Area models for multiplication throughout the K-12 curriculum

Let’s take a look at area models, shall we?

My thesis today is that area models should be ubiquitous across the entire curriculum because mathematics is a sense making discipline. As math educators, we ought to encourage our students to take every opportunity to visualize their mathematics in an effort to illuminate, explain, prove, and bring intuition.

So let’s take a walk through the K-12 math curriculum and highlight the use of area models as they might apply to arithmeticalgebra, and calculus.

base-ten-blocks

Arithmetic

Students experience area models for the first time in elementary school as they work to visualize multi-digit multiplication. This can also be used for division as well, just running the logic in reverse–that is, seeking an unknown “side length” rather than an unknown area. And Base Ten Blocks can be used to help students understand the building blocks of our number system.

Here’s how you might work out 27\times 54:

27\times 54 = (20+7)(50+4)=(20)(50)+(20)(4)+(7)(50)+(7)(4)

area-model-multiplication

27\times 54=1000+80+350+28=1458

The advantage of using a visual model like this is that you can easily see your calculation and explain why constituent calculations, taken together, faithfully produce the desired result. If you do a “man on the street” interview with most users or purveyors of the standard algorithm, you would almost certainly not get crystal clear explanations for why it produces results. For a further discussion of area models for multi-digit multiplication, see this article, or read Jo Boaler’s now famous book Mathematical Mindsets.

Algebra

In middle school, as students first encounter algebra, they may use area models to support their algebraic reasoning around multiplying polynomials. And in an Algebra 2 course they may learn about polynomial division and support their thinking using an area model in the same way they used area models to do division in elementary school. Here Algebra Tiles can be used as physical manipulatives to support student learning.

Here’s how you might work out (x+4)(2x+3):

(x+4)(2x+3)=(x)(2x)+(x)(3)+(4)(2x)+(4)(3)

area-model-polynomials

(x+4)(2x+3)=2x^2+3x+8x+12=2x^2+11x+12

Notice also that if you let x=10, you obtain the following result from arithmetic:

14\times 23 = 200+110+12=322

The Common Core places special emphasis on making such connections. I agree with this effort, even though I can also commiserate with fellow math teachers who say things like, “My Precalculus students still use the box method for multiplying polynomials!” We definitely want to move our students toward fluency, but perhaps we should wait for them to realize that they don’t need their visual models. Eventually most students figure out on their own that it would be more efficient to do without the models.

Calculus

Later in high school, as students first study calculus, area models can be used to bring understanding to the Product Rule–a result that is often memorized without any understanding. Even the usual “textbook proof” justifies but does not illuminate.

Here’s an informal proof of the Product Rule using an area model:

The “change in” the quantity L\cdot W can be thought of as the change in the area of a rectangle with side lengths L and W. That is, let A=LW. As we change L and W by amounts \Delta L and \Delta W, we are wondering how the overall area changes (that is, what is \Delta A?).

If the side length L increases by \Delta L, the new side length is L+\Delta L. Similarly, the width is now W+\Delta W. It follows that the new area is:

A+\Delta A=(L+\Delta L)(W+\Delta W)=LW+L\Delta W+W\Delta L+\Delta L\Delta W

area-model-product-rule

Keeping in mind that A=LW, we can subtract this quantity from both sides to obtain:

\Delta A=L\Delta W+W\Delta L+\Delta L\Delta W

Dividing through by \Delta x gives:

\frac{\Delta A}{\Delta x}=L\cdot\frac{\Delta W}{\Delta x}+W\cdot\frac{\Delta L}{\Delta x}+\frac{\Delta L}{\Delta x} \frac{\Delta W}{\Delta x} \Delta x

And taking limits as \Delta x\to 0 gives the desired result:

\frac{dA}{dx}=L\cdot\frac{dW}{dx}+W\cdot\frac{dL}{dx}

Conclusion

If you’re like me, you once looked down on area models as being for those who can’t handle the “real” algebra. But if we take that view, there’s a lot of sense-making that we’re missing out on. Area models are an important tool in our tool belt for bringing clarity and connections to our math students.

