Archive for June, 2009

Math Teachers at Play #10 and Time Savers

June 29, 2009 On Friday, right on schedule, Math Teachers at Play #10 was posted at Homeschool Math Blog.  There’s good information on Wolfram Alpha and homework and optical illusions and a host of other interesting things.

The discussion on homework had me remembering one of my favorite Teaching Time Savers.  Jane Murphy Wilburne explained here how she dealt with homework.

As the students entered class the next day, they would list the page number and problem number of the problems they could not solve, on the front board in a designated area. If the problem was already listed, they placed a check (√) next to it. Once the class started, they were not allowed to record problem numbers at the board. Other students, who were successful in solving these problems, immediately went to the board when they entered the class, indicated that they would solve one of the listed problems, and worked it out in detail. When they finished they signed their name to the problem.

I tried this out in calculus, a course where I assign homework every day, and it worked fabulously.    It didn’t eliminate grading, but it did cut way down on how much time I spent going over homework in class.  The only change I’ve made is that I now  insist that people write down what they had trouble with, so instead of writing just #23 they’d write “#23:  couldn’t get started” or “#23:  what do you do after [this step]?” or “#23:  I got x2 but the book says x3.”  And I count answering the questions correctly as extra credit.  (In perspective, homework is 10% of a person’s grade in calculus; usually each assignment gets 5 points, and there are probably 35 assignments over the semester.  I’ve ended up giving 1 extra credit point for each time a person added correctly, with a cap of earning 100% total for homework.)

I haven’t done this in all courses, but I do use it in every calculus-type course (where the problems tend to be shorter) and it’s gone really well.  So that’s my two cents.

Multiplying on an Abacus

June 25, 2009 Continuing with different ways to multiply, here’s just the one

(18) The Abacus.  To start, you need to know how to write numbers on the Chinese abacus.  Numbers are written with the 1s column on the far right, then the 10s column, then the 100s column, etc.  and you only look at which beads are touching the center bar.  Beads on the top count as 5, while beads on the bottom count as 1.  Here’s the number 358: Notice that you could get by with each rod only having 1 bead on the top and 4 on the bottom.  Some have that; it’s called a 1/4 abacus, and I believe it’s been popular in Japan for about 60 or 70 years.  It’s called a soroban.  The 2/5 abacus (which is what I have) is used in China, and it’s called a Saun-pan.  It just makes certain carrying easier, plus it’s what I could find.

On to adding!  Some of this is easy:  If you have a number like 358 and you want to add 100, you just move one of the 100 beads [third column from the right]. But what if it’s not that simple?  If you have a number like 358 and you want to add 18, you could do it bead by bead (Slow.   Boring.  Lots of regrouping.)  or you could just add 2 tens and then remove 2 ones.  That’s the more authentic way. Likewise, if you had 358 and you wanted to add 4, you could add a five bead and subtract 1 unit, then regroup, or you could add 1 ten bead and subtract the five and a one bead.    [Is that even clear?  You can find java versions online to follow along, like this one although it uses two decimal places on the far right instead of starting with the 1s.] On to multiplication!  For single digits, well, you pretty much just need to know all your basic multiplication facts.  No way around it.  But this process will work for anything beyond that.

Let’s find 87×625.  Pick one of the numbers (87) and put it on the far left-hand side of the abacus.  Since 87 is a 2 digit number, we want to leave 3 blank spaces that’s (2+1) on the far right, and we’ll put 625 after that (where by “after” I guess I mean “before”, to the left).  Just look at the picture and it should be clear. (Pause.)  Before we get started on the process, it’s helpful to take a look at the big picture.  We’re going to put our answer on the far right, which is why we needed those blank spaces.  But our answer will be more than 3 digits long, so we’re going to need more space.  We’ll create the extra space as we move along.  We’ll first multiply 87 by the 5 of 625 then we’ll get rid of that 5, freeing up an extra column.  then we’ll multiply 87 by the 2 of 625, freeing up yet another column, and finally we’ll multiply the 87 by the 6.

Let’s get started.

We want to multiply 87 by the 5.  We’ll multiply the digit 8 and then the digit 7 [left to right].  Yes, the is backwards from how we’re multiplying 625 [right to left], but that just keeps it fun.   Whenever we multiply by the 8 of 87 we’ll put the answer two columns over, and whenever we multiply by the 7 of 87 we’ll put the answer three columns over.   [No matter how many digits that number is, when you multiply by the largest place-value you put the answer two columns over; the next place value goes three columns over; the next would go four columns over, etc.]

So to start, we know that 8×5 is 40.  Since we’re using the 8, we’ll put the answer two columns over from the 5.  We always use two for the left-most column. Now we’ll move to 7×5.  This is 35, and we put the answer three columns over from the 5. Since we’re done multiplying 87 by 5, we can remove the 5. Notice the number on the far right, 435, is just 87×5.

