, or with a little prodding — go back to the algebraic situation and ask them if they can NOW see what’s wrong. Very often they can.
When I was studying to become a teacher long ago, I learned a key idea about how people learn: to learn anything we need to relate new information to something we already know and understand. In essence, that’s what I am suggesting we do when students make procedural mistakes in algebra. Students know arithmetic (hopefully!). So use that. Put the algebra in an arithmetic context. Usually students will see the problem in the arithmetic setting. Then they’ll realize that algebra, even though it looks different than arithmetic, still works a lot like arithmetic. Getting that idea across will help students greatly as they continue in their study of algebra.
In any case, feel free to give this a shot, and see how it works for you. And feel free to share any feedback on how this works for you.
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Online Algebra now Available from Author
of the Algebra Survival Guide

Josh is now offering services online through SKYPE:
algebra tutoring sessions, and entire Algebra 1 classes.

If interested, please send email to: josh@SingingTurtle.com,
telling what kind of help you are looking for.

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Singing Turtle PRODUCTS

To see the line of Singing Turtle products for math education,
visit: http://www.singingturtle.com/pages/PARENTS_new.html

You can buy all of our books on Amazon.com
If you do, please consider writing a review.

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Well, that’s all for this month, though I will continue to blog and send out tweets on items of interest.

Have a great month, and I’ll be back in February.

— Josh

Tutor Tales — Algebra/Arithmetic Connections

Here’s an idea I came up with today for helping students understand more deeply the mistakes they make in algebra.

One thing that makes algebra difficult is that students have, basically, no sense as to whether something is true — or not — when they look at algebra. They have virtually no intuition about this. However, they do have intuition as to whether or not things are correct in arithmetic.

But we can use this idea to help students understand algebra. For example, we can use this approach to help students understand what is “wrong” when they make mistakes in algebra.

For example, let’s say that a student makes the following mistake:

3x  –  4  –  2x  =  12

+ 2x +  2x

5x    –   4             =   12

What is the student is doing wrong? The student is adding 2x to the same side of the equation two times, instead of adding it to both sides of the equation.

How can we help the student see that this is wrong?

Change it to an arithmetic situation. Ask them is the following makes sense:

9    +     3   =    12

– 3 – 3

6                 =   12

They will see that this is wrong because they know that the addition is wrong. What is more, they will get the general idea that it makes no sense to subtract 3 twice from the same side of the equation.

This mistake — in the algebra — will make little sense unless you do something like this, something they can grasp.

Challenge Problem – the ANSWER

Hi everyone,

Here is the answer to yesterday’s challenge problem, the probability problem about catching fish.

First, there are 6 ways to catch the three fish in three casts, getting exactly one trout, one carp and one bass.

You could get the fish in any of these six orders:

TCB / TBC / CTB / CBT / BTC / BCT

Next you find the probability of getting one of these possibilities. Let’s take the first one, TCB.

Keep in mind that after you catch the first fish, the number of fish left goes down by one, to 11; after catching the second fish, the number of fish goes down to 10.

The probability for the TCB possibility is calculated by multiplying:  6/12 x 4/11 x 2/10 = 2/55

When you think about the five other ways to catch the fish, you’ll see that the order of the numerators changes, but the denominators remain 12, 11, 10. So the probability for catching the three fish in any of the six ways is always the same:  2/55.

To get the probability for all six catches, just multiply the probability of one catch by 6:

6/1 x 2/55  =  12/55

And that is the answer: the probability of catching exactly one trout, one carp and one bass is 12/55, which works out to about 21.8%, meaning that this should happen a little more than 1/5 of the time.

I didn’t see anyone submit any answers, but feel free to send them in. Remember that I will post only the correct answers, so no one has to worry about seeing an incorrect answer posted.

Have a great day!

—  Josh

Interactive Challenge Problem — Send in responses

O.K., time to wake up the ol’ brain cells.

Here’s a little challenge problem. I’ll post the problem today, and then I’ll post the answer the next day. My thought is that this would be relevant for many kinds of people.

Teachers can use this as a fun class-opener. Homeschoolers can use it to start their math studies. And anyone who enjoys math can use it to sharpen math skills. So enjoy.