Okay, so last question: Base Ten Blocks exist, and Algebra Tiles exist. What do you think? Shall we manufacture and sell Calculus DX Tiles © ? 🙂

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Extraneous Solutions – Part 2 of 3

Solving an Equation as a Sequence of Equation Replacement Operations

Part 1 was so long because I wanted to be extremely thorough and to present things to an audience that perhaps hadn’t thought much about the logic of equation solving at all. Since we’re now all experts, perhaps it’s worth it to summarize everything very succinctly.

Given an equation in one free variable, we want to find the solution set. To do this, we replace that equation with an equivalent equation whose solution set is more obvious.

(1) 8x - 5 = 5x + 1

(2) 8x = 5x + 6

(3) 3x = 6

(4) x = 2

If in the transition from (1)-(2), from (2)-(3), and from (3)-(4) we are careful to replace each equation with an equivalent equation, then by the transitivity of equivalence, the original equation and terminal equation are guaranteed to be equivalent. Since the solution set of the terminal equation is obvious, we know the solution set of the original equation, as well. Thus solving an equation requires establishing that certain equation replacement operations are indeed equivalence preserving and having the creativity and experience to know which ones to apply and in what order.

What are the Equivalence-Preserving Operations on Equations?

If a = b, then f(a) = f(b) for any well-defined function f. If a and b are expressions containing a free-variable, then any value of that variable which satisfies a = b will also satisfyf(a) = f(b). In other words, if you find it useful, feel free to replace any equation with a new equation which is the result of applying any function to both sides of the original equation. Any solution to the original equation will also be a solution to the new equation.

If the function f is also one-to-one, then by definition, f(a) = f(b) \Rightarrow a = b so any solution of f(a) = f(b) will also be a solution to a = b. Thus applying f to both sides of an equation is equivalence-preserving. If f is not one-to-one, then in general, the operation is not equivalence-preserving.

In solving equation (1), we applied f(n) = n + 5, g(n) = n - 5x and h(n) = n/2 in that order. Since all three of the functions are one-to-one, we are assured that (1) and (4) are equivalent. If we had cause to apply a non-one-to-one function, then we should be vigilant for extraneous solution.

A More Interesting Example

Consider

(5) \sqrt{6x-2} - \sqrt{x+1} = 2

As I mentioned in the other post, these square roots are begging to be squared, but since there are two of them, one squaring will not be enough. Even though it’s not necessary to do so, it’s helpful to move one radical expression to the other side.

(6) \sqrt{6x-2} = 2 + \sqrt{x+1}

(7) 6x - 2 = 4 + 4\sqrt{x + 1} + x + 1 We squared!

(8) 5x - 7 = 4\sqrt{x+1}

(9) 25x^2 - 70x + 49 = 16x + 16 We squared again!

(10) 25x^2 - 86x + 33 = 0

(11) (25x - 11)(x - 3) = 0

So x \in \{\frac{11}{25}, 3\}

Since in the transition from (6)-(7) and again in the transition from (8)-(9) we had reason to apply the non-one-to-one function f(n) = n^2, we should be vigilant for extraneous solutions. [Note: since both sides of (6) are necessarily positive, applying f(n) = n^2 is equivalence-preserving, so no extraneous roots will be created there.] By checking back in the original equation, we see that 3 is a solution, but \frac{11}{25} is not. I am more or less content to leave it at that. But some may ask for more clarity as to exactly what happened and when, so let’s indulge them.