We’ll do the same thing, multiplying 87 by 2.  First we find 8×2 and put that answer two columns over from the 2.  [I’m going to do that by adding 20 instead of 16, and then subtracting 4.] Then we’ll multiply 7×2 and put that answer three columns over from the 2.  [Here I added 15 instead of 14, and then I subtracted 1.] We’re done multiplying by 2, so we can remove it.  Except I forgot to take a picture with it removed.  If you ignore that 2 of 62 (formerly 625), you can see that the number on the far right is now 2175.  Sure enough, this is 87×25.

Nearly done, we’ll multiply the 8 of 87 by 6 and put the answer two columns over from 6.  Notice that this 6 originally stood for 600 (in 625), but we don’t need to keep track of which place value the 6 originally had.  By always putting the single-digit products two or three (or four for longer numbers) over, the place value takes care of itself. And finally, we multiply the 7 of 87 by the 6 and put the answer three columns over from 6. And remove the 6 because we’re done multiplying by it.  Here’s the final answer: Yes, it’s 54,375.  And a quick check shows that   I made a mistake so I had to redo all of the photos I took, and then I made another mistake and had to retake more photos, but on the third time It’s right!  And despite how long it takes to explain, it really isn’t too bad to do by hand even if you’re learning it for the first time.

June 19, 2009 The Spring issue of our department newsletter is up!  Which is good, because spring ends pretty soon, so we really were working against a deadline.   [Fortunately, Batman and I are the editors so if we do miss a deadline, nothing actually happens.]  Most of the information is local to our college, but there”s some information about summer research programs and conferences, and a bit of advice from folk in the financial world during Career Night.    We change the name of the newsletter each quarter (harkening back to its start three years ago when we couldn’t figure out a name), and this issue is called The Wiley Wiles after, of course, Andrew Wiles.  The best part of naming a newsletter after him is that we could include a picture of our former Chair’s program from the 1996 Joint Mathematics Meetings in Orlando, which Andrew Wiles kindly signed: Pretty cool, huh?

In case you’d like to do some math this weekend, here are the Problems from the newsletter.  Answers are welcome in the comments!

Problem 3.3.1: What are the next two numbers in the sequence 1, 8, 72, 46,
512, 612, …?

Problem 3.3.2: Choose a positive real number x and compute 100x2, x3, and 1.05x, then arrange the three results from least to greatest. How many orders are possible?

Problem 3.3.3: Express |x| in terms of the maximum function, and express max(x,y) as an absolute value.

Problem 3.3.4: What is the area of the largest semicircle that can be inscribed in a unit square?

Calculus Demonstration: 3D printing

June 18, 2009 I love the concept of 3D printing.  Of course, I also really enjoy teaching Calc II when we talk about slicing and shell formulas and volumes of revolution, because I remember the AHA! moment when I suddenly put together what the formulas were all describing and it all made perfect sense.  (Sadly, this moment came when I was a senior studying for comps, several years after actually taking the class, but still, AHA! moments are glorious.)

In 3D printing, objects are built from the bottom up, cross section by cross section, the same way you’re supposed to envision the pieces when you calculate volumes by slicing.  This article in the Christian Science Monitor last week likens it to building with legos, although my experience with legos is that separate sections are constructed and then put together (you build the walls, then add the furniture, then the roof); that concept might work with printing too, where you print separate components and then put them together.  And what’s amazing is that you can print some pretty complicated things with moving parts.

So what is used for the printing?  The article above describes a layer of powder being put down and the printing is actually done by spraying glue instead of ink.  Wikipedia also describes printers that build with a liquid gel.  But my favorite is printing done with candy.

That’s right:  candy.  Not surprisingly, the CandyFab 6000 and it’s earlier prototypes are made by the folk at Evil Mad Scientist Laboratories.  Here’s an example of 3D printing: Isn’t that amazing?  Here’s the machine it was done on: You can see it making some 3D Candy above.  They keep making improvements, as you can see in the two dodecahedrons below (printed at different times): Here’s another fancy piece o’ printing: And finally, here’s a Möbius Strip.  I thought it was a cake at first but no, it’s eight pounds of sugar. Now that’s some pretty neat printing.  You can even see the “slices” on the surface.

All photos are licensed under Creative Commons; the photos link to their home on Flickr, and you can find even more here, plus more information here.

The Fourth Bunch of Ways to Multiply

June 17, 2009 Just three more ways today, although with all the ones that have been suggested I think we’ll get to the 25!

(15) Multiplication with Log Tables.  People had already been using trig tables to multiply, but when logarithms were discovered they became THE way to multiply numbers.  The idea behind log tables is that logarithms turn multiplication (nasty) into addition (fun!) without having to derive a bunch off trig formulas.  In particular, we’re going to use the fact that log(x·y)=log(x)+log(y).

Here’s how we can find 875×978 with logarithm tables.  We’ll start by writing the two numbers in Scientific Notation:  875 would become 8.75×102 and 978 would become 9.78×102.   We’re going to multiply 8.75 and 9.78, and then adjust that product by the appropriate power of 10 (in this case, 104).

