A brief rant

In light of the on-going controversy regarding admission to New York City’s selective high schools, there has been some discussion of bringing test prep to the middle school classroom.  At first, test prep in the math classroom may sound like an awful idea.  Should we take what is frequently, sad to say, a joyless, stultifying slog and add in a large helping of stress-inducing rote drill to prepare for the SHSAT, PSAT, SAT, what have you?  That can’t be a good idea. And yet…

Maybe we should work on fixing the joyless, stultifying part.  I have taught (oh, lord) a ton of SAT prep.  None of it involves rote drills.  It does involve a number of ideas and methods that absolutely help make math more joyful.  And many, many students have come into my SAT class announcing that they don’t like math only to later report that they “kind of like” SAT math.   So what is happening in SAT class that isn’t happening often enough in school?  Here are some possible answers:

1. SAT math is subversive.  And being subversive is fun.  You don’t have to solve the problem the “right” way!  You can play, guess, draw pictures, make lists…and if it works, you get full credit!  School math is often presented like some deranged cooking class where everyone learns to cook the same dish in the exact same way. No invention, no experimentation, no surprises. (And it’s rarely anything anyone wants to eat anyway.)  We should be encouraging mathematical improvisation and creativity – even when it doesn’t “work”.  After all, how often do we really care where the vertex of the parabola is?  Isn’t the more valuable experience the search?  A math problem that only has one solution path isn’t really a “problem” — it’s more like a “chore”.  No one likes doing chores.

2. SAT math rewards trial and error.  School math often treats trial and error as a tool for the weak.  But trial and error is a fabulous way to begin a problem when you don’t see an algebraic path.  It is not merely an algebra-evasion technique (though on the SAT, it can certainly work that way). Trial and error can serve as an on-ramp to the algebraic solution.  I have used trial and error this way with students in SAT class, physics, precalc, calculus, and multivariable calc.  To keep this rant brief (well…) I will expand on and demonstrate this idea in a separate post.  But to give you the flavor:

1. You take a guess (or pick an answer if it is a multiple choice test)

2. You mess around with your guess, applying whatever line of reasoning you can come up with to show that your guess is wrong (as it usually will be).

3. You replace your guess with a variable and then use your earlier reasoning as a template to come up with the algebraic equation or equations that will lead to an answer.

I have shown this method to students as young as 6^th grade and to colleagues who have been teaching for decades.  I can tell you that most of them got a kick out of this.  Again, it feels subversive. And who doesn’t like the feeling that you are getting away with something?  But really, you’re not. You are just separating the problem into two phases, doing the reasoning before you attempt to apply the algebra.  But it only works if you are willing to take a guess. “Could this be the answer?” is an amazingly helpful question, and not just on a standardized test.

3. SAT math rewards making up numbers for variables.  This one is an SAT classic that should really be a staple of school math as well.  Many students see an algebraic equation as a sequence of letters and symbols, probably including an “=” somewhere in there.  We want them to see it as a statement, a sentence written in the language of algebra, one that might be true if the variables have the “right” values. 

This gives us an efficient way of finding whether two expressions are equivalent.  Make up numbers (ideally, slightly “unusual” ones), put them into both expressions, and see if they come out equal.  And yes, this method can generate false positives. But it never generates false negatives – if you get two different numeric answers, then your two algebraic expressions were definitely not equivalent.

This idea has many uses.  First of all, we can teach students that arithmetic is a testing bench for algebra.  Any time you are not certain whether something is allowed, try it with numbers!  We all have taught plenty of students who believe that (x + y)² = x² + y².  So let x =4 and y = 6 and see what happens!

Secondly, it inoculates students from the classic standardized test question: “which of the following expressions is equivalent to…” , a type of question that deserves to be subverted.

And thirdly, once again it gives us a way to separate the reasoning from the algebra.  You can make up numbers for the variables, work out an answer, and then put your variables back into the answer choices.  It’s probably the SAT method that seems most devious. Even in physics class, often my beginning students are flummoxed by questions like:

A car has mass m and speed v.  A second car has half the mass and three times the speed. Find the ratio of the second car’s kinetic energy to the first car’s kinetic energy.

The problem is not that these students don’t know the formula for kinetic energy. It’s that they have not yet developed the ability to think about the ratios and proportions involved.  When they are fluent, they will be able to look at the formula, K = ½ mv² and think through the problem with ease.  But while they are still developing, isn’t it nice to have an alternative scaffold available? 

Make up numbers for m and v and calculate the kinetic energy.

Cut the m in half  and triple the v and recalculate the kinetic energy.

Find the ratio of your two answers.

Beginners are often surprised that this works, and even skeptical, thinking that there was something special about the numbers they made up. So let them make up a new set of numbers.    It seems like magic. And again, it serves as an on-ramp to the algebraic discussion.