I will now list each equation in reverse order along with its solution set:

(11) (25x - 11)(x - 3) = 0                            \{\frac{11}{25}, 3\}

(10) 25x^2 - 86x + 33 = 0                             \{\frac{11}{25}, 3\}

(9) 25x^2 - 70x + 49 = 16x + 16                 \{\frac{11}{25}, 3\}

(8) 5x - 7 = 4\sqrt{x+1}                                     \{3\}

Since 5\cdot\frac{11}{25} - 7 = \frac{11}{5} - \frac{35}{5} = -\frac{24}{5} \neq 4\sqrt{\frac{11}{25} + 1} = 4\sqrt{\frac{11}{25} + \frac{25}{25}} = 4\sqrt{\frac{36}{25}} = 4\cdot\frac{6}{5} = \frac{24}{5}

So we have isolated the precise moment when the extraneous solution x = \frac{11}{25} is created and it appears exactly where we would expect it, in the transition from (8) to (9) as we replaced (8) with the result of applying the non-one-to-one function f(n) = n^2 to both sides.

More specifically, if x = \frac{11}{25}, (8) reads - \frac{24}{5} = \frac{24}{5}, which is false, but (9) reads (- \frac{24}{5})^2 = ( \frac{24}{5})^2, which is true. For this particular value of x, we squared both sides and replaced a false statement with a true statement. In retrospect, we can say that x =\frac{11}{25} is not a solution to (8) or to any previous equation in the solving sequence, but is a solution to (9) and thus to all subsequent equations in the solving sequence.

(7) 6x - 2 = 4 + 4\sqrt{x + 1} + x + 1                          \{3\}

Since both sides of (7) are positive when x = 3, it does not surprise us that,

(6) \sqrt{6x-2} = 2 + \sqrt{x+1}                               \{3\}

(5) \sqrt{6x-2} - \sqrt{x+1} = 2                                \{3\}

By fully analyzing the logic behind each step of our equation replacement sequence, we not only:

  • confirm that x = 3 is a solution and that x = \frac{11}{25} is not and
  • understand that squaring both sides may produce an extraneous solution

but also

  • isolate the precise step in the solving sequence in which this extraneous solution was created answering the why, how, and when for this problem
  • confirm that the non-solution status of x = \frac{11}{25} is not merely due to an error of algebra or arithmetic, but is a direct result of that fact that this value produces an equation (8) of the form a = -a

That last point is crucial in distinguishing the phenomenon of extraneous roots from the phenomenon of user error in algebra or arithmetic. If our equation solving sequence consists solely of equivalence-preserving operations, we do not even need to check to see if solutions to our terminal equation are also solutions to our original equation. If we do decide to check, perhaps out of an abundance of caution, and find a discrepancy, then user error must be to blame.

On the other hand, if a solver does employ solution-set-enlarging operations in the solving sequence and finds that a solution to the terminal equation is not a solution to the original equation, is this because the solution is extraneous or due to user error? One could perform an analysis like I did above and confirm that the non-solution is not due to user error, but instead to the logic of the process.

Extraneous Solutions – Part 1 of 3?

Disclaimer

Within my small inner circle of math teachers, the mystery of extraneous solutions seems to be the issue of the year. I have so much to say on this topic (algebraic, logical, pedagogical, historical, linguistic) that I don’t really know where to begin. My only disclaimer is that I’m not really sure if this topic is all that important.

Solving an Equation with a Radical Expression

Consider the following equation:

(1) 2\sqrt{x+8} +5 = 11

One hardly needs algebra skills or prior knowledge to solve this, but prior experience suggests trying to isolate x.

(2) 2\sqrt{x+8} = 6 (we subtract 5 from both sides)

(3) \sqrt{x+8} = 3 (we divide both sides by 2)

Now, if the square root of something is 3, then that something must be 9, so it immediately follows that

(4) x+8 = 9

(5) x = 1 (we subtract 8 from both sides)

Squaring Both Sides

In my transition from (3) to (4), I used a bit of reasoning. Some conversational common sense told me that “if the square root of something is 3, then that something must be 9”. But that logic is usually just reduced to an algebraic procedure: “squaring both sides”. If we square both sides of equation (3), we get equation (4).

On the one hand, this seems like a natural move. Since the meaning of \sqrt{a} is “the (positive) quantity which when squared is a“, the expression \sqrt{a} is practically begging us to square it. Only then can we recover what lies inside. A quantity “which when squared is a” is like a genie “which when summoned will grant three wishes”. In both cases you know exactly what to do next.