The next step for the multiplication is to look in your Table of Common [Base 10] Logarithms.  If you don’t have a copy handy, you can look here.  It turns out that log(8.75) is 0.9420081 [that’s our log(x)] and log(9.78) is 0.9903389 [that’s our log(y)]. Now we’ll add those together to get 1.9323470.  This must be the log of our product!  So we work backwards with the table, looking for the number whose log is 1.932347.  Unfortunately, the table only gives results between 0 and 1, so initially it seems like we’re stymied, but we can be sneaky and subtract 1.  We’ll come back to that in a moment.

So now we’re looking for a number whose log is 0.932347. Looking back at our table, we see that log(8.55)=0.9319661 and log(8.56)=0.9324738, so the number we’re looking for must be between 8.55 and 8.56.  We can pick 8.56, which is the closer number, or do a little interpolation.  If we round, we’re looking for 0.9323 and instead we got 0.9320 and 0.9325.  The number we wanted was about 60% of the way from the smaller to the larger, and so instead of choosing 8.55 or 8.56 we could go about 60% of the distance between them, and guess that the product was 8.556.  Let’s do that.  [We could be even more accurate if we used a calculator to figure the exact percentage, but that seems to defeat the purpose of using the log table to multiply.]

So now we have a product of 8.556, but clearly that’s not exactly right.  We first have to account for the fact that we subtracted 1.  Notice that 1+log(blah) is the same as log(10)+log(blah) [because we’re using the common logarithm], and THAT is the same as log(10·blah).  Here log(blah) is 0.9323470, and we just saw that blah was approximately 8.556, so the product that we want — the product of 8.75 and 9.78 — is approximately 85.56.

We’re almost there!  Remember how originally we wanted the product of 875 and 978 but we first wrote those in Scientific Notation?  We need to adjust our answer of 85.56 by 104, leading us to the conclusion that 875×978 is approximately  (drum roll please) 855600.  The correct answer is 855750, so we’re certainly in the right ballpark and more accurate than we were with the trig tables, but, well, it’s a wonder to me that this method was so popular for so long when so many other ways are even more accurate.  Maybe I’m missing something.

(I’ve multiplied with log tables before, but refreshed my memory with this site on the Obsolete Skills Wiki, which also explains how to Get off the couch to Change Channels on the TV set and how to Make Change in [old] Shilling and Pence.)

(16) Slide Rule.  This takes the ideas of the log table, but bypasses the actual looking up.  Instead, the numbers are scaled on the slide rule in such a way that you don’t have to do much at all.  William Oughtred is credited in many places (online) with the slide rule, and the date 1622 shows up, so this happened just a few years after the invention of logarithms.  Pretty quick thinking!

Let’s do a simple example first:  2.5×3.   You can use an actual slide rule, or use the java version here.    The slide rule looks complicated because it can do a lot of things, but we’re mostly going to be looking at the bottom of the slidey part in the middle (C) and the fixed part at the bottom (D).

To multiply 2.5 by 3, you start with2.5  and align the 1 on C with the 2.5 on D.  Then look for 3 on C, and right below it will be the product (in this case, 7.5): See how once you’ve aligned the 1 with the first factor 2.5, you can move the slider doohickey to the 3 on C so that you can see what’s right below it?

Now let’s look at a more complicated example:  back to 875×978.  As with log tables, we’ll start by writing the numbers in scientific notation (8.75×102 and 9.78×102) and then we’ll multiply 8.75 and 9.78 and adjust our final answer.

Normally we’d align the 1 of C with 8.75 beneath it on D, and then start looking to the right.  But 8.75 is so large, we quickly run out of room: So we’re sneaky, and instead of aligning the 1 (on C) with the 8.75, we align 10 (on C) with 8.75.  We’ll have to adjust by multiplying by an extra power of 10 at the end, kind of like we did with the log tables before (and, really, it’s not a coincidence that this happened in both calculations). So we’ve aligned the 10 on C with the 8.75 on the bottom.  Then we move the slider over to 9.78 on C, and look below to see the product!  The product looks like, ummm, it seems to be a bit over 8.55 but not yet at 8.56, so we’ll say 8.555.

As our final step, we need to multiply by 10 because we aligned with the 10 instead of with the 1 (as discussed above), and also by 104 because we’d had to write the numbers in scientific notation.  This means our final answer is 8.555×105, or 855500.  As we’ve seen before, it’s not exact but it is pretty close and it’s a lot faster than looking up in tables!

(17) The Gunter scale.  This is the precursor to the slide rule, invented by Edmund Gunter, and it didn’t slide at all but it was BIG:  two feet long was standard.  It was a big ole piece of wood with a logarithm scale on it, and if you wanted to measure a product like 2.5×3 you’d measure the physical distances to 2.5 and to 3, add them together, and see where you ended up on the scale. Essentially it was a Slide Rule that didn’t slide:  you had to do that part by hand.