It would also be fun to see how many people get it right, so please send in your answers as comments. I will post only the correct answers.

The problem:

A lake contains exactly 12 fish: 6 trout, 4 carp, and 2 bass.

On any given cast, you catch exactly one fish, and no kind of fish is biting any more than any other kind. (i.e.:  Your odds of catching fish are governed by mathematical probability alone.)

What is the probability that in three casts you will catch exactly one trout, one carp and one bass?

To get credit, you must provide the correct answer and show how you solved the problem.

Conquering Proportions, Part 1

Proportions can seem intimidating, but actually they are one of the easiest kinds of word problems to solve. In this series I’ll offer a number of tips that help you conquer this type of algebra word problem. Search “Conquering Proportions” to see the whole series.

First, a cool shortcut that you can use whenever you have to solve an algebraic proportion …

In working with algebraic proportions, I’m amazed to see that few students know how to do a certain kind of canceling that allows them to find the solution more quickly and efficiently.

So I thought I’d share the trick, for anyone who has never seen it.

Of course, given a problem like:  6/x  =  24/32

we all know that we can cancel vertically with the two numbers in the fraction on the right, to get:

6/x  =  3/4

Then you can cross-multiply to get:

3x  =  24, and then x  =  8

In other words, everyone knows that you can cancel vertically when you have a proportion, just as you can cancel vertically for any fraction.

What not everyone realizes is that there’s another way you can cancel when solving proportions — horizontally.

— What? you say.

— Horizontally, I say. And no, I’m not joshing.

For example,  in the problem:  7/4  =  21/x

you can cancel horizontally with the two numbers in the numerator: the 7 and the 21. These reduce to 1 and 3.

The problem then becomes:

1/4  =  3/x

Cross-multiplying, you get the answer in one quick step:   x = 12.

What’s really convenient is that you can also cancel both vertically and horizontally in the same problem. For example, in

6/x  =  42/28

you could first cancel horizontally, to get:

1/x  = 7/28

Then you can cancel vertically, to get:

1/x  =  1/4

Cross-multiplying, you get the answer in just a step:  x = 4

I find that when students cancel before cross-multiplying, they are more apt to get the right answer, and to become less frustrated, for the numbers they deal with remain small.

For example, in the last problem, if the student had not canceled at all, he would have a cross-multiplication mess of:

6 x 28 = 42x

That sort of problem just opens up the door to arithmetic mistakes. But canceling before cross-multiplying closes that door since it makes the numbers smaller and easier to manage.

So now you get a chance to practice.

First use horizontal cancelling to get the answer to these
proportions. Those who’d like an added challenge might like to try them in their head:

a)   x/12  =  3/4

b)  3/7  =  x/35

c)   z/48  =  7/12

d)  y/56  =  7/8

Now go really wild. Use both horizontal and vertical canceling to make quick work of these proportions:

e)  x/9  =  16/36

f)   x/22  =  30/66

g)  32/56  =  y/14

h)  13/q  =  65/35

And here are the answers to these problems:

a)  x  =  9

b)  x  =  15

c)   z  =  28

d)   y  =  49

e)   x  =  4

f)   x  =  10

g)   y  =  8

h)  q  =  7

Conquering Mixture Problems — Answers

Answers to Mixture Problems

In my last post I provided three mixture problems for all of you to do.

Here again are the problems, with the answers to them italicized.

1.  Kendra starts with 10 liters of a 40% antifreeze solution. How many liters of pure antifreeze would she need to add to end up with a solution that is 60% antifreeze?

Kendra would need to add 5 liters of pure antifreeze.

2.  Keith the chemist has a solution that is 25 quarts of 20% Boric Acid. How many quarts of 70% Boric Acid would Keith need to add to end up with a solution that is 50% Boric Acid?

Keith would need to add 37.5 quarts of 70% Boric Acid.

3.  Erin has a 2-liter solution that is 15% alcohol. How much pure alcohol would she need to add to it to end up with a solution that is 40% alcohol?

Erin would need to add 5/6 of a quart.