Unfortunately, squaring both sides of an equation is problematic. If a = b is true, then a^2 = b^2 is also true. But the converse does not hold. If a^2 = b^2, we cannot conclude that a = b, because opposites have the same square.

This leads to problems when solving an equation if one squares both sides indiscriminately.

A Silly Equation Leads to Extraneous Solutions

Consider the equation,

(6) x = 4

This is an equation with one free variable. It’s a statement, but it’s a statement whose truth is impossible to determine. So it’s not quite a proposition. Logicians would call it a predicate. Linguistically, it’s comparable to a sentence with an unresolved anaphor. If someone begins a conversation with the sentence “He is 4 years old”, then without context we can’t process it. Depending on who “he” refers to, the sentence may be true or false. The goal of solving an equation is to find the solution set, the set of all values for the free variable(s) which make the sentence true.

Equation (6) is only true if x has value 4. So the solution set is \left\{4 \right\} . But if we square both sides for some reason…

(7) x^2 = 16 has solution set \left\{4, -4\right\}

We began with x = 4, “did some algebra”, and ended up with x^2 = 16. By inspection, -4 is a solution to x^2 = 16, but not to the original equation which we were solving, so we call -4 an “extraneous solution”. [Extraneous – irrelevant or unrelated to the subject being dealt with]

Note that the appearance of the extraneous solution in the algebra of (6)-(7) did not involve the square root operation at all. But this example was also a bit silly because no one would square both sides when presented with equation (6), so let’s look at a slightly less silly example.

Another Radical Equation

(8) 2\sqrt{x+8} + 5 = -1

(9) 2\sqrt{x+8} = -6

(10) \sqrt{x+8} = -3

People paying attention might stop here and conclude (correctly) that (10) has no solutions, since the square root of a number can not be negative. Closer inspection of the logic of the algebraic operations in (8)-(10) enables us to conclude that the original equation (8) has no solutions either. Since a = b \iff a - 5 = b -5, any solution to (8) will also be a solution to (9) and vice versa. Since a = b \iff a/2 = b/2, any solution to (9) will also be a solution to (10) and vice versa. So equations (8), (9), and (10) are all “equivalent” in the sense that they have the same solution set.

But what if the equation solver does not notice this fact about (10) and decides to square both sides to get at that information hidden inside the square root?

(11) x+8 = 9

(12) x = 1

Again we have an extraneous solution. x = 1 is a solution to (12), but not to the original equation (8). Where did everything go wrong? By the previous logic, (8), (9), and (10) are all equivalent. (11) and (12) are also equivalent. So the extraneous solution somehow arose in the transition from (10) to (11), by squaring both sides.

So unlike subtracting 5 from both sides or dividing both sides by 2, squaring both sides is not an equivalence-preserving operation. But we tolerate this operation because the implication goes in the direction that matters. If a = b, then a^2 = b^2, so if a and b are expressions containing a free variable x, any value of x that makes a = b true will also make a^2 = b^2 true.

In other words, squaring both sides can only enlarge the solution set. So if one is vigilant when squaring both sides to the possible creation of extraneous solutions, and is willing to test solutions to the terminal equation back into the original equation, the process of squaring both sides is innocent and unproblematic.

Those Who are Still Not Satisfied

Still there are some who are not satisfied with this explanation: “Why does this happen? What is really going on? Where do the extraneous solutions come from? What do they mean?”

One source of the problem is the square root operation itself. \sqrt{a} is, by the conventional definition, the positive quantity which when squared is a. The reason that we have to stress the positive quantity is that there are always two real numbers that when squared equal any given positive real number. There are a few slightly different ways of making this same point. The operation of squaring a number erases the evidence of whether that number was positive or negative, so information is lost and we are not able to reverse the squaring process.