(There’s a picture here and more information about Gunter here, including the fact that he coined the terms for Cosine and Cotangent, though Cosecant is a bit older.  I hadn’t known about this method until Pat Ballew brought it up in a recent comment:  thanks Pat!)

The Third Bunch of Ways to Multiply

June 16, 2009 Let’s multiply!  (The First Bunch had Egyptian, Babylonian, lattice, and current methods, and the Second Bunch had Greek, Napier and Genaille-Lucas, plus links to Vertically and Crosswise Multiplication and finger multiplication.)

We’ll just do two more today, starting with one I forgot we’d already talked about:

(13) The Method of the Cups, which was described by Friar Juan Diez in 1556 (see  this post from last year).  It’s done digit by digit, but multi-digit numbers aren’t always written on a single line: if you want a glimpse of what it’s like, here’s the image for 875×978. (14) Prosthaphaeresis, or Multiplication with Trig Tables. This method dates back to the late 1500s, although logarithms essentially rendered this method obsolete.

Here’s the basic idea:  remember the sum and difference formulas for sine and cosine?  We’ll use those.  For example, since: $cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$

and $cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$

then by adding those together, we get $cos(a+b)+cos(a-b) = 2cos(a)cos(b)$ $cos(a)cos(b)=\frac{1}{2}cos(a+b)+\frac{1}{2}cos(a-b)$

And that’s the formula that we’ll use, although by also looking at sine we could just have easily come up with a different formula that had a mixture of sines and cosines.    When the last step of dividing by 2 is ignored, the resulting formulas are called the Werner formulas.

Here’s how we’ll do it.  Suppose we want to multiply 875 and 978, because those were the numbers that appeared in the previous example.  We’re going to be using a table of cosine values (like this one).  This table actually shows both sine and cosine, but cosine is read on the right. Notice that 0.9997 is cos(1.4°), for example — we’re adding 0.2° to the angle for each column, going right to left.

Cosine values always fall between 0 and 1, so we need to scale our numbers to be between 0 and 1.  Thus 875 becomes 0.875 and 978 becomes 0.978, and we’ll have to multiply our final answer by 103·103=106.

Remember that our formula is $cos(a)cos(b)=\frac{1}{2}cos(a+b)+\frac{1}{2}cos(a-b)$.  In this case, $cos(a)$ is 0.875 and $cos(b)$ is 0.978.  We need to find a and b. Remember that we read the angles on the right.  Since 0.875 is between 0.8746 and 0.8763, our angle a must be between 28.8° and 29°.  We’ll use 29°, since that gives the best estimate. Likewise, we need to find b knowing that $cos(b)=0.978$.  From the chart we can see that b is between 12.0° and 12.2°, but basically it’s 12°.  Yay!  We’re halfway there!

We have a and b,  so the sum $a+b$ is 41° and the difference $a-b$ is 17°.  Now we need to know $cos(a+b)$ and $cos(a-b)$. This part is easy!  The cosine of 41° is just 0.7547 and the cosine of 17° is 0.9563.

Now we can plug all this into our formula!  Remember how 0.875 was $cos(a)$ and 0.978 was $cos(b)$?  Using: $cos(a)cos(b)=\frac{1}{2}cos(a+b)+\frac{1}{2}cos(a-b)$

we get $0.875 \cdot 0.978 = \frac{1}{2}\cdot 0.7547 + \frac{1}{2} \cdot 0.9563$

which simplifies down to 0.8555.  Multiplying that by the 106 from earlier gives 875×978 as approximately 855500.   We’re only off by 250, which feels like a lot but it’s less than ½% of the answer, so in the grand scheme of things it’s pretty good, and with more accurate tables we could have done even better.

Next up:  Log tables and slide rules!

A few fun language puzzles

June 15, 2009 I *meant* to get another post up on Multiplication today, but wasn’t able to start finish that.  Instead, here are some  punctuation puzzles!

Nelson Rich, my first department chair, sent this along to me about ten years ago.  It turns out it’s over 60 years old, and was first used in 1947 by Hans Reichenbach.

The challenge:  Add punctuation below to make the following sentence correct:

While looking this up (I couldn’t remember how many “had”s there were, I discovered two similar problems.

Add punctuation to the following sentence so that it is clear what is means.  (It is supposedly correct as written, although some extra words seem to be implied.)  It was first used in 1972 by William J. Rapaport who is a prof at the University of…you guessed it…Buffalo.

Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo.

And finally, add punctuation to the following sentence(s):

That that is is that that is not is not is that it it is

Wasn’t that one fun?  It dates back to 1953, to Brewer’s Dictionary of Phrase and Fable by Ebenezer Cobham Brewer.

The stature of books, published under the GNU-FDL by Lienhard Shultz, is part of the Walk of Ideas of Berlin-Mitte.