Two more points and I’ll call it a rant

1. None of what I am recommending will take time away from your curriculum. These are attitudes and strategies that can overlay whatever you are currently teaching. Still, take a hard look: it isn’t all joyful now and it isn’t all as essential as you may think.  But we can argue about that another time.  I would teach these methods even if the SAT and its ilk were to vanish.  It’s a more fun way to do math.

2. Test prep in middle school is not going to magically bring diversity to the elite high schools. It might help, but I would guess that it won’t be nearly enough.  We are going to have to do more.

File this under “math I should have realized”.

Even after all these years, I still find logarithms to be challenging, compelling, and surprising.  This post is about an idea that snuck up on me this morning. And it snuck up even though:

• I have written: Log₂ 8 = 3 because 2³ = 8 on many whiteboards and many review sheets. 
• I have learned (and taught!) the classic collection of assorted log properties
• I have read David Mermin’s lovely piece “Logarithms!” (and blogged about it here.)
• I have written a chapter on exponential functions and logarithms in “Advanced Math for Young Students”.

And yet, there is a basic idea about logarithms that I never really thought about (until this morning).  So here it is:

The integer part of the logarithm of an integer is one less than the number of digits the integer has when it is represented in the same base as the logarithm itself.

You could write:

Floor[Log_b  N] + 1 = the number of N’s digits when N is represented in base b.

In other words, the logarithm counts the digits — and not just in base-10. You just have to round down and add 1.  I would like to explain this, if for no other reason than to convince myself that it is true! 

Starting with base 10:

Let’s look at some common (base 10) logarithms that we can find without a calculator:

Log 1 = 0               …because 10⁰ = 1

Log 10 = 1            …because 10¹ = 10

Log 100 = 2          …because 10² = 100

Log 1000 = 3        …because 10³ = 1000

Log 10000 = 4     …because 10⁴ = 10000

I think you get the idea.  If we construct a number, say N, by writing a 1 followed by k zeroes, then the common (base 10) logarithm, log N = k.  And the number will have k + 1 digits: the k zeroes and the 1 in front of them. So for each of these, adding 1 to the common logarithm gives you the number of digits.

Now let’s look at one that you do need a calculator for. I’ll report three decimal places:

Log 375 = 2.574   …because 10^2.574 = 374.973 ≈ 375

And it makes sense: 375 is more than 100 so its common logarithm is greater than 2.  But 375 is also less than 1000 so its common logarithm is less than 3.  It is more than the smallest 3 digit number and less than the smallest 4 digit number.    And if you take the common logarithm, which was 2.574, round it down and then add 1, you get the number of digits.  Nice.

Let’s try another (again reporting 3 decimal places):

Log (8787) = 3.944   … because 10^3.944  ≈ 8787 (though with a little more round-off error, but still…)

And this makes sense too: 8787 is more than 1000 so its common log is greater than 3.  But it is less than 10000 so its common log is less than 4.  In fact, it is closer to 10000 so its log is close to 4, but still less.  And if we round down 3.944 and add 1, again we get the number of digits in the number we started with.

OK, I admit that I kind-of knew that already.  But for some reason, this next step caught me by surprise…let’s move away from base 10 and see what happens.

Other bases and other logarithms

On and off this year, I have been playing with James Tanton’s invention, “Exploding Dots”.  If you have not yet seen this, you really should take a look.  You can get started here. But really, Google it – those dots are not just exploding, they are proliferating!

One thing that Exploding Dots teaches you almost immediately and organically is how to express numbers in other bases.  And something interesting happens when you look at the base 2 logarithms as you look at the base 2 representations…

 So if a number N happens to be an integer power of 2, say 2^k, then we know two things:

1. Its base-2 logarithm, log₂ N = k

2. Its base-2 representation will consist of k zeroes and a leading 1.  And that makes k + 1 digits.

Now let’s get some help from a calculator.

Log₂ 5 = 2.322 …because 2^2.322  ≈ 5

Does it make sense? Well, since 5 is more than 4, we expect its base-2 logarithm to be more than 2.  But 5 is less than 8, so we expect its base-2 logarithm to be less than 3.  So this looks right.

But we are not done! If we explode some dots, we can find the base 2 representation.

5 = 1 0 1 and that makes 3 digits – which is also what we get when we round down the base-2 logarithm and then add 1! 

OK, let’s check a big number:

500 expressed in base 2 is 1 1 1 1 0 1 0 0 (which you might want to check…)

I see 9 digits.

Log₂ 500 = 8.966 (and yes, it checks out: 2^8.966  ≈ 500)

Round it down and add 1…YES!

So if you want to know the number of digits in the base-b representation of a number, find the base-b logarithm, round it down to its floor and then add one.  The logarithm is the digit counter!