We can also phrase the phenomenon in the language of functions. Since squaring is a common and useful mathematical practice, information will often come to us squared and we’ll need an un-squaring process to unpack that information. f(x) = x^2, for all the reasons just mentioned, is not a one-to-one function, so strictly speaking, it is not invertible. But un-squaring is too important, so we persevere. As with all non-one-to-one functions, we first restrict the domain of f(x) = x^2 to [0, \infty) to make it one-to-one. This inverse, f^{-1}(x) = \sqrt{x} thus has a positive range and so the convention that \sqrt{a} \geq 0 is born. So every use of the square root symbol comes with the proviso that we mean the positive root, not the negative root. We inevitably lose track of this information when squaring both sides.

[Note: Students can easily lose track of these conventions. After a lot of practice solving quadratic equations, moving from x^2 = 9 effortlessly to x = \pm 3, students will often start to report that \sqrt{9} = \pm 3.]

The convention that we choose the positive root is totally arbitrary. In a world in which we restricted the domain of  f(x) = x^2 to (-\infty, 0] before inverting, \sqrt{9} would be -3. In that world, x = 1 is a perfectly good solution to 2\sqrt{x+8} + 5 = -1, not extraneous at all.

A Trigonometric Equation which Yields an Extraneous Solution

For parallelism, consider the (somewhat artificial) equation:

(13) \arccos(2x-1) = \frac{4\pi}{3}

Like in (10), careful and observant solvers might notice that the range of the \arccos(x) function is [0, \pi] and correctly conclude that the equation has no solutions. But there seems to be a lot going on inside that \arccos expression, so many will rush ahead and try to unpack it by “cosineing”. Indeed, since a=b \Rightarrow \cos(a) = \cos(b), this seems innocent.

(14) 2x - 1 = -\frac{1}{2}

(15) 2x = \frac{1}{2}

(16) x = \frac{1}{4}

But x = \frac{1}{4} is an extraneous solution since \arccos(-\frac{1}{2}) = \frac{2\pi}{3} not \frac{4\pi}{3}.

The explanation for this extraneous solution will be similar to the logic we used above. If a = b, then \cos(a) = \cos(b), so if a and b are expressions containing a free variable x, any value of x that makes a = b true will also make \cos(a) = \cos(b) true. So we will not lose any solutions by “taking the cosine of both sides”. But as the cosine function is not one-to-one, \cos(a) = \cos(b) does not imply that a = b. So taking the cosine of both sides, just like squaring both sides, can enlarge the solution set.

The above paragraph explains why extraneous solutions could appear in the solution of (13), but maybe not why they do appear. For that, we again must look to the presence of the \arccos function. Since \cos is not one-to-one, we had to arbitrarily restrict its domain to [0, \pi] prior to inverting. So every use of the \arccos symbol comes with its own proviso that we are referring to a number in a particular interval of values. In a world in which we had restricted the domain of \cos to [\pi, 2\pi] prior to inverting, x = \frac{1}{4} would be a perfectly good solution to \arccos(2x-1) = \frac{4\pi}{3}, not extraneous at all.

The above examples seem to suggest that one can avoid dealing with extraneous solutions by carefully examining one’s equations at each step. But in practice, this really isn’t possible. I saved the fun examples for the end, but as this post is already way way too long, they will have to wait for a bit later.

-Will Rose

Thanks

Thanks to John Chase for letting me guest post on his blog. Thanks to James Key for encouraging me again and again to think about extraneous solutions.

Proving identities – what’s your philosophy?

What happens in your classroom when you give students the following task?

Prove 1+\frac{1}{\cos{\theta}}=\frac{\tan^2{\theta}}{\sec{\theta}-1}.

Sometimes the command is Verify or Show instead of Prove, but the intent is the same.

kreispuzzel-1713170_960_720

Two non-examples

Here are two ways that a student might work the problem.