Math Teachers at Play #9 is up at HB

June 12, 2009 Or, rather, Math Teachers at Play #9 is up at Homeschool Bytes (who have hosted MTaP before as well!)  The theme is Game Time, and the post has several games and magic tricks.

Speaking of games, did you know that they have a bunch of math games at coolmath?  You can play Mancala (although it seems like there are a million variations to that), the Tower of Hanoi, and lots of other stuff.  Watch out for the sound effects, because that can get a little overwhelming at times.   (Fifteen minutes later:  I just finished a game of Sudoku, helped out by the 8-9 year olds in the living room.  Why didn’t I play with this site before?  I think Sudoku is one of the best ways to teach Proof by Contradiction, since once you get past the infinitude of the natural numbers, the infinitude of the primes, and the square root of ___ is irrational it’s harder to come up with good examples.

(Incidentally, in courses where I teach that the square root of 2 is irrational I usually make it homework to show that the square root of 3 is irrational.  Once they’re comfortable with that, I have them prove that the square root of 4 is irrational, and to find the flaw in the proof.  This proves to be really challenging to the sophomore math majors.)

I think I’ll end my stream of consciousness here.

The Second Bunch of Ways to Multiply

June 11, 2009 Hey, let’s do some more multiplication!   (See Ways (1)(6) here.)  I promised some that wouldn’t be taught in school (at least not as a practical way to multiply), so that’s where I’ll start.

(7) Greek Multiplication.  This is really different from other kinds of multiplication because it’s based on Geometry.  The Ancient Greeks certainly had the concept of discrete quantities, but in a lot of their mathematics numbers were interpreted as lengths; to multiply a and b like an Ancient Greek, you start with the quantities a and b and also the reference point 1.

Draw two intersecting lines and, from the point of intersection mark lengths 1 and b on one line and length a on the other line, and then draw a line between 1 and a. Starting from b, draw a line that is parallel to the line you just drew.  (I could really have said “construct” because all of this can be done with a straightedge and compass.) See how the length x is where that line you just drew intersects the line with a?  It turns out that x is the length of the product ab!    It’s actually pretty easy to show this:  the triangle with sides x and b is similar to the triangle with sides a and 1 [they have one angle in common, and because their third side are parallel the remaining pairs of angles are equal as well].  Since they are similar, x/b must equal a/1, and cross multiplication gives x equal to ab.

One nice feature of this is that you can actually see how if b<1 then the positions of 1 and b will be switched, and the product ab will be less than a.

(8) Napier Rods.  These were first published in 1617 in the Rabdologia by John Napier, a Scottish mathematician who is more famous for introducing the idea of logarithms.  Napier Rods, which I just realized are supposed to be called Napier‘s Rods (or Napier’s Bones) are like a portable version of Grid Multiplication, so if they were allowed on Standardized Testing it might take care of that whole time issue.  (Then again, one could ask why these would be allowed and not, say, a calculator.  So I guess this also falls under the cool just because it’s cool category of multiplication.)

Godzilla is here to show how they work.  He’s going to demonstrate how to multiply 3558  by 274.

Start by getting yourself  some Napier Rods. Do you need to see those up close?  Here they are: Those are still hard to see aren’t they?  That’s because I don’t know how to focus my camera.  And THAT’S why I love Wikipedia, folks.  Here’s a drawing that (someone?  I can’t tell who) posted on Wiki under GNU-FDL. So each rod (or bone) has a digit on top, and the multiples of that digit are written underneath.

Since Godzilla wants to multiply 3558 by 274, we’ll start by picking up rods for 3, 5, 5, and 8 and lining them up.  [I printed out two copies of this set of Napier rods since the digit 5 appeared more than once.]  There’s a rod with nothing at the top:  this is called the Index Rod, and it can be put at the right or left.  It just helps you keep track of the rows. We’re going to multiply this number by 274, so we’ll start by looking in Row 2. See how that looks like a little tiny grid from lattice multiplication?  If you add along the diagonals, it shows that 3558×2=07116.

Now let’s multiply by the 7 of 274.  We’ll look in Row 7. This shows us (after adding along the diagonals) that 3558×7=24906.  We had to do a little carrying here.

Finally, we’ll multiply 3558 by the 4 of 274.  Look in Row 4: So 3558×4=14232.

Now that we have all the pieces we need, we add, keeping track of the place value (by staggering on the left): Yay!  We have our answer!  If you prefer to play with a java version that allows you to switch the base, you can find it here at Cut The Knot.

(9) Genaille-Lucas Rulers.  These take the basic concept of the Napier Rods and modify it using shaded triangles in a way that completely eliminates adding and carrying within each row (though you’ll still have to add to get the total at the very end).  They were invented by Frenchmen Henri Genaille and Edouaird Lucas just over a century ago.  Godzilla is using paper versions that Brian Borchers created. He’s already lined up the 3558 to do the same multiplication (3558×274) as before.  Let’s take a closer look.  Notice that with this set, the Index piece is aligned on the left. OK, we’re going to use the same basic idea as before, where we’ll look in rows 2, 7, and 4.  Let’s start with Row 2: This row should give us 3558×2.  To read it, start on the RIGHT with the number on the very top.  That’s a 6.  Then you read towards the LEFT, following the little gray triangles.  The final number is 7116.