Conquering Mixture Problems — Practice

In my last two blogs I showed how to solve mixture problems. So now I want to give you some practice, so you can become an expert at solving these kinds of problems.

The answers will be stated in the next blog.

1.  Kendra starts with 10 liters of a 40% antifreeze solution. How many liters of pure antifreeze would she need to add to end up with a mixture that is 60% antifreeze?

2.  Keith the chemist has a mixture that is 25 quarts of 20% Boric Acid. How many quarts of 70% Boric Acid would Keith need to add to end up with a mixture that is 50% Boric Acid?

3.  Erin has a 2-liter mixture that is 15% alcohol. How much pure alcohol would she need to add to it to end up with a solution that is 40% alcohol?

Conquering Mixture Problems, Part 2

If you read my last post, Conquering “Mixture” Problems, Part 1, you’re “sitting pretty” for solving these word problems. If you haven’t, go for it, and you’ll kiss those “mixture-problem blues” away.

In this post we will get to the heart of the matter  —  setting up and solving mixture problems. Here’s the problem we’ll solve:

Tuning up your car in late fall, you realize that you need to adjust the antifreeze level with the cold weather just around the corner. With some fancy equipment, you find that you have 12.0 liters of a 40% antifreeze solution, but you need to wind up with a 60% solution. So you need to figure out how much pure antifreeze to add to the 40% solution to wind up with the 60% solution.

At first glance it appears that you would need to be a genius to solve such a problem. How could anyone figure out anything so tricky?

But as it turns out … you are a genius — because you just learned how to use “master equations”  through my last post  —  so not to worry.

As you learned, master equations give you a super-clear outline for solving word problems. In the last post you learned two key facts about mixture problems:

1) There is a simple formula for calculating the amount of stuff you are interested in, the equation:  Stuff =  (Concentration)  x  (Volume of liquid). Or, more simply:
S  =  C  x  V

2)  For most mixture problems the master equation is this:

(Original Amount of Stuff)  +  (Amount of Stuff Added)  =  (Amount of Stuff at End)

In such word problems, the stuff we’re interested in — what we’re keeping track of — is the antifreeze. That being the case, our master equation will look like this:

(Original Amount of ANTIFREEZE)  +  (Amount of ANTIFREEZE Added)  =  Amount of ANTIFREEZE at End)

Now here is where the first part comes in. The equation  S  =  C   x  V   gives us a nifty way of calculating the stuff for all three parts of the equation. Let’s take this one step at a time.

(Original Amount of Antifreeze):  Couldn’t be easier! The problem tells us that we start with 12 liters of a 40% solution. Using S =  C  x   V, we get:

(Original Amount of Antifreeze)  =  (.4)  x  (12.0)

(Amount of Antifreeze Added):  The problem asks HOW MUCH pure antifreeze to add. Since this is the unknown, we’ll let V  =  the volume of pure antifreeze that we’ll add. And since this is pure antifreeze, the concentration is 100% which, written as a decimal number, is just 1.0. Plugging in to S  =  C  x  V, we get:

(Amount of Antifreeze Added)  =   (1.0) x (V)

(Amount of Antifreeze at End):  We know two facts about the antifreeze at the end.
1)   Its volume will be the sum of what we start with, (12) and what we add (V). So the end volume will be (12 + V).  We also know that our target concentration at the end is 60%, and we convert this to the decimal number, .6. Again, plugging in to S  =  C  x  V, we get:

(Amount of Antifreeze at End)  =  (.6) x (12  +  V)

Now going back to our master equation:

(Original Amount of Antifreeze)  +  (Amount of Antifreeze Added)  =  (Amount of Antifreeze at end),

we plug in the three quantities we just found, and we get this:

(.4)  x  (12.0)  +  (1.0)  x  (V)  =  (.6)  x  (12.0  +  V)

We can solve this algebraically for V, as follows:

4.8  +  V    =    7.2  +  .6V

.4V  =  2.4

V  =  6

Since we let the letter V stand for the amount of pure antifreeze we add, this means that we add exactly 6.0 liters. And that is the solution; it is no harder than this.

So take a little time to let this sink in, and in my next post I’ll give you a few practice problems, to let you sharpen your skills.