Method 1

1+\frac{1}{\cos{\theta}}=\frac{\tan^2{\theta}}{\sec{\theta}-1}

1+\sec{\theta}=\frac{\tan^2{\theta}}{\sec{\theta}-1}

(\sec{\theta}-1)(1+\sec{\theta})=\tan^2{\theta}

\sec^2{\theta}-1=\tan^2{\theta}

\tan^2{\theta}=\tan^2{\theta}

Method 2

1+\frac{1}{\cos{\theta}}=\frac{\tan^2{\theta}}{\sec{\theta}-1}

1+\sec{\theta}=\frac{\sec^2{\theta}-1}{\sec{\theta}-1}

1+\sec{\theta}=\frac{(\sec{\theta}-1)(\sec{\theta}+1)}{\sec{\theta}-1}

\sec{\theta}+1=\sec{\theta}+1

How do you feel about these methods? In my opinion, both methods represent a fundamental misunderstanding of the prompt. Method 1 is especially grotesque, but Method 2 also leaves a lot to be desired. Let me explain. And if you think the above methods are perfectly fine, please be patient and hear me out.

This is the crux of the issue:

The prompt was to prove the statement. But if the first line of our work is the very thing we’re out to prove, then we are already assuming the thing we want to prove. We’re Begging the Question.

It’s as if someone demands,

“Prove Statement X, please!”

and we reply,

“Well, let’s first start by assuming that Statement X is true.”

This is nonsense.

What went wrong?

So what is the proper way to engage this proof? Let’s roll back a bit.

The error in these approaches seems to stem from a desire to perform algebraic operations on both sides of an equation in the same way that you might if you were solving an equation.

When we “do algebra” and write Equation B below another Equation A without any words, we always mean that Equation A implies Equation B. That is, when we write

Equation A

Equation B

Equation C

etc…

we mean that Equation C follows from Equation B, which follows from Equation A.

Some might claim that each line should be equivalent to the last. But, again, when we “do algebra” by performing algebraic manipulations to both sides of an equation to transform it from equation A into equation B, we always mean A\Rightarrow B, we don’t mean A\iff B. Take, for example, the following algebra which results in an extraneous solution:

\sqrt{x+2}=x

(\sqrt{x+2})^2=x^2

x+2=x^2

0=x^2-x-2

0=(x-2)(x+1)

x=2 \text{ or } x=-1

In this example, each line follows from the previous, however reversing the logic doesn’t work. But we accept that this is the usual way we do algebra (A\Rightarrow B\Rightarrow C\Rightarrow \cdots). Here the last line doesn’t hold because only one solution satisfies the original equation (x=2). Remember that our logic is still flawless, though. Our logic just says that IF \sqrt{x+2}=x for a given xTHEN (\sqrt{x+2})^2=x^2.

As we move through the algebra line by line, we either preserve the solution set or increase its size. In the case above, the solution set for the original equation is {2}, and as we go to line 2 and beyond, the solution set is {2,-1}.

For more, James Tanton has a nice article about extraneous solutions and why they arise, which I highly recommend.

So if this is the universal way we interpret algebraic work, which is what I argue, then it is wrong to construct an argument of the form A\Rightarrow B\Rightarrow C in order to prove statement A is true from premise C. The argument begs the question.

Both Method 1 and Method 2 make this mistake.

 

How does a proof go again?

I want to actually make a more general statement. The argument I gave above regarding how we “do algebra” is actually how we present any sort of deductive argument. We always present such an argument in order, where later statements are supported by earlier statements.

ANY time we see a sequence of statements (not just equations) A, B, C that is being put forward as a proof, if logical connectives are missing, the mathematical community agrees that “\Rightarrow” is the missing logical connection.

That is, if we see the proof A,B,C as a proof of statement C from premise A, we assume that the argument really means A\Rightarrow B\Rightarrow C.

This is usually the interpretation in the typical two-column proof, as well. We just provide the next step with a supporting theorem/definition/axiom, but we don’t also go out of our way to say “oh, and line #7 follows from the previous lines.”

Example: Given a non-empty set E with lower bound a and upper bound b, show that a\leq b.