Now we’ll look at 3558×7: Again, you start with the number in the TOP of the right-hand column (which is a 6).  Follow the little gray triangles to the next number, to get that 3558×7=24906.

Finally, here is Row 4: As before when we start from the top of the rightmost column and read to the left we see that 3558×4=14232.

This is a completely mindless way to multiply until you get to the end, where you align your answers (staggered on the right): And we have the same answer as before!  And right here is a picture of some of the rods from 1885 in France.  (They and Napier’s Bones were actually more like square prisms, with a different Ruler printed on each side so that each Rod could show one of four different sides depending on what digits you need.)

This post is already rather long, so I’ll end by briefly mentioning three final methods:

(10) Vertically and Crosswise Multiplication (which TwoPi wrote about just over a year ago).  TwoPi tells me that it should probably be called Trachtenberg multiplication, since it was created by the Ukrainian engineer Jakow Trachtenberg while he was imprisioned in a concentration camp in WWII.  (There are widespread claims that this technique is of Vedic origins, but there is little historical evidence to support that.)

Edited to add: Whoops, Trachtenberg might have come up with it on his own, but he wasn’t the first.  It was published in Italy over 500 years ago and was known as Crocetta.  (See Pat’s comment below.)

Edited again (7/8) to add: Looking in “Capitalism and Arithmetic”, I saw that this is attributed even further back, to the Indian Lilavati of around 1150.

(11) and (12) Speaking of multiplication we’ve already talked about, in this post from New Year’s Eve 2007 I wrote about multiplying by 9 on your fingers (I’ll call that #11) and also two Medieval ways to multiply on your fingers.  They’re pretty similar, but one works for multiplying any two of {5, 6, 7, 8, 9, 10} and the other works for multiplying any two of {10, 11, 12, 13, 14, 15}.  Together I’ll count those as #12.

That’s it for now!  I can think of three more ways offhand (abacus, slide rule, and trig functions) so I’ll write up something about those in the next couple days.  This is fun!

The First Bunch of Ways to Multiply

June 10, 2009 I blithely mentioned in yesterday’s post that I only knew about 13 ways to multiply [“only” because it would be great to write a book called Twenty-five ways to Multiply], and then Jason asked me to list them.  I was originally going to list them all, but then I started describing them which is taking a lot longer and I won’t have time to watch the next episode of Heroes on Netflix prepare for a committee meeting if I do that all today so I’ll do it in steps and you can read all about multiplication for a few days!   I’ll see if I can get Godzilla do to some demonstrations of the more complicated methods tomorrow.

These are carefully ordering according to The Order In Which I Thought of Them.

(1) Doubling and Halving, as described yesterday (used in Egypt, Ethiopia, and presumably Russia).  This is one of my favorite methods, because it’s surprising that it works.

(2) Duplation, which is a variant on the above method, in which you start with 1 and one of the the numbers and double both.  For example, to multiply 14 and 12 you’d start with 1 & 12, and double both until the left-hand column was going to be bigger than the other number, in this case 14.  That’s not at all clear, is it?  Here’s what I mean for 14×12:

1 & 12
2 & 24
4 & 48
8 & 96
16 & <— Oh, I can stop because 16 is bigger than 14.

Since 14 can be written as 8+4+2, put a little mark by those rows:
1 & 12
*2 & 24
*4 & 48
*8 & 96

and add up the corresponding numbers on the right:  96+48+24=168.  And there’s your product!  What you’re really doing is adding the appropriate doubles (96 is 8 12s, 48 is 4 12s, and 24 is 2 12s so when you add them you get 14 12s, as you wanted).  This was also used by the Ancient Egyptians, and it’s referred to on the video from yesterday.

One thing that neat about these methods is that all they use is adding, subtracting, doubling, and/or halving.   The folk who used them had to remember the process, but they didn’t have to memorize 45 separate single digit multiplication facts.

(3) and (4) Mesopotamian Multiplication.  Like the Egyptians, the Babylonians broke down multiplication into addition, subtraction, and halving.  They had one more trick, though:  they had tables that contained the squares of all the numbers from 1 to 60 [which was all they really needed, since they used a Base 60 system].   So to multiply a and b they used one of two formulas: $\frac{\left( a+b \right) ^2 - a^2-b^2}{2}$ or $\frac{\left( a + b \right) ^2 - \left( a - b \right) ^2}{4}$

In other words, to find 14×12, to use the first formula you’d look up 262, 142, and 122 in your table, subtract the last two numbers (196 and 144)  from the first one (676), and then cut that answer in half.  With the second formula you’d look up 262 and 22 in your table, take the difference, and cut that in half twice.

This method totally amuses me because although you can check the algebra to make sure it works (indeed, I usually introduce this by asking students to come up with a formula for ab that uses only addition, subtraction, doubling, halving, and a2, b2, (a+b)2, and/or (ab)2), it’s really an ancient plug-and-chug method.  You don’t have to think at all about why it works, you just do it.

Incidentally, if you walpha a number, Wolfram Alpha will give it to you in Babylonian symbols.  Here’s  34. But watch out: it’s sometimes wrong.

Edited 7/8 to add: Although I read in one book (I’m not sure which now; it’s been several years) that this was done by the Babylonians, Victor J. Katz’s A History of Mathematics doesn’t match that.   I’m not quite sure what to believe now.

(5) Grid or lattice multiplication.  A lot of schools are teaching this today because, although it takes a little while to draw the boxes, you do all your single digit multiplication first and then all your addition.  I’ve heard people complain about it being a new-fangled method, and I like to point out that it’s actually about a thousand years old (Doesn’t that sound snarky?  I try to say it in a friendly way.  I probably fail.).

It’s possible that it originated in India (I’ve seen sources claim this, but I think it’s not considered absolutely certain); it certainly appeared in Arabic books before it made its way to Western Europe.

Here’s how it works.  Suppose you want to multiply 345 and 12, which are conveniently the numbers that I found in this Wikipedia example.  You have a 3-digit number and a 2-digit number, so you make a 3×2 array of boxes.  Put the first number (345) on the top, and the second number (12) on the side.  One of those is the multiplier and one is the multiplicand, but I can never remember which is which. The picture above also shows the next step:  divide each little square in half, and then multiply each pair of digits.  For example, 5×2 is 10, so you write the 1 above the diagonal and the 0 below.

For the next step, you add along the diagonals, carrying as necessary. The reason this works is that the digaonals automatially take care of place value.  For example, look at the third diagonal, with the purple arrow (that ends up in the hundreds place).  It has the 4 of 4×1, which was really 40×10=400.  It has the 0 in the tens place of 4×2, which really would stand for the hundreds place of 40×2.  It has the 6 of 3×2, which really comes from 300×2=600.  Plus it has anything that was carried from the previous diagonal.

Writing the total flat along the bottom makes more sense if you’re teaching it to kids, but in some older books people wrote the answer on the bottom and left-hand side, like on this example from page 23 of The Treviso Arithmetic (Arte dell’Abbaco), an Italian textbook [published in the town of Treviso in Northeastern Italy] from 1478.  This example shows that 934×314=293,276. When I do this in class I often decorate the sides when I’m done, because I’ve seen that done before.  Then it looks like this: (5½) There’s a variation of the above method in which the diagonals go in the other direction but you write the number on the right “upside down”.   Here’s an example from the same page of the Trevisio, again showing 934×314=293,276. I didn’t want to count this as a separate method, because it has all of the same principles as the previous grid.  But I wanted to mention it because if you look at it, it’s a small step to getting the next version of 934×314, shown below, in which the boxes above have had the diagonals removed but you still add along the diagonal (although the total is written at the bottom): This is actually a little bit confusing because with the diagonals removed, the numbers don’t quite line up the way they’re supposed to.   It’d be clearer if the rows were shifted a bit, like this: And THAT looks an awful lot like “traditional” multiplication, where in this case “traditional” means “the way I learned multiplication in in 1978”.  But this isn’t a new-fangled method  either:  this “traditional” multiplication ALSO appeared in the Trevisio!  And that leads us to….

(6) Multiplication like I first learned it, like this: Phew!  That was more lattice multiplication than you ever wanted to see, wasn’t it?  But isn’t it neat how all those pictures (all on the same two pages) just lead into one another, even though they actually were in the reverse order in the original book?

Incidentally, while we’re on the subject of The Trevisio it might amuse you to learn that this 500 year old book also contains the problem “If 17 men build 2 houses in 9 days, how many days will it take 20 men to build 5 houses?”.  If you’re wanting to get yourself a full copy, here’s a link to all 24 Mb of it!  Or, if you don’t read Italian, you can learn a little more about it in this 1996 column by Ivars Peterson.

Next up, some different ways that they don’t teach in school.

Edited to add: Here is The Second Bunch of Ways!

Ethiopian Multiplication

June 9, 2009 One of our recent (oh my goodness has it really been seven years???) grads just sent me this Youtube video of Ethiopian Multiplication, with a note that this reminded her of History of Mathematics.  Which, of course, made me totally happy.

This method of multiplication is also called Egyptian Multiplication (because it was done in Egypt) and Russian Peasant Multiplication (although the Peasant part might be intended as a bit of a pejorative).

Here’s the basic idea:  Suppose you want to multiply two numbers like 14 and 12.  You could use your fingers, of course, but here’s another way:

Start with the two numbers on top.  Halve one, ignoring any remainders or fractions, and double the other, stopping when you get to 1.

14 & 12

7 & 24

3 & 48  [See how I ignored the fact that halving 7 leaves 1 left over?]