1. E is non-empty and a and b are lower and upper bounds for E. (given)
2. Set E contains at least one element x. (definition of non-empty)
3. a\leq x and x\leq b. (definitions of lower and upper bound)
4. a\leq b. (transitive property of inequality)

Notice I never say that one line follows from the next. And also notice that it would be a mistake to interpret the logical connectives as biconditional.

The path of righteousness

I encourage my students to work with only ONE side of the expression and manipulate it independently, in its own little dark box, and when it comes out into the light, if it looks the same as the other side, you’ve proved the equivalence of the expressions.

For example, to show that \log\left(\frac{1}{t-2}\right)-\log\left(\frac{10}{t}\right)=-1+\log\left(\frac{t}{t-2}\right) for t>2, I would expect this kind of work for “full credit”:

\text{LHS }=\log\left(\frac{1}{t-2}\right)-\log\left(\frac{10}{t}\right)

=-\log(t-2)-\log(10)+\log(t)

=-\log(10)+\log(t)-\log(t-2)

= -1 + \log\left(\frac{t}{t-2}\right)

=\text{ RHS}

Interestingly, I WOULD also accept an argument of the form A\iff B\iff C as justification for conclusion A from premise C, but I would want a student to say “A is true if and only if B is true, which is true if and only if C is true.” Even though it provides a valid proof, I discourage students from using this somewhat cumbersome construction.

So let’s return to the original problem and show a few ways a student could do it correctly.

Three examples

Method A – A direct proof by manipulating only one side

\text{LHS}=\frac{\tan^2{\theta}}{\sec{\theta}-1}

=\frac{\sec^2{\theta}-1}{\sec{\theta}-1}

=\frac{(\sec{\theta}-1)(\sec{\theta}+1)}{\sec{\theta}-1}

=\sec{\theta}+1

=\text{RHS}

Method B – A proof starting with a known equality

\tan^2{\theta}=\tan^2{\theta}

\sec^2{\theta}-1=\tan^2{\theta}

(\sec{\theta}-1)(1+\sec{\theta})=\tan^2{\theta}

1+\sec{\theta}=\frac{\tan^2{\theta}}{\sec{\theta}-1}

1+\frac{1}{\cos{\theta}}=\frac{\tan^2{\theta}}{\sec{\theta}-1}

Method C – Carefully specifying biconditional implications

1+\frac{1}{\cos{\theta}}=\frac{\tan^2{\theta}}{\sec{\theta}-1}

\text{if and only if}

1+\sec{\theta}=\frac{\tan^2{\theta}}{\sec{\theta}-1}

\text{if and only if}

(\sec{\theta}-1)(1+\sec{\theta})=\tan^2{\theta}

\text{if and only if}

\sec^2{\theta}-1=\tan^2{\theta}

\text{if and only if}

\tan^2{\theta}=\tan^2{\theta}

While all of these are now technically correct, I think we all prefer Method A. The other methods are cool too. But please, please, promise me you won’t use Methods 1 or 2 which I presented in my introduction.

In conclusion

Some might argue that the heavy criticism I’ve leveled against Methods 1 and 2 is nitpicking. But I disagree. This kind of careful reasoning is exactly the business of mathematicians. It’s not good enough to just produce “answers,” our job is to produce good reasoning. Mathematics, remember, is a sense-making discipline.

Thanks for staying with me to the end of this long-winded post. Can you tell I’ve had this conversation with a lot of students over the last ten years?

Further reading

  1. Dave Richeson has a similar rant with a similar thesis here.
  2. This article was originally inspired by this recent post on Patrick Honner’s blog. A bunch of us fought about this topic in the comments, and in the end, Patrick encouraged me to write my own post on the subject. So here I am. Thanks for pushing me in the right direction, Mr. Honner!

 

Challenge Problems

Want to enrich your Precalculus course with difficult problems? Look no further!

Very-Difficult-Mazes-Coloring-Page-1I teach a high-octane version of Precalculus to students in our magnet program. Our course, like most Precalculus courses, covers a very wide variety of topics. As often as possible, I like to give them more difficult problems that enrich the material from the book. These documents are a work in progress, but feel free to steal them (just email me a copy if you improve them!):

If you want solutions for any of these, shoot me an email.