1 & 96  <— Stop here.

Now look at the numbers on the right.  Some are across from an even number: in this case, 12 is across from the original 14.  Ignore those, and add the rest.  So we’ll add 24, 48, and 96, which were across from odd numbers, and get 168.  And that’s the product!  Isn’t that cool?

(I think it would be fantastic to write a book called 25 ways to multiply.  I only have about 13 at the moment, though.)

Here’s the video!

Number photo from gokuro.

The Illusion of Winning — or vice versa

June 8, 2009 (Because “The Winning of Illusion” just didn’t sound as good.)

The Winners of the 2009 Best Visual Illusion of the Year Contest have been announced!    There’s a ball that drops straight down, but if you look to the side while it drops it appears to fall at a different angle, a  dove that appears to change color depending on the background, a pair of facesthat are identical except for the coloring (the one with more contrast between the face and eyes/mouth appears female, while the one with less contrast appears male), and more.

As a bonus, Arthur Shapiro, one of the creators of the dropping ball, has several other illusions up on his blog.  These can be posted for non-profit educational use, but I couldn’t get the html code to work.  Bummer.

Since a post on illusions would not be complete without a couple illusions, here are a few.  This first one is called Sander’s Parallelogram or the Sander Illusion (after creator Matthew Luckiesh author Friedrich Sander), and the two blue diagonals are the same length.  Seriously. And finally, here’s a grid illusion.  There are white dots in the middle, but black spots seem to appear: And finally, here’s one in which the bar in the middle is the same shade of gray throughout, but looks like it’s changing color (courtesy of Dodek, published under GNU-FDL). (Contest Winners found via New Scientist.)

Carnival of Mathematics #53 is up!

June 5, 2009 As promised earlier, the Carnival of Mathematics #53 is up at The Math Less Traveled.    It has a healthy number of posts on topics from brain exercises to hyperbolic models to Sangaku and GeoGebra.    The Carnival may be  a bit unpredictable these days, but it’s good to see that it’s just as fun to read!

Speaking of Carnivals, we went to one tonight — a real live one.  And we did the Cake Walk (and won some vivid cupcakes that might be interesting to view under a black light to see if they would glow), but before doing it I did a quick estimate as to whether it was better for four of us to play in a single 10-person game, or to spread it out over more than one game [two going in one round and two going in another].  This is similar to the Box Top problem, but has a different answer because the number of people per round is fixed:  the expected number of wins is the same whether or not we all go in the same round, but we’re more likely to win at least once if we all go at the same time.   In particular, although we lose the possibility of winning more than once by going in the same round, that’s offset by an increase in the probability that we’ll win at least once.  (I find it interesting that it has a different answer than the Box Top problem, which I still don’t feel 100% settled about.)

Probability and Spirographs

June 4, 2009 It would be great if I wrote a post actually combining probability and spirographs, but that’s not what this is.  This is two completely different topics, joined together  by the fact that they both elicited conversation during or after dinner last night.

The Probability Problem:
Suppose your school collects Box Tops, and to encourage you to turn them in, for each 10 you turn in each month you’re entered into a drawing for a Webkinz (so if you turn in 20, you’re entered twice).  The least well formed question is:  if you were able to generate at least one group of 10 per month, is it better to enter them once a month, or to save them all until the end in the hopes of maximizing the opportunity for that single month?  Feel free to put answers in the comments!

I have an idea as to the answer to this in simplest terms, and also an idea as to the answer in practical terms.  In reality, we work on the system of turning them in whenever we remember, which is something between the two extremes.

The Spirograph Site:
I was surfing the web, and found a site called Spirograph Math (which is actually part of a larger site, but this is the game that occupied me).  You have one circle going around the other, and you can trace the design like a spirograph.  You can also pause, change the color, etc.  It draws pretty pictures like this: I think you could use math as an excuse for playing with it.  For example, can you predict how many times the second circle will go around the first before repeating, or how many lines of symmetry the final figure will have?  (One thing to note:  the Pen Position is set relative to the center of Radius B:  if the Pen Position matches Radius B, it means the Pen is on the edge of the second circle, larger and it’s on the outside of that circle, and smaller and it’s on the inside.)

WordPress does the math right

June 3, 2009 This is hardly worthy of a blog post (but, really, if I waited for stuff that was there would be like 3 posts a month), but it still really amused me.

WordPress lets you look at stats for individual posts, which is fun because you can see which posts were the most popular over time (Scoring March Madness by a landslide, thanks to Basketball Guy).  There’s also a column for the % change each week, so if 2 let’s say 200 people look at that particular post one week, and 3 300 people look at it the next week, the percentage increase is +50%, from (300-200)/200.  But what happens when you first post?  The previous week clearly no one looked at the post, so finding the percentage increase causes problems.  But not for WordPress! The percentage increase (in this case of the views of Godzilla makes a hexaflexagon) is  .  That little attention to detail makes me happy.