These aren’t 100% polished by any means, but I’m sharing them anyway! Long live the spirit of sharing :-).

By the way, many of these problems are collected from other sources but I’m too far removed from those sources to properly attribute the problem-creator. My sincere apologies!

Guess who!

In an effort to share more of my resources through this blog, here’s another installment.

This time I’m sharing a little worksheet that I created called Guess Who? It’s a short activity–a warm up, or an exit card–and students should be able to do it in 5 minutes or so. I do this in my Precalculus class at the beginning of the year, but depending on the timing and the context, it could be appropriate in an Algebra 2 or Calculus class as well.

The functions and the questions have a one-to-one correspondence and there is a unique solution to the worksheet.

These 12 functions might seem a bit strange, but they are the “12 basic functions” named by our Precalculus textbook authors.

Here are two additional activities that can go with a discussion of functions and their properties:

  1. I have all the functions printed out on 8.5″x11″ paper and backed with colored paper so they look nice. I get twelve volunteers to go up to the front and hold the functions. Then we can play all sorts of games. We can ask all the functions that have an asymptote to step forward. We can ask all the odd functions to step forward. Which functions are bounded? Which functions are always increasing? Which functions have a range of all real numbers? But we can also play a guessing game: A student in the audience picks a function and writes it down without telling everyone. The other students in the audience ask yes-no questions about their function, like “Is your function continuous for all real numbers?” Each time, functions that don’t qualify step back and only a few functions remain. This is repeated until the chosen function is the only one that remains.
  2. Another fun game idea comes from one of my colleagues. I love this: Have a bunch of “name tags” made up for all your students. The name tags will be one of the 12 basic functions and students will wear these on their backs, without knowing what their function is. They then have to walk around the room and ask other students yes-no questions about the features of their function until they can identify which function they are. I think I’ve played a version of this with celebrities or something. But it’s perfect for the math classroom, too!

Okay, that’s my contribution to the MTBoS for the day :-).

How do you expand √(a+b)?

This is a question that was recently asked on Quora:

it’s easy to expand
(a+b)^2 = a^2+2ab+b^2 or
(a+b)^3=a^3+3ab^2+3a^2b+b^3
or some other (a+b)^n but what about (a+b)^{1/2} aka. \sqrt{a+b}

Here’s my answer:

Just have Wolfram|Alpha do it for you :-).

But if you were on a desert island without access to Wolfram Alpha, here’s how you might think it through:

Are you already comfortable with the Binomial Theorem? Here it is again, but stated in a particular way that I think we’ll like.

\left(x+1\right)^r=1+rx+\frac{r(r-1)}{2!}x^2+\frac{r(r-1)(r-2)}{3!}x^3+\cdots

Look at it and make sure you understand it, and verify that it really is equivalent to the formulation of the Binomial Theorem you know.

Now, for the big trick. It turns out the above statement holds true not for just r=1,2,3,\ldots but for all real r. The only catch is that this often results in an infinite series. (These series results can also be obtained by Taylor expansion.)

In particular, it works for r=1/2:

\left(x+1\right)^{1/2}=1+\frac{1}{2}x+\frac{1/2(1/2-1)}{2!}x^2+\cdots

\left(x+1\right)^{1/2}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+\cdots

Now, rewriting your original expression (a + b)^{\frac{1}{2}} as \sqrt{b}\left(a/b+1\right)^{1/2} gives

\sqrt{b}\left(1+\frac{1}{2}\left(\frac{a}{b}\right)-\frac{1}{8}\left(\frac{a}{b}\right)^2+\frac{1}{16}\left(\frac{a}{b}\right)^3+\cdots\right)

=\sqrt{b}+\frac{a}{2\sqrt{b}}-\frac{a^2}{8b^{3/2}}+\frac{a^3}{16b^{5/2}}+\cdots

which is the same result Wolfram Alpha will spit back.

Hope that